這不行？有什麼建議麼？

Question

我有我正在爲音樂評分的這個腳本，沒有錯誤，而且我得到'Data Inserted'，我知道我還沒有逃脫數據，但是有沒有問題在這個腳本中可能會阻止它將插入的數據放入數據庫中？

<? 
    include($_SERVER['DOCUMENT_ROOT'].'/assets/global/scripts/connect.php'); 
    $songname = $_GET['songname']; 
    $artist = $_GET['artist']; 
    $ratenum = 1; 
    $chkquery = "SELECT * FROM hotmuze_music WHERE songname='$songname'"; 
    $plusOneQuery = "SELECT * FROM hotmuze_music WHERE songname='$songname'"; 
    $updateQuery = "UPDATE hotmuze_music SET rating='$rating2' WHERE songname='$songname'"; 
    $checkdata = mysql_query($chkquery); 
    $checkrows = mysql_num_rows($checkdata); 
    if($checkrows==0) 
    { 
    $insquery = "INSERT INTO hotmuze_music (id, songname, artist, sex, genre, rating, promoted) VALUES('', '$songname', '$artist', '', '$genre' '$ratenum')"; 
    $insdata = mysql_query($insquery); 
    } 


    if($checkrows!=0) 
    { 
    $plusData = mysql_query($plusOneQuery); 
    } 

    if(mysql_num_rows($plusData)!=0) 
    { 
    $result = mysql_fetch_assoc($plusData); 
    $rating = $result['ratng']; 
    $rating2 = $rating+1; 
    mysql_query($updateQuery); 
    echo "Data Inserted"; 
    } 
    ?> 

謝謝！ :)

Answer 1

你得到Column count doesn't match value count at row 1的錯誤是由你的INSERT語句失蹤1個人更DB柱造成的。

(id, songname, artist, sex, genre, rating, promoted) 7個領域，你只給值6這是VALUES('', '$songname', '$artist', '', '$genre' '$ratenum')";

Answer 2

你錯過了一個逗號，價值提升領域，以及嘗試：

 
$insquery = "INSERT INTO hotmuze_music (id, songname, artist, sex, genre, rating, promoted) VALUES('', '$songname', '$artist', '', '$genre', '$ratenum', '$promoted')"; 

Answer 3

您在腳本引用變量（$rating2）的頂部定義$updateQuery其中尚未確定。嘗試最後一行改爲：

if(mysql_num_rows($plusData)!=0) 
{ 
    $result = mysql_fetch_assoc($plusData); 
    $rating = $result['rating']; 
    $rating2 = $rating+1; 
    mysql_query("UPDATE hotmuze_music SET rating='$rating2' WHERE songname='$songname'"); 
    echo "Data Inserted"; 
}