Question

我有一个用于获取两个输入的表单：user_avatar和user_backgroundpicture。我想从单个表单更新（如果是现有用户）或插入（如果是第一次注册）。

这是我的代码：

<?php
ini_set("display_errors",1);
if(isset($_POST))
{
require '../_inc/db.php';
$Destination = '../Backgroundimages';
if(!isset($_FILES['BackgroundImageFile']) || !is_uploaded_file($_FILES['BackgroundImageFile']['tmp_name']))
{
    //die('Something went wrong with Upload!');
    $BackgroundNewImageName= 'background4.jpg';

    move_uploaded_file($_FILES['BackgroundImageFile']['tmp_name'], "$Destination/$BackgroundNewImageName");
}
else{
$RandomNum   = rand(0, 9999999999);

$ImageName      = str_replace(' ','-',strtolower($_FILES['BackgroundImageFile']['name']));
$ImageType      = $_FILES['BackgroundImageFile']['type']; //"image/png", image/jpeg etc.

$ImageExt = substr($ImageName, strrpos($ImageName, '.'));
$ImageExt = str_replace('.','',$ImageExt);

$ImageName      = preg_replace("/\.[^.\s]{3,4}$/", "", $ImageName);

//Create new image name (with random number added).
$BackgroundNewImageName = $ImageName.'-'.$RandomNum.'.'.$ImageExt;

move_uploaded_file($_FILES['BackgroundImageFile']['tmp_name'], "$Destination/$BackgroundNewImageName");

}



   require 'authenticationforupload.php';



      $sql1="UPDATE user SET user_backgroundpicture='$BackgroundNewImageName' WHERE user_username = '$user_username'";


$sql2="INSERT INTO user (user_backgroundpicture) VALUES ('$BackgroundNewImageName') WHERE user_username = '$user_username'";

$result = mysql_query("SELECT * FROM user WHERE user_username = '$user_username'");
if( mysql_num_rows($result) > 0) {
mysql_query($sql1)or die(mysql_error());
header('location:../input.php?profile=updated');
}
else{
mysql_query($sql2)or die(mysql_error());
header('location:../input.php?profile=notupdated');

}  








    $Destination = '../uploads';
if(!isset($_FILES['ImageFile']) || !is_uploaded_file($_FILES['ImageFile']['tmp_name']))
{
    //die('Something went wrong with Upload!');
    $NewImageName= 'default.png';

    move_uploaded_file($_FILES['ImageFile']['tmp_name'], "$Destination/$NewImageName");
}
else{
$RandomNum   = rand(0, 9999999999);

$ImageName      = str_replace(' ','-',strtolower($_FILES['ImageFile']['name']));
$ImageType      = $_FILES['ImageFile']['type']; //"image/png", image/jpeg etc.

$ImageExt = substr($ImageName, strrpos($ImageName, '.'));
$ImageExt = str_replace('.','',$ImageExt);

$ImageName      = preg_replace("/\.[^.\s]{3,4}$/", "", $ImageName);

//Create new image name (with random number added).
$NewImageName = $ImageName.'-'.$RandomNum.'.'.$ImageExt;

move_uploaded_file($_FILES['ImageFile']['tmp_name'], "$Destination/$NewImageName");


}



       $sql5="UPDATE user SET user_avatar='$NewImageName' WHERE user_username = '$user_username'";


$sql6="INSERT INTO user (user_avatar) VALUES ('$NewImageName') WHERE user_username = '$user_username'";

$result = mysql_query("SELECT * FROM user WHERE user_username = '$user_username'");
if( mysql_num_rows($result) > 0) {
mysql_query($sql5)or die(mysql_error());
header('location:../input.php?profile=updated');
}
else{
mysql_query($sql6)or die(mysql_error());
header('location:../input.php?profile=notupdated');

}  
 ?>

当上传user_avatar和user_backgroundpicture时，表单正常工作。如果上传了任何一个，而另一个保持不变，则sql的操作只会删除未给出输入的条目。

我想要的是，如果给出一个输入，则只应将该输入插入数据库。另一个应保持不变。

Answer 1

你可以试试这个：

/** Updates the database only if the file field isn't empty. **/
    if( mysql_num_rows($result) > 0) {
        if(!empty($_FILES['BackgroundImageFile'])){ // ALTERNATIVES: ($_FILES['BackgroundImageFile']['name']=='') or ($_FILES["BackgroundImageFile"]["error"] == 4)
            mysql_query($sql1)or die(mysql_error());
            header('location:../input.php?profile=updated');
        }
    } else {

<强>提示：

避免使用mysql_*个功能。
尝试重写代码，使其不那么混乱。

现有用户的Mysql更新/插入

1 个答案: