dongmao3131 2013-12-01 18:32
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mysqli查询语句不起作用

I have written a mysqli statement for advanced search of my website. the generated query does not work. here is the generated query:

 SELECT title, stars, pic_main, id, province FROM ads WHERE 'province' = 8 AND( title LIKE %شریف% OR content LIKE %شریف% OR keywords LIKE %شریف% OR title LIKE %شریف%)

And here is the code that I use to execute the query:

if($transorder = $site_db->query($statement))
{
    echo "True";
    $i=0;
    while($row_obj = $transorder->fetch_object())
    {
        $item[$i]['id'] = $row_obj->id;
        $item[$i]['pic_main'] = $row_obj->pic_main;
        $item[$i]['title'] = $row_obj->title;
        $item[$i]['province'] = $row_obj->province;
        $item[$i]['stars'] = $row_obj->stars;       
        $i++;
    }
}
else
{
    echo "False";
}
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1条回答 默认 最新

  • dsvs50005 2013-12-01 18:33
    关注

    You need quotes around text in your LIKE statements, also I notice you are using title 2 times. If you use single quote (province), MySQL might expect you are comparing the String 'province' with the value 8 you should use ` around the 'province':

     SELECT title, stars, pic_main, id, province 
     FROM ads 
     WHERE `province` = 8 
           AND ( title LIKE '%شریف%' 
                 OR content LIKE '%شریف%' 
                 OR keywords LIKE '%شریف%' 
                 OR title LIKE '%شریف%')
    

    To find out what error MySQL is returning you can use echo $site_db->error;.

    The Solution was

    $site_con->set_charset("utf8");
    

    Because of the use of the use of none default characters of the MySQL connection. So the connection should be set to utf8

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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