如何從表中選擇所有，並跳過某個值的某一行

Question

即時嘗試選擇全部並跳過某一行。

這行有一些語法錯誤，我不知道我做錯了什麼。

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

更多的代碼

$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; 
     } 

感謝。

Answer 1

您需要使用WHERE只有一次：

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

此外，設置%約$的數據，如果你想搜索包含$數據串詞：

bbookschool LIKE '%$data%' 

PS：我相信time也是reserved word，所以加上引號（`）。

Answer 2

更改查詢這一個：

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

Answer 3

試試這個

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

Answer 4

試試這個。

SELECT * FROM ads WHERE LIKE '％$數據％' bbookschool和bbookname！= '測試' ORDER BY time遞減