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;
}
?>`
