drexlz0623 2011-10-10 18:01
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mysql like语句没有按预期工作

I have a table with 4 record.

Records: 1) arup Sarma
         2) Mitali Sarma
         3) Nisha
         4) haren Sarma

And I used the below SQL statement to get records from a search box.

$sql = "SELECT id,name FROM ".user_table." WHERE name LIKE '%$q' LIMIT 5";

But this retrieve all records from the table. Even if I type a non-existence word (eg.: hgasd or anything), it shows all the 4 record above. Where is the problem ? plz any advice..

This is my full code:

$q = ucwords(addslashes($_POST['q']));
$sql = "SELECT id,name FROM ".user_table." WHERE name LIKE '%".$q."' LIMIT 5";
$rsd = mysql_query($sql);
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3条回答 默认 最新

  • doucan1979 2011-10-10 18:05
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    Your query is fine. Your problem is that $q does not have any value or you are appending the value incorrectly to your query, so you are effectively doing:

    "SELECT id,name FROM ".user_table." WHERE name LIKE '%' LIMIT 5";
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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