Question

我试图选择所有并跳过一些行。

这一行有一些语法错误，我不知道我做错了什么。

SELECT * FROM ads WHERE bbookschool like '$data' && WHERE bbookname != 'TEST' ORDER BY time desc

更多代码

$data = preg_replace ('#[^a-zA-Z0.-_% ]#i', '', $_POST['data']);
$Result = mysql_query("SELECT * FROM ads WHERE bbookschool like '$data' && WHERE bbookname != 'TEST0160' ORDER BY time desc")
 or die (mysql_error());
while($row = mysql_fetch_array($Result)){
require 'book_ad.php';
$adCondition = (!empty($row['bbookname'])) ? $ad : '';  
echo $adCondition;   
         }

感谢。

Answer 1

您只需使用WHERE一次：

SELECT
    *
FROM
    ads
WHERE
    bbookschool LIKE '$data'
    && bbookname != 'TEST'
ORDER BY
    `time` DESC;

另外，如果要在$ data中搜索包含字符串的单词，请在$ data周围设置%：

bbookschool LIKE '%$data%'

PS：我相信time也是reserved word，所以在它周围加上引号（`）。

Answer 2

更改此查询：

SELECT * FROM ads WHERE bbookschool like '$data' AND bbookname != 'TEST' ORDER BY time desc

Answer 3

试试这个

SELECT * FROM ads WHERE `bbookschool` like '%$data%' && bbookname != 'TEST' ORDER BY `time` desc

Answer 4

试试这个。

SELECT * FROM ads在哪里bbookschool喜欢＆＃39;％$ data％＆＃39;和bbookname！=＆＃39; TEST＆＃39; ORDER BY time desc

如何从表中选择所有并跳过具有特定值的某一行

4 个答案: