$query = "
SELECT distinct count(*) as num
FROM table1 ge
INNER JOIN table 2 ad
ON ge.ID = ad.ID
WHERE ge.ID = ad.ID AND ge.field LIKE '".mysql_real_escape_string($_REQUEST['search'])."%'
AND ge.field not like '%word%'
order by ge.field ASC"
或者
$query = "
SELECT *
FROM table1 ge
INNER JOIN table 2 ad
ON ge.ID = ad.ID
WHERE ge.ID = ad.ID AND ge.field LIKE '".mysql_real_escape_string($_REQUEST['search'])."%'
AND ge.field not like '%word%'
order by ge.field ASC"
似乎无法从此查询中获得正确的结果,任何带有 $word
的记录都不应显示?非常感谢任何帮助...
谢谢
最佳答案
最好的方法是调试变量的值并尝试直接在数据库上执行。
看起来您尝试使用 'word'
作为字符串,而不是将其用作变量。
$query = "
SELECT *
FROM table1 ge
INNER JOIN table 2 ad
ON ge.ID = ad.ID
WHERE ge.ID = ad.ID AND ge.field LIKE '".mysql_real_escape_string($_REQUEST['search'])."%'
AND ge.field not like '%".$word."%'
order by ge.field ASC"
关于mysql - 不像 MySQL select 语句不返回正确的结果,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33926121/