$query = "
SELECT distinct count(*) as num
FROM table1 ge
INNER JOIN table 2 ad
ON ge.ID = ad.ID
WHERE ge.ID = ad.ID AND ge.field LIKE '".mysql_real_escape_string($_REQUEST['search'])."%'
AND ge.field not like '%word%'
order by ge.field ASC"
$query = "
SELECT *
FROM table1 ge
INNER JOIN table 2 ad
ON ge.ID = ad.ID
WHERE ge.ID = ad.ID AND ge.field LIKE '".mysql_real_escape_string($_REQUEST['search'])."%'
AND ge.field not like '%word%'
order by ge.field ASC"
似乎无法摆脱此查询正确的结果,不应该显示与$word
任何记录?任何帮助非常赞赏...
感谢
'$ word'是变量还是字符串''word''? –
btw您的第一个查询看起来很奇怪,您可以在没有'group by'的情况下执行'COUNT()',但是您不需要'DISTINCT'。所以,如果你是担心你缺少多个结果'集团BY' –