خرید بک لینک

Vote count: 0

So i am trying to display the name of the picture file, when the Movie_ID is equal to 12, written in the database, though my SELECT query is coming back with an error and i am not sure where it goes wrong. Here is the php code i am using.

'<?php

    $con=mysqli_coect("fdb3.awardspace.net","2113371_imbd","password1","2113371_imbd");
    $result = mysqli_query($con, "SELECT PictureName FROM `Movie` WHERE Movie_ID = 12");
    if (mysqli_coect_ero())
    {
            print "Failed to coect to MySQl:" . mysqli_error();
    }
    if (!$check1_res) {
printf("Error: %sn", mysqli_error($con));
exit();
}


           while($row = mysqli_fetch_array($result)){
            //echo $row['image'];
            $PictureName =$row['PictureName'];
            echo $PictureName;
    }
    ?>`

asked 34 secs ago

برچسب: نویسنده: استخدام کار تاريخ: سه شنبه 11 خرداد 1395 ساعت: 7:11

صفحه بندی