成功印刷多次

my code running correctly but if $file have three item then after successful insert success message print three time.But I need to print message for one time at all.

for($i = 1; $i <= count($file); $i++){
        $insert=mysqli_query($con,"INSERT INTO song (song_name, song_url, album,artist)
        VALUES ('".current(explode(".", $file))."', '".$file."','".$album."','".$artist."')");

        if($insert){            
            print'Success';         
        }else{
            print''.mysql_error().'';           
        }
    }

Write it this way:

$success = true;
for($i = 1; $i <= count($file); $i++)
{
    $insert=mysqli_query($con,"INSERT INTO song (song_name, song_url, album,artist)
    VALUES ('".current(explode(".", $file))."', '".$file."','".$album."','".$artist."')");

    if(!$insert)
    {   $success = false;
        print''.mysql_error().'';
    }
}
if($success){
    print'Success';
}

This only prints "Success" only, if all queriey succeeded, and it will only print it once.

You should consider mysqli_* instead of mysql_*, as this functions are already deprecated due to security issues.

try this: (you also wrote mysql_error instead of mysqli_error, this fixes it as well.)

$success = 0;
for($i = 1; $i <= count($file); $i++)
{
    $insert=mysqli_query($con,"INSERT INTO song (song_name, song_url, album,artist)
    VALUES ('".current(explode(".", $file))."', '".$file."','".$album."','".$artist."')");


    if($insert)
    {
        $success = 1;
    }
    else
    {
        print''.mysqli_error($con).'';
        $success = 0;
        break;
    }


}
if($success)
print'Success';

Stop using mysql_* functions, they are deprecated.

Try This:

$error = false;
for($i = 1; $i <= count($file); $i++){
        $insert=mysqli_query($con,"INSERT INTO song (song_name, song_url, album,artist)
        VALUES ('".current(explode(".", $file))."', '".$file."','".$album."','".$artist."')");

        if($insert){    
            $error = true;    
        }
        else{
            $error = false;
            print''.mysql_error().'';    
        } 
    }
if($error) {
    print 'Success';
}

use in this way

function abc(){
    $flag = "";    
    for($i = 1; $i <= count($file); $i++)
        {
            $insert=mysqli_query($con,"INSERT INTO song (song_name, song_url, album,artist)
            VALUES ('".current(explode(".", $file))."', '".$file."','".$album."','".$artist."')");
        if($insert)
        {
          $flag = true;
        }
        else
        {
        print''.mysql_error().''; return;
        }
    }
    if($flag == true){
       print 'success';
    }
}

You could do something like the following:

$errors = array();

for ($i = 1, $n = count($file); $i <= $n; ++$i)
{

  $insert = mysqli_query($con, $query);

  if (!$insert)
  {
    $errors[] = mysqli_error($con);
  }

}

if (empty($errors))
{
  echo 'Success';
}

else
{
  echo 'Errors:<br><br>' . implode('<br>', $errors);
}

Also note that I fixed your call to mysql_error() to the correct mysqli_error() function.