0

我有这个代码：

mysql_query("
   SELECT id as tc_id, firstname, surname, position, @concat := CONCAT(firstname,' ',surname) 
   FROM tc 
   WHERE (@concat LIKE '%". mysql_real_escape_string($_REQUEST['search_term']) ."%');");

search_term 只是指文本输入的值。

当我在本地运行 sequel pro 中的 SELECT 语句时——我的 mysql 程序——它工作得很好，选择了适当的行。但是，当我通过我的 PHP 页面在本地运行查询时，它不会返回任何行，即使它应该返回。帮助任何人？

4

3 回答 3

1

WHERE不适用于计算列，因此您的查询本身将不起作用。

它可能有效，因为您在本地@concat声明了一个变量。如果您在特定条件下运行两次查询，它似乎也可以工作，因为您的查询实际上可能分配了一个@concat变量。

你想要的是

SELECT id as tc_id, firstname, surname, position,
    CONCAT(firstname,' ',surname)  AS concat
FROM tc HAVING concat LIKE '%<YOUR SEARCH TERM>%';

作为测试：

-- Declare a minimum table to match the query
CREATE TABLE tc (id integer, firstname varchar(20), surname varchar(20), position integer);

INSERT INTO tc (firstname, surname) VALUES ('alfa', 'bravo');

-- Your query...

SELECT id as tc_id, firstname, surname, position,
    @concat := CONCAT(firstname,' ',surname) FROM tc WHERE (@concat LIKE '%alfa%');

-- ...returns nothing

Empty set (0.00 sec)

-- The proper query works.

SELECT id as tc_id, firstname, surname, position, CONCAT(firstname,' ',surname)  AS concat FROM tc HAVING concat LIKE '%alfa%';
+-------+-----------+---------+----------+------------+
| tc_id | firstname | surname | position | concat     |
+-------+-----------+---------+----------+------------+
|  NULL | alfa      | bravo   |     NULL | alfa bravo |
+-------+-----------+---------+----------+------------+
1 row in set (0.00 sec)

-- But if I declare a @concat variable    

SELECT @concat := 'alfa';
+-------------------+
| @concat := 'alfa' |
+-------------------+
| alfa              |
+-------------------+
1 row in set (0.00 sec)

-- Then your query SEEMS to work.

mysql> SELECT id as tc_id, firstname, surname, position, @concat := CONCAT(firstname,' ',surname)  FROM tc  WHERE (@concat LIKE '%alfa%');
+-------+-----------+---------+----------+------------------------------------------+
| tc_id | firstname | surname | position | @concat := CONCAT(firstname,' ',surname) |
+-------+-----------+---------+----------+------------------------------------------+
|  NULL | alfa      | bravo   |     NULL | alfa bravo                               |
+-------+-----------+---------+----------+------------------------------------------+
1 row in set (0.00 sec)

-- "SEEMS" because the select query isn't actually working:

UPDATE tc SET firstname = 'delta';
Query OK, 1 row affected (0.28 sec)
Rows matched: 1  Changed: 1  Warnings: 0

-- Having renamed the only row to "delta", a search for "alpha" should fail,
-- but since @concat still holds 'alpha', then the query matches ALL rows:

mysql> SELECT id as tc_id, firstname, surname, position, @concat := CONCAT(firstname,'    ',surname)  FROM tc  WHERE (@concat LIKE '%alfa%');
+-------+-----------+---------+----------+------------------------------------------+
| tc_id | firstname | surname | position | @concat := CONCAT(firstname,' ',surname) |
+-------+-----------+---------+----------+------------------------------------------+
|  NULL | delta     | bravo   |     NULL | delta bravo                              |
+-------+-----------+---------+----------+------------------------------------------+
1 row in set (0.00 sec)
于 2012-09-17T10:39:47.607 回答
0

试试这个：

mysql_query("
   SELECT id as tc_id, firstname, surname, position, CONCAT(firstname,' ',surname) as concatt 
   FROM tc 
   WHERE (CONCAT(firstname,' ',surname) LIKE '%". mysql_real_escape_string($_REQUEST['search_term']) ."%');");
于 2012-09-17T10:39:14.060 回答
0

这是因为您的 Sequel Pro 将查询语句翻译成 MySql 查询，而 MySql 不明白“where @concat LIKE ...”是什么意思。

尝试将 PHP 代码上的 where 子句更改为“where CONCAT(firstname,'',surname) LIKE '%...”

完整代码如下：

mysql_query("SELECT id as tc_id, firstname, surname, position, CONCAT(firstname,' ',surname) FROM tc WHERE (CONCAT(firstname,' ',surname) LIKE '%". mysql_real_escape_string($_REQUEST['search_term']) ."%');");
于 2012-09-17T10:43:47.703 回答