PHP File Upload загрузить файл, даже если оператор не является правильным

Question

HTML:

<form action="" method="post" enctype="multipart/form-data"> 
    Select file to upload: 
    <input type="file" name="fileToUpload" id="fileToUpload"> 
    <input type="password" name="filepassword" id="filepassword"> 
    <input type="submit" value="Upload File" name="submit"> 
</form> 

PHP:

<?php 
if(isset($_POST["submit"])) { 
    $target_dir = "uploads/"; 
    $target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]); 
    $valueOne =trim($_POST["filepassword"]); 
    if($valueOne != "1212"){ 
     move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file); 
     echo "file uploaded successfully !"; 
    } 
    else{ 
     echo "file is not entered !"; 
    } 
} 
?> 

Независимо от того, что код внутри если (проверка) заявление, файл получить закачано

Answer 1

Перепишите код, как показано ниже: -

if(!empty($_POST["submit"]) && !empty($_POST["filepassword"]) && !empty($_FILES['fileToUpload'])){ 
    $valueOne =trim($_POST["filepassword"]); 
    $target_dir = "uploads/"; 
    $target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]); 
    if($valueOne == "1212"){ // if password is 1212 then file will upload. 
    // make sure you want to check == or != 
      move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file); 
      echo "file uploaded successfully !"; 
    } 
    else{ 
      echo "file is not entered !"; 
    } 
} 

Answer 2

Просто поменять код внутри, если и еще

if($valueOne != "1212"){ 
    echo "file is not entered !"; 
    } 
    else{ 
    move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file); 
    echo "file uploaded successfully !"; 
    }