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Home/ Questions/Q 8258593
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Editorial Team
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Editorial Team
Asked: June 8, 20262026-06-08T02:35:43+00:00 2026-06-08T02:35:43+00:00

I have a code which kinda works, but not really i can’t figure out

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I have a code which kinda works, but not really i can’t figure out why, what im trying to do is check inside the database if the URL is already there, if it is let the user know, if its not the go ahead and add it.

The code also makes sure that the field is not empty. However it seems like it checks to see if the url is already there, but if its not adding to the database anymore. Also the duplicate check seems like sometimes it works sometimes it doesn’t so its kinda buggy. Any pointers would be great. Thank you.

    if(isset($_GET['site_url']) ){

    $url= $_GET['site_url'];


    $dupe = mysql_query("SELECT * FROM $tbl_name WHERE URL='$url'");
    $num_rows = mysql_num_rows($dupe);
    if ($num_rows) {
    echo 'Error! Already on our database!';
    }
    else {
    $insertSite_sql = "INSERT INTO $tbl_name (URL) VALUES('$url')";
    echo $url;
    echo ' added to the database!';

    }

}
else {
echo 'Error! Please fill all fileds!';
}
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-08T02:35:44+00:00Added an answer on June 8, 2026 at 2:35 am

    You should take PhpMyCoder’s advice on the UNIQUE field type.

    Also, you’re not printing any errors.

    Make sure you have or die (mysql_error()); at the end of your mysql_* function(s) to print errors.

    You also shouldn’t even be using mysql_* functions. Take a look at PDO or MySQLi instead.

    You’re also not executing the insert query…

    Try this code:

    if(isset($_GET['site_url']) ){
    
    $url= $_GET['site_url'];
    
    
    $dupe = mysql_query("SELECT * FROM $tbl_name WHERE URL='$url'") or die (mysql_error());
    $num_rows = mysql_num_rows($dupe);
    if ($num_rows > 0) {
    echo 'Error! Already on our database!';
    }
    else {
    $insertSite_sql = "INSERT INTO $tbl_name (URL) VALUES('$url')";
    mysql_query($insertSite_sql) or die (mysql_error());
    echo $url;
    echo ' added to the database!';
    
    }
    
    }
    else {
    echo 'Error! Please fill all fileds!';
    }
    
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