Php 现有用户的Mysql更新/插入

我有一个表单，用于获取两个输入：user\u avatar和user\u backgroundpicture。我想从单个表单中更新（如果是现有用户）或插入（如果是首次注册）

这是我的密码：

<?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');

}  
 ?>

您可以尝试以下方法：

/** 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 {


提示：


不要使用
mysql.*
函数
试着重写你的代码，使它不那么混乱
还是不行！只有在同时给出用户头像和用户背景图片时才有效。如果只给出一个输入，那么另一个输入将变为空。您对这两个实例都做了建议的更改，对吗？近
mysql\u query（$sql5）
也？
if（mysql\u num\u rows（$result）>0）{if（！empty（$\u FILES['BackgroundImageFile']）{//备选方案：（$\u FILES['BackgroundImageFile']['name']==''）或（$\u FILES[“BackgroundImageFile”[“error”]==4）mysql\u query（$sql1）或die（mysql\u error（））；头（'location:../input.php？profile=updated'）；}}else{mysql_query（$sql2）或die（mysql_error（））；header（'location:../input.php？profile=notupdated'）；}
if（mysql_num_rows（$result）>0{if（！empty（$_FILES['ImageFile']）{//备选方案：（$_FILES['BackgroundImageFile']['name'==''）或（$)u FILES[“BackgroundImageFile”][“error”]==4）mysql_查询（$sql5）或die（mysql_error（））；头（'location:../input.php？profile=updated'）；}}其他{mysql_查询（$sql6）或die（mysql_error（））；头（'location:../input.php？profile=notupdated'）；}

您尝试过任何替代方案吗？

替代方案：（$\u文件['BackgroundImageFile']['name']=''）或（$\u文件['BackgroundImageFile'][“error”]==4）

？我也不确定是否有许多

头（）

存在。按照您现在的代码方式，脚本将发生冲突。