Ошибка 1248 mysql

I cant for the life of me figure out why this is erroring

SELECT * FROM 
SELECT 
c.city_id,
p.prop_ynow_id,
p.propertyid,
p.prop_add_value,
p.name,
picture,
ifnull(p.address,'') as`location`,
ifnull(city,'')as`city`,
ifnull(ShortCut,'') as `state`,
ifnull(p.zip,'') as `zip`,
min(if(pr.minrent = 0,99999999,pr.minrent)) as minrent, 
max(pr.maxrent) as maxrent,
'' as service,
hood_id,
ifnull(p.phone,'') as `phone`,
latitude,
longitude,
min(CAST(pu.fullBath AS UNSIGNED)) as`minbath`,
max(CAST(pu.fullBath AS UNSIGNED)) as`maxbath`,
min(CAST(pu.Bed AS UNSIGNED)) as`minbed` ,
max(CAST(pu.Bed AS UNSIGNED)) as`maxbed`,
'' as url,
'' as source_id,
'' as source_name,
'' as addresscode,
'' as citycode,
'' as ctime,
'' as paid,
'' as similar_url,
'' as created_at,
'' as updated_at,
'' as city_name,
'' as service_listing_id

FROM 
wiki_city_list c join propertyinfo p on c.city_id=p.city 
join ynow_rentwiki.Property_Unitlayout pu on p.prop_ynow_id=pu.P_Ident
join (SELECT CAST(substring_index(if(Rent >0 ,Rent,RentLow),'.',1) AS UNSIGNED) as minrent, CAST(substring_index(if(Rent >0,Rent,Renthigh),'.',1) AS UNSIGNED) as maxrent,PRE_Ident,P_Ident,UNL_Ident,RTY_Ident from ynow_rentwiki.Property_rents where P_Ident in (3958, 4576, 4577) and (Rent!='' or (Rentlow!='' and Renthigh!='')) )  as pr on pu.UNL_Ident=pr.UNL_Ident
join state s on (p.state = s.stateid OR p.state = s.ShortCut ) 
WHERE 
pu.Status='Active'
and p.delete_date='0000-00-00'

GROUP BY 
c.city_id, p.prop_ynow_id

UNION 
SELECT 
c.city_id,
p.prop_ynow_id,
p.propertyid,
p.prop_add_value,
p.name,
picture,
ifnull(p.address,'') as`location`,
ifnull(city,'')as`city`,
ifnull(ShortCut,'') as `state`,
ifnull(p.zip,'') as `zip`,
min(if(pr.minrent = 0,99999999,pr.minrent)) as minrent, 
max(pr.maxrent) as maxrent,
'' as service,
hood_id,
ifnull(p.phone,'') as `phone`,
latitude,
longitude,
min(CAST(pu.fullBath AS UNSIGNED)) as`minbath`,
max(CAST(pu.fullBath AS UNSIGNED)) as`maxbath`,
min(CAST(pu.Bed AS UNSIGNED)) as`minbed` ,
max(CAST(pu.Bed AS UNSIGNED)) as`maxbed`,
'' as url,
'' as source_id,
'' as source_name,
'' as addresscode,
'' as citycode,
'' as ctime,
'' as paid,
'' as similar_url,
'' as created_at,
'' as updated_at,
'' as city_name,
'' as service_listing_id

FROM 
wiki_city_list c join propertyinfo p on c.city_id=p.city 
join ynow_rentwiki.Property_Unitlayout pu on p.prop_ynow_id=pu.P_Ident
join (SELECT CAST(substring_index(if(Rent >0 ,Rent,RentLow),'.',1) AS UNSIGNED) as minrent, CAST(substring_index(if(Rent >0,Rent,Renthigh),'.',1) AS UNSIGNED) as maxrent,PRE_Ident,P_Ident,UNL_Ident,RTY_Ident from ynow_rentwiki.Property_rents where P_Ident in (9744) and (Rent!='' or (Rentlow!='' and Renthigh!='')) )  as pr on pu.UNL_Ident=pr.UNL_Ident
join state s on (p.state = s.stateid OR p.state = s.ShortCut ) 
WHERE 
pu.Status='Active'
and p.delete_date='0000-00-00'

GROUP BY 
c.city_id, p.prop_ynow_id

UNION 
SELECT  
'' as prop_ynow_id, 
id as propertyid, 
0 as prop_add_value,
t.name as name,
'' as picture,  
t.address as location,
t.city as city, 
s.ShortCut as state, 
t.zip as zip,   
CAST(REPLACE(REPLACE(t.price,'$',''),',','') as UNSIGNED) as minrent, 
'' as maxrent,
t.service as service, 
'' as hood_id, 
'' as phone, 
t.latitude as latitude, 
t.longitude as longitude, 
t.bathrooms as minbath, 
'' as maxbath, 
t.bedrooms as minbed,
'' as maxbed,   
t.url as url,   
t.source_id as source_id, 
t.source_name as source_name, 
t.addresscode as addresscode, 
t.citycode as citycode, 
t.ctime as ctime, 
t.paid as paid,
t.similar_url as similar_url, 
t.created_at as created_at, 
t.updated_at as updated_at, 
SUBSTRING_INDEX(c.city_name,'_',1) as city_name,    
t.service_listing_id as service_listing_id

FROM LBCPrimary.third_party_properties as t, LBCPrimary.wiki_city_list as c, LBCPrimary.state as s
WHERE 
t.city in ( '230' ) 
and 
address <> '' and 
t.city = c.city_id and 
c.city_state = s.stateid

order by t.ctime 
desc
limit 46 as a limit 0,50

[SQL] SELECT * from   
(  
select e.account from employee e  
UNION  
SELECT u.account from `user` u  
UNION  
SELECT a.account from agent a  
)  
  
[Err] 1248 - Every derived table must have its own alias  

 Это предложение означает, что каждая производная таблица должна иметь собственный псевдоним

Эта ошибка обычно возникает в многотабличных запросах или подзапросах, поскольку во вложенных запросах результат подзапроса используется в качестве производной таблицы для запроса верхнего уровня, поэтому результат подзапроса должен иметь псевдоним.

В приведенном выше примере измените оператор запроса:

SELECT * from   
(  
select e.account from employee e  
UNION  
SELECT u.account from `user` u  
UNION  
SELECT a.account from agent a  
)as total  

 Как показано выше, добавление предложения в качестве итога после подзапроса эквивалентно присвоению псевдонима производной таблицы набора результатов подзапроса как итога, и проблема решена.

“Every derived table must have its own alias” is an error that often occurs when you try to run a query in MySQL. You see this error when you use the FROM command and basically try to generate a new, derived table with your subquery.

If you’re having trouble understanding which line of code is causing the problem or what you can do to get rid of this error, don’t worry; we have you covered. In this article, we’ll see the problem in more detail and all the solutions you can try to remove this error.

This alias error occurs when you use the FROM command to generate a new table using the subquery. If this derived table doesn’t have an alias, i.e, a name reference, you’ll see an error that says every derived table must have its own alias with an error code 1248.

– Absence of Alais

Keep in mind that a subquery and derived tables are not the same; a derived table vs subquery comparison will show you that there are many differences between the two. Also, don’t forget that a table must have a primary key.

Let’s take a look at a derived table example to reproduce the error and get a better understanding of why it actually occurs. Consider this SQL code:

SELECT city,
COUNT (*) AS count_of_order,
SUM(total_amount) AS total_order_sum
FROM (
SELECT o.order_id,
o.city,
o.total_amount
FROM orders o
INNER JOIN consumer c ON o.consumer_id = c.consumer_id
)
GROUP BY city;

This SQL query basically finds three things: the total sum of the order, the total number of orders, and the city from the consumer and order tables.

If you further break down this query, you’ll notice that there’s an inner and an outer query. The former is used to get data from customers and orders. This is the inner query:

SELECT o.order_id,
o.city,
o.total_amount
FROM orders o
INNER JOIN consumer c ON o.consumer_id = c.consumer_id

The join here is done on a common field to get the data we need, and the table that results from this inner query is the derived table.

Meanwhile, the purpose of the outer query is to select just a few columns from the subquery. So instead of the table, the data is fetched from the results produced by this subquery. Here’s the outer query:

SELECT city,
COUNT (*) AS count_of_order,
SUM(total_amount) AS total_order_sum
FROM (
// inner query
)
GROUP BY city;

The outer query shows the city, includes two aggregate functions and combines the result by city. But if you run this query in MySQL, you’ll get the error 1248, which states that every derived table must have its own alias.

Now that we’ve reproduced this error and understood the derived table SQL syntax, hopefully, you better understand why we’re seeing the error – it’s because we haven’t given any alias name to the derived table we’ve produced with this SQL query.

How To Fix Every Derived Table Must Have Its Own Alias

All you need to do to solve the error is add an alias right after the closing bracket. Your alias can be anything you want it to be. It can be a word, just a random few letters, or even a single letter.

You’ve now understood that you’re getting this error in MySQL because each derived table or subquery, which is a SELECT query, must include an alias after it. The example discussed above doesn’t include any alias for the subquery that we’ve run.

Note that the alias for your subquery will be the same as the column alias and comes after the closing bracket of the FROM clause of the subquery. It’s common practice to use sub for subquery if it’s a small query, but you can also use something more self-descriptive if you have a longer query. So if we fix the SQL query above, this is what it will look like:

SELECT city,
COUNT (*) AS count_of_order,
SUM(total_amount) AS total_order_sum
FROM (
SELECT o.order_id,
o.city,
o.total_amount
FROM orders o
INNER JOIN consumer c ON o.consumer_id = c.consumer_id
) order_info
GROUP BY city;

What’s different in this code is that we’ve added “order_info” right next to the closing bracket after the FROM subquery in the second-last line.

This is our alias for the derived table. Note that you only need to provide this alias in MySQL; other vendors don’t need it. And if you run the query now, you won’t get the error anymore.

– Optional Method: Using the as Keyword

You’ll often see people suggesting that you should use the AS keyword to add the alias but you should know that it’s optional; you don’t necessarily need to add it for the error to be resolved but adding it adds a little bit of clarity to the SQL query.

So it’s really up to you. If we use the same SQL query used above and add the AS keyword to it, here’s what it’ll look like:

SELECT city,
COUNT (*) AS count_of_order,
SUM(total_amount) AS total_order_sum
FROM (
SELECT o.order_id,
o.city,
o.total_amount
FROM orders o
INNER JOIN consumer c ON o.consumer_id = c.consumer_id
) AS order_info
GROUP BY city;

The alias, in this case, is “AS order_info” in the second-last line. That’s all there is to it.

FAQs

1. Why Every Derived Table Must Have Its Own Alias?

When you use a stand-alone subquery in the FROM clause of your SELECT statement, it’s also known as a derived table. A stand-alone query is essentially a subquery that can be executed independently of your outer query. But unlike a subquery, you need to give the derived table an alias. This is so that you can refer to its name later on in the same query.

2. What Is the Purpose of a Table Alias?

The main purpose of using table aliases is to rename tables in a certain SQL statement. Renaming the table is just a temporary change and doesn’t change the name of the table in the database. Similarly, you can rename the columns of a table using column aliases only within the context of that specific SQL query.

3. Is Aliasing Tables With One Letter a Good Practice in SQL?

Generally, this is not a good practice, especially since code is communication. Using code, you communicate with the underlying engine or compiler and you communicate with your future yourself and others who will see your code.

In the case of the former, what you name doesn’t matter since SQL compilers don’t care what you name the table. But for the sake of clarity for others, it’s better if you use self-descriptive naming conventions, even for aliasing tables.

Conclusion

After reading this article, we’re sure you’ll now be able to solve the MySQL error that says every derived table must have its own alias. To make sure you can easily solve this problem, let’s quickly go over the key points we’ve discussed in this article:

• The reason you’re seeing this error is that you haven’t given an alias when using a subquery with the FROM command to generate a new (derived) table.
• A derived table is essentially a SELECT subquery in the FROM clause used in the main query.
• Every derived table must have an alias, which is basically a name reference you can use to refer to the table later in the query.
• Solving this error is very easy and just requires you to add an alias after the closing bracket of the FROM statement in the subquery.
• You can also add the AS keyword before the alias, but this is optional, and you won’t see an error if you add an alias without it.

An alias can be anything – a word, a letter, or a few letters. That’s all you really need to know to solve this MySQL error!

What do I do wrong with the update in mysql?

I tried many different ways and can not make it work.

Update is made on the same table. Corrected the sql as shown but I still get an described in later comment

update auctions A 
SET A.active = -1 
WHERE A.auction_id IN
(
    SELECT auction_id 
    FROM 
    (
        SELECT B.auction_id FROM
        table auctions 
        WHERE B.auction_id = A.auction_id AND B.active = 0 AND B.ended_on < "2019-04-18" AND B.ended_on > "2018-01-06" AND B.item_id 
        not IN 
                (
                    SELECT item_id 
                    FROM 
                    (
                        SELECT C.item_id from auctions C 
                        WHERE C.active = 1 
                        AND C.item_id = B.item_id
                   )    AS temp_c
             )

    )    AS temp_b
);

INSERT INTO `auctions` (`auction_id`, `item_id`, `active`, `created_by`, `started_on`, `buy_price`, `prefs`, `ended_on`, `bids`) VALUES (7333209574, 20354, 1, 2, '2019-08-23 16:12:51', NULL, 'a:23', NULL, 0);
INSERT INTO `auctions` (`auction_id`, `item_id`, `active`, `created_by`, `started_on`, `buy_price`, `prefs`, `ended_on`, `bids`) VALUES (7333209575, 20354, 0, 2, '2018-03-13 16:12:51', NULL, 'a:23', '2018-03-23 16:30:31', 0);
INSERT INTO `auctions` (`auction_id`, `item_id`, `active`, `created_by`, `started_on`, `buy_price`, `prefs`, `ended_on`, `bids`) VALUES (7333209576, 20752, 0, 2, '2018-02-13 16:12:51', NULL, 'a:23', '2018-02-23 16:30:31', 0);
INSERT INTO `auctions` (`auction_id`, `item_id`, `active`, `created_by`, `started_on`, `buy_price`, `prefs`, `ended_on`, `bids`) VALUES (7333209577, 20752, 0, 2, '2018-02-13 16:12:51', NULL, 'a:23', '2018-02-23 16:30:31', 0);
INSERT INTO `auctions` (`auction_id`, `item_id`, `active`, `created_by`, `started_on`, `buy_price`, `prefs`, `ended_on`, `bids`) VALUES (7333209577, 20752, 0, 2, '2018-06-13 16:12:51', NULL, 'a:23', '2018-06-23 16:30:31', 0);


CREATE TABLE `auctions` (
    `auction_id` BIGINT(20) NOT NULL,
    `item_id` INT(11) NOT NULL,
    `active` TINYINT(4) NULL DEFAULT '1',
    `created_by` INT(11) NULL DEFAULT NULL,
    `started_on` DATETIME NULL DEFAULT NULL,
    `buy_price` DOUBLE NULL DEFAULT NULL,
    `prefs` TEXT NULL COLLATE 'utf8_unicode_ci',
    `ended_on` DATETIME NULL DEFAULT NULL,
    `bids` INT(11) NOT NULL DEFAULT '0',
    PRIMARY KEY (`auction_id`),
    INDEX `item_id` (`item_id`),
    INDEX `created_by` (`created_by`),
    CONSTRAINT `auctions_ibfk_1` FOREIGN KEY (`item_id`) REFERENCES `data_1` (`id`),
    CONSTRAINT `auctions_ibfk_2` FOREIGN KEY (`created_by`) REFERENCES `login` (`id`)
)
COLLATE='utf8_unicode_ci'
ENGINE=InnoDB
;

Here is the select statement that contains the correct output for an update statement. Your answer contains auction_id that item_id has active = 1 but it shouldn’t.

SELECT * from auctions WHERE active = 0 AND ended_on < "2019-04-18" AND ended_on > "2018-01-06" AND item_id NOT IN (SELECT item_id FROM auctions WHERE active = 1);

Here is the output of EXPLAIN in the form of an INSERT

INSERT INTO `NieznanaTabela` (`id`, `select_type`, `table`, `type`, `possible_keys`, `key`, `key_len`, `ref`, `rows`, `Extra`) VALUES (1, 'SIMPLE', 'A', 'ALL', 'item_id', NULL, NULL, NULL, 20554, 'Using where');
INSERT INTO `NieznanaTabela` (`id`, `select_type`, `table`, `type`, `possible_keys`, `key`, `key_len`, `ref`, `rows`, `Extra`) VALUES (1, 'SIMPLE', 'B', 'ref', 'item_id', 'item_id', '4', 'dbauction.A.item_id', 10, 'Using where');

i have problem here,
when im trying to select with query below, i got error message

Error Code : 1248 Every derived table must have its own alias

SELECT B.BRANCH_CODE, B.BRANCH_NAME, C.COMPANY_CODE, C.NAME, TSK.DATE_CREATE,
  CASE TB.BULK_TYPE
    WHEN 1 THEN 'Bulk Transfer'
    WHEN 2 THEN 'Bulk Payment'
    ELSE 'Payroll'
  END AS TRX_METHOD_E, 
  CASE TB.BULK_TYPE
    WHEN 1 THEN 'Bulk Transfer'
    WHEN 2 THEN 'Bulk Pembayaran Tagihan'
    ELSE 'Pembayaran Gaji'
  END AS TRX_METHOD_I,
  TB.TOTAL_RECORD,
  TB.ACCOUNT_NO,
  TSK.TRX_TOTAL,
  TC.TOTAL_CHARGE,
  DATE(TSK.DATE_TRX) AS DATE_TRX,
  TB.REF_ID,
  CASE
    WHEN TSK.TRX_COUNT_SUCCESS = TSK.TRX_COUNT_ALL THEN 'All Success'
    WHEN TSK.TRX_COUNT_FAIL = TSK.TRX_COUNT_ALL THEN 'All Failed'
    WHEN TSK.TRX_COUNT_SUCCESS > 0 AND TSK.TRX_COUNT_FAIL > 0 THEN 'Partial Success (' || TSK.TRX_COUNT_SUCCESS || '/' || TSK.TRX_COUNT_ALL || ')'
    ELSE 'Pending Execution'
  END AS TRX_STATUS_E,
  CASE
    WHEN TSK.TRX_COUNT_SUCCESS = TSK.TRX_COUNT_ALL THEN 'Berhasil Semua'
    WHEN TSK.TRX_COUNT_FAIL = TSK.TRX_COUNT_ALL THEN 'Gagal Semua'
    WHEN TSK.TRX_COUNT_SUCCESS > 0 AND TSK.TRX_COUNT_FAIL > 0 THEN 'Berhasil Sebagian (' || TSK.TRX_COUNT_SUCCESS || '/' || TSK.TRX_COUNT_ALL || ')'
    ELSE 'Tunggu Eksekusi'
  END AS TRX_STATUS_I  
FROM CB_TASK_BULKS TB
  JOIN CB_TASKS TSK ON TSK.REF_ID = TB.REF_ID
  JOIN CB_COMPANIES C ON C.COMPANY_ID = TSK.COMPANY_ID
  JOIN CB_BRANCHES B ON B.BRANCH_CODE = C.BRANCH_CODE,
(
  SELECT REF_ID, SUM(CHARGE) AS TOTAL_CHARGE
  FROM
  (
    SELECT XTB.REF_ID, SUM(CHARGE) AS CHARGE
    FROM CB_TRANSFERS XT
    JOIN CB_TASK_BULK_DETAILS XTBD ON XTBD.BULK_DETAIL_ID = XT.BULK_DETAIL_ID
    JOIN CB_TASK_BULKS XTB ON XTB.REF_ID = XTBD.REF_ID
    GROUP BY XTB.REF_ID

    UNION

    SELECT XTB2.REF_ID, SUM(CHARGE) AS CHARGE
    FROM CB_TRANSFERS_DOM XTD
    JOIN CB_TASK_BULK_DETAILS XTBD2 ON XTBD2.BULK_DETAIL_ID = XTD.BULK_DETAIL_ID
    JOIN CB_TASK_BULKS XTB2 ON XTB2.REF_ID = XTBD.REF_ID
    GROUP BY XTB.REF_ID

    UNION

    SELECT XTB3.REF_ID, SUM(CHARGE) AS CHARGE
    FROM CB_PAYMENTS XP
    JOIN CB_TASK_BULK_DETAILS XTBD3 ON XTBD3.BULK_DETAIL_ID = XP.BULK_DETAIL_ID
    JOIN CB_TASK_BULKS XTB3 ON XTB3.REF_ID = XTBD.REF_ID
    GROUP BY XTB.REF_ID
  )
  GROUP BY REF_ID
) TC
WHERE TC.REF_ID = TSK.REF_ID
AND (TSK.TRX_COUNT_SUCCESS > 0 OR TSK.TRX_COUNT_FAIL > 0);

can somebody help me to find my problem here,
i think i’ve given all tables with his own «name» like TB for CB_BULKS_DATA (example)

If you see MySQL SQL error 1248, this tutorial will help you.

Updated

Speed up your computer today with this simple download.

You must provide an alias for all your inline mood / view tables in MySQL, even if they are nested. In your code, you are using a view in the following view, but you seem to have forgotten to mention the alias for your company’s internal view.

MYSQL ERROR 1248 (42000): Each view must have its own nickname.

I could not understand why this was a mistake

  SELECT * FROMCHOOSEc.city_id,p.prop_ynow_id,Object id p.,p.prop_add_value,name,Drawing,ifnull (p.address, '') as 'place',ifnull (city, '') as 'ville',ifnull (shortcut, '') as "State",ifnull (p.zip, '') as `zip`,min (if (pr.minrent implies 0.9999999, pr.minrent)) as minrent,max (pr.maxrent) as maxrent,'' during the service,Hood_id,ifnull (p.phone, '') as 'phone',Huge,Longitude,min (CAST (pu.fullBath AS UNSIGNED)) as `minbath`,max (CAST (pu.fullBath AS UNSIGNED)) as `maxbath`,min (CAST (pu.Bed AS UNSIGNED)) as `minbed`,max (CAST (pu.Bed AS UNSIGNED)) as `maxbed`,'' as a URL,'as source_id,'Source name,'' as well as the address code,'' as area code,almost like ctime,'' as paid,like Similar_url,like created_at,mainly because update_at,as the name of the city,'' as service_listing_idFROMwiki_city_list they will join Propertyinfo p at c.city_id = p.cityjoin ynow_rentwiki.Property_Unitlayout Please take away p.prop_ynow_id = pu.P_Identjoin (SELECT CAST (substring_index (if (Rent> 0, Rent, RentLow), '.', 1) AS UNSIGNED) as minrent, CAST (substring_index (if (Loyer> 0, Rent, Renthigh), '.', 1 )) AS UNSIGNED) as maxrent, PRE_Ident, P_Ident, UNL_Ident, RTY_Ident from ynow_rentwiki.Property_rents, where P_Ident in 4576, (3958, 4577) and (Rent! = '' And also (low! = '' And Renthigh! = ' ')))) as page rank on pu.UNL_Ident = pr.UNL_IdentJoin state s on (p.state matches s.stateid OR p.state = s.ShortCut)ORpu.Status = 'Active'and p.delete_date = '0000-00-00'BY GROUPc.city_id,p.prop_ynow_idUNIONCHOOSEc.city_id,p.prop_ynow_id,Object id p.,p.prop_add_value,name,Drawing,ifnull (p.address, '') as 'place',ifnull (city, '') as 'ville',ifnull (shortcut, '') as "State",ifnull (p.zip, '') as `zip`,min (if (pr.minrent matches 0.99999999, pr.minrent)) as minrent,max (pr.maxrent) as maxrent,'for the reason that the serviceHood_id,ifnull (p.phone, '') as 'phone',Huge,Longitude,min (CAST (pu.fullBath AS UNSIGNED)) as `minbath`,max (CAST (pu.fullBath AS UNSIGNED)) as `maxbath`,min (CAST (pu.Bed AS UNSIGNED)) as `minbed`,max (CAST (pu.Bed AS UNSIGNED)) as `maxbed`,'' as a URL,'as source_id,'Source name,'' although as an address code'' as area code,like ctime,'' paid,like Similar_url,'like created_at,'' the_update_at,as the name of the city,'' as service_listing_idFROMwiki_city_list c participate in Propertyinfo p at c.city_id = p.cityjoin ynow_rentwiki.Property_Unitlayout Please take away p.prop_ynow_id = pu.P_Identjoin (SELECT CAST (substring_index (if (Rent> 0, Rent, RentLow), '.', 1) AS UNSIGNED) as minrent, CAST (substring_index (if (Loyer> 0, Rent, Renthigh), '.', 1 )) AS UNSIGNED) as maxrent, PRE_Ident, P_Ident, UNL_Ident, RTY_Ident from ynow_rentwiki Where property_rents P_Ident is in (9744) and (Rent! = '' Or (Rentlow! = '' And Renthigh! = ''))) As pr to pu.UNL_Ident = pr.UNL_IdentJoin point s at (p.state = s.stateid OR p.state = s.ShortCut)ORpu.Status = 'Active'and p.delete_date = '0000-00-00'BY GROUPc.city_id, p.prop_ynow_idUNIONSELECT AS''prop_ynow_id,id as a property identifier,0 as prop_add_value,t.name as a name,'' as an image,t. address as location,T. a city as a city,see ShortCut as a state,t.zip is essentially zip,CAST (REPLACE (REPLACE (t.price, '$', ''), ',', '') as UNSIGNED) although minrent,like Maxrent,t. service as a service,'' available as Hood_id,like a phone,t latitude in latitude,t. longitude as longitude,i.e. there are bathrooms in minbad,like maxbad,t. the room is like a bed,like maxbed,t.url as a URL,t.source_id as source_id,t.source_name in the view that source_name,t.addresscode as an address code,t.citycode as city code,t.ctime is ctime,i.e. paid as paid,t.similar_url, although Similar_url,t.created_at as created_at,t.updated_at as updated_at,SUBSTRING_INDEX (c.city_name, '_', 1) mainly because city_name,t.service_listing_id is essentially service_listing_idDE LBCPrimary.third_party_properties as t, LBCPrimary.wiki_city_list as, LBCPrimary.state as sORt.stadt in ('230')andAddress <> '', etc.t.city = c.city_id andc.city_state = s.stateidorder at t.ctimedescriptionLimit 46 as limit 0.50 

You

Write any type of query in MySQL and get the big error “1248: Each view must have its own alias”? If so, then this is a good, reliable, and simple error that can be corrected.

How To FixMySQL “Each View Must Have Its Own Alias” Error

So you have a specific bug in MySQL. How to solve this problem?

The short answer you want is to give your subqueries an alias in your company’s SELECT statement. The alias for adding the FROM term subquery after the closing curly brace.

This is not required in 3rd party SQL, but MySQL requires you to create an alias for your current subqueries.

Derived Table Example

A Derived Chart is a SELECT subquery in the FROM clause of your main query. It helps, like a table, because it contains articles and returns rows. For example, it might be the old location of the table in some kind of FROM clause. It is often recognized as an inline view or simply as a specific subquery.

  SELECT client_ville,COUNT (*) AS order_count,SUM (order_amount) AS order_sumFROM (  SELECT o.order_id,  o.Kunde_Stadt,  o.order_amount ORDERS  INNER JOIN member c ON o.customer_id = c.customer_id)GROUP BY Customer_City;  

This query finds the location, the number of orders associated with the order invoices, and the total Accounts of orders from tables of orders and customers.

The request has an outer request and a body request. An internal query is used to find data about orders and customers:

  SELECT o.order_id, o.customer_city, o.order_amountORDERSINNER JOIN customer f ON o.customer_id = c.customer_id  

In this case, the data is initially taken from two tables, which are combined into a common row of work. This is a “derived table”.

The outer topic selects multiple columns prior to this subquery. Marketing information does not come from an array, but from the results of this subquery.

  SELECT client_ville,COUNT (*) AS order_count,SUM (order_amount) AS order_sumFROM (...)GROUP BY Customer_City;  

It shows customer_city, two combined skills, and city groups.

  SELECT client_ville,COUNT (*) AS order_count,SUM (order_amount) AS order_sumFROM (  SELECT o.order_id,  o.Kunde_Stadt,  o.order_amount Delivery of AB from  INTERNAL customer CONNECTION c ON o.customer_id = c.customer_id)GROUP BY Customer_City;  

Error 1248: Each view must have its own alias

Solution For “Every View Must Have An Alias”

The reason for this error The point is that in MySQL, every dining view (a subquery that is a SELECT query) must have an alias.

In the sample query, there was no alias for your subquery here. The alias of the absolute subquery is the same as the alias of the channel. These are the closing parentheses of the FROM clause for a general subquery. B

So, in this example, we just need to add an alias after the closing parenthesis to fix this error.

An alias can be almost anything that interests you: a single letter, a word, several letters. I often use the word “under” when it is a small query (for a subquery), and something more descriptive when it is a longer specific query.

  SELECT client_ville,COUNT (*) AS order_count,SUM (order_amount) AS order_sumFROM (  SELECT o.order_id,  o.Kunde_Stadt,  o.order_amount ORDERS  INNER JOIN customer c ON o.customer_id implies c.customer_id) underGROUP BY Customer_City;  

Notice how the word “sub” is added on the penultimate line after the parentheses are closed? This is your alias for a subquery or view. This alias is required in MySQL but ignored Other suppliers.

This query should work and no longer receive the “Each table retrieved must have its own alias” error message.

Optional: Add The AS Keyword

You can add the AS keyword as this is a great optional part of adding an alias, but it’s up to you. If it still works:

  SELECT client_ville,COUNT (*) AS order_count,SUM (order_amount) AS order_sumFROM (  SELECT o.order_id,  o.Kunde_Stadt,  o.order_amount ORDERS  INTERNAL customer CONNECTION c ON o.customer_id = c.customer_id) AS-SubGROUP BY Customer_City;  

You can fix this derived table alias error in the same way. Add an alias to your subquery.

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Error 1248 De Mysql Sql
MySQL SQL 오류 1248
Ошибка SQL SQL 1248
Błąd MySQL Sql 1248
Erreur SQL SQL 1248
MySQL-SQL-Fehler 1248
Mysql SQL-fel 1248
Mysql SQL-fout 1248
Errore SQL MySQL 1248
Erro SQL 1248 Do MySQL