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I have a simple query:
SELECT u_name AS user_name FROM users WHERE user_name = "john";
I get Unknown Column 'user_name' in where clause.
Can I not refer to 'user_name' in other parts of the statement even after select 'u_name as user_name'?
SQL is evaluated backwards, from right to left.
So the where clause is parsed and evaluate prior to the select clause.
Because of this the aliasing of u_name to user_name has not yet occurred.
7,34242148
10.3k1672132
See the following MySQL manual page:
"A select_expr can be given an alias
using AS alias_name. The alias is used
as the expression's column name and
can be used in GROUP BY, ORDER BY, or
HAVING clauses."
161k31228283
What about:
SELECT u_name AS user_name FROM users HAVING user_name = "john";
33.2k33682
select u_name as user_name from users where u_name = "john";
Think of it like this, your where clause evaluates first, to determine which rows (or joined rows) need to be returned.
Once the where clause is executed, the select clause runs for it.
To put it a better way, imagine this:
select distinct(u_name) as user_name from users where u_name = "john";
You can't reference the first half without the second.
Where always gets evaluated first, then the select clause.
If you're trying to perform a query like the following (find all the nodes with at least one attachment) where you've used a SELECT statement to create a new field which doesn't actually exist in the database, and try to use the alias for that result you'll run into the same problem:
SELECT nodes.*, (SELECT (COUNT(*) FROM attachments
WHERE attachments.nodeid = nodes.id) AS attachmentcount
FROM nodes
WHERE attachmentcount & 0;
You'll get an error "Unknown column 'attachmentcount' in WHERE clause".
Solution is actually fairly simple - just replace the alias with the statement which produces the alias, eg:
SELECT nodes.*, (SELECT (COUNT(*) FROM attachments
WHERE attachments.nodeid = nodes.id) AS attachmentcount
FROM nodes
WHERE (SELECT (COUNT(*) FROM attachments WHERE attachments.nodeid = nodes.id) & 0;
You'll still get the alias returned, but now SQL shouldn't bork at the unknown alias.
corrected:
SELECT u_name AS user_name FROM users WHERE u_name = 'john';
45k12105187
SELECT u_name AS user_name
u_name = "john";
SELECT user_name
SELECT u_name AS user_name
u_name = "john";
The latter ought to be the same as the former if the RDBMS supports predicate pushing into the in-line view.
34.4k43255
Your defined alias are not welcomed by the WHERE clause you have to use the HAVING clause for this
SELECT u_name AS user_name FROM users HAVING user_name = "john";
OR you can directly use the original column name with the WHERE
SELECT u_name AS user_name FROM users WHERE u_name = "john";
Same as you have the result in user defined alias as a result of subquery or any calculation it will be accessed by the HAVING clause not by the WHERE
SELECT u_name AS user_name ,
(SELECT last_name FROM users2 WHERE id=users.id) as user_last_name
FROM users
WHERE u_name = "john" HAVING user_last_name ='smith'
32.8k73160
No you need to select it with correct name. If you gave the table you select from an alias you can use that though.
5,602104571
No you cannot. user_name is doesn't exist until return time.
13.2k44247
Unknown column in WHERE clause caused by lines 1 and 2 and resolved by line 3:
$sql = "SELECT * FROM users WHERE username =".$userN
$sql = "SELECT * FROM users WHERE username =".$userName."";
$sql = "SELECT * FROM users WHERE username ='".$userName."'";
2,08242338
May be it helps.
SET @somevar := '';
SELECT @somevar AS user_name FROM users WHERE (@somevar := `u_name`) = "john";
BUT MAKE SURE WHAT YOU DO!
Indexes are NOT USED here
There will be scanned FULL TABLE - you hasn't specified the LIMIT 1 part
So, - THIS QUERY WILL BE SLLLOOOOOOW on huge tables.
But, may be it helps in some cases
While you can alias your tables within your query (i.e., "SELECT u.username FROM"), you have to use the actual names of the columns you're referencing. AS only impacts how the fields are returned.
3,85412344
Not as far as I know in MS-SQL 2000/5.
I've fallen foul of this in the past.
SELECT user_name
SELECT name AS user_name
user_name = "john"
2,08242338
I had the same problem, I found this useful.
mysql_query("SELECT * FROM `users` WHERE `user_name`='$user'");
remember to put $user in ' ' single quotes.
Just had this problem.
Make sure there is no space in the name of the entity in the database.
e.g. ' user_name' instead of 'user_name'
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＞＞＞正弦交流电经过匝数比为n1n2=101的变压器与电阻R、交流电压表V、..
正弦交流电经过匝数比为n1n2=101的变压器与电阻R、交流电压表V、交流电流表A按如图甲所示方式连接，R=10Ω．图乙是R两端电压u随时间变化的图象，Um=102&V，则下列说法中正确的是（　　）A．通过R的电流iR随时间t变化的规律是iR=2cos&100πt&AB．电流表A的读数为0.1&AC．电流表A的读数为210&AD．电压表V的读数为Um=102&V
题型：多选题难度：偏易来源：不详
A、正弦式电流且电阻R的Um=102&V，则有效值U=10V．所以通过电阻的电流为I=UR=1A，因此通过R的电流iR随时间t变化的规律是iR=2cos100πt（A）．故A正确；B、由正弦式电流且电阻R的Um=102&V，则有有效值U=10V．所以通过电阻的电流为I=UR=1A，又由于电流表读出的是有效值，再由匝数比为n1n2=101的变压器，得电流表A的读数为0.1A．故B正确，C错误；D、由正弦式电流且电阻R的Um=102&V，则有有效值U=10V．所以通过电阻的电流为I=UR1A，又由于电压表读出的是有效值，得电压表A的读数为10V．故D错误；故选AB
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变压器的结构和原理
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常见变压器:
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