Question

这里我试图使用从其他表中提取的信息将数据输入到表播放列表中。在执行它之后，我收到了“成功”消息，但实际上没有插入数据。任何人都可以指出错误吗？

 <?php

$connect= mysql_connect("localhost","root","") or die('couldnt connect to database');

mysql_select_db("music") or die("couldn't find the database");

        $regis['id']=$_GET['id'];
        $regis['username']=$_GET['username'];

           $query="SELECT * FROM songs WHERE id ='id'";
                       $quer="SELECT * FROM images WHERE id ='id'";
            $q="SELECT * FROM users WHERE username ='username'";




           $query=mysql_query($query);
          $quer=mysql_query($quer);

            $q=mysql_query($q);



           $row=mysql_fetch_assoc($query);

          $regis['id']=$row['id'];
           $regis['title']=$row['title'];

          $regis['artist']=$row['artist'];

        $rows=mysql_fetch_assoc($q);
         $regis['link']=$rows['link'];


      $r=mysql_fetch_assoc($q);
         $regis['uid']=$r['uid'];


        $sql="insert into playlist (songid,song,image,album,uid) values('{$regis['id']}','{$regis['title']}','{$regis['link']}','{$regis['artist']}','{$regis['uid']}')";
        $res=mysql_query($sql);
        if($res)
        {
            echo "<script type='text/javascript'>";
            echo 'alert("addition to your playlist was succesful")';
            echo "</script>";

            }
        else
        {   
            echo "<script type='text/javascript'>";
            echo 'alert("'.mysql_error().'Please try again")';
            echo "</script>";
        }

＆GT;

Answer 1

if(mysql_num_rows($res)>0){
  //query inserted at least one row in db
}

即使显示成功的消息，数据也不会插入到表中

1 个答案: