douzi7890 2015-11-07 09:23
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MySQL表没有更新没有错误

I have this little block of PHP code used to update a SQL table. However, it should, because right now it's not doing anything, nor is it producing any errors:

$sql = "INSERT INTO `wordpress`.`thor_members` ($key) VALUES('$_POST[$key]') WHERE ID = '$_POST[ID]'";
var_dump($sql);
mysqli_real_escape_string($conn, $sql);

if(mysqli_error($conn)){
    var_dump(mysqli_error($conn));
}

The var_dump actually prints MySQL lines that work perfectly fine if I run them into PHPMyAdmin.

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  • droxy80248 2015-11-07 09:37
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    $keyVal = mysqli_real_escape_string($conn, $_POST[$key]);
    $id     = intval(mysqli_real_escape_string($conn, $_POST['ID']));
    // I'm assuming $_POST['ID'] is an int?    
    
    $sql    = "UPDATE `wordpress`.`thor_members` SET $key = '$keyVal' WHERE ID = $id";
    
    mysqli_query($conn, $sql);
    
    if(mysqli_error($conn)){
        var_dump(mysqli_error($conn));
    }
    

    This should work.

    If $key comes from an unknown source, you should escape that too but id you set that in your own code before, this should be enough.

    However, you should really look up prepared statements instead.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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