AJAX / Javascript与上传PHP问题

Question

I have been working on making a form that uploads an image through ajax with php doing the upload. For some background I am having an issue where the xhr.readyState and .Status seem to run and upload.php seems to run (in safari it goes and runs upload.php it seems to me... I have not used the ajax/javascript debugger too much) but it will run my else statement (i believe becuase it goes to status 2 and then 4) and then run xhr.responseText. Unfortunately it does not seem to run the upload.php right (which may be a php problem). I went ahead and replaced the php with a print statement which when running alert(xhr.response.Text) comes up with a blank alert statement and I am at a loss where the issue is. Below is my form code, ajax code and the php code i'm trying to get a response from.

Does anyone see the issue I am having? I am new to AJAX and have not worked with PHP in some time. I am working on upping my ajax/javascript skills but from my view it does not seem to be getting the ajax correctly.

HTML CODE

> <form action="scripts/upload.php" method="post"
> enctype="multipart/form-data">
>       <label>Select a File to Upload</label> <input id="upload_file" type="file" name="upload_file" /> 
>       <input type="button" onclick="loadFile()" value="Upload"  />    </form>

JAVASCRIPT/AJAX CODE

//LOAD IMAGES THROUGH JAVASCRIPT WINDOW.ONLOAD = INIT();

$(document).ready(function() {
  // Handler for .ready() called.

  $('.heading').click(function() {
      if($(this).find('li').is(':visible')) {   
        $(this).find('li').slideUp('slow'); 
    }else {
      $('.heading').find('li').slideUp('slow');
      $(this).find('li').slideDown('slow'); 
    }
  })
});

var xhr = createRequest();

function loadFile() {
//retrieve the FileList object from the referenced element ID
    var myFileList = document.getElementById('upload_file').files;

    //grab the first file object from the filelist
    var myFile = myFileList[0];

    //set some variables containing attributes of file
    var myFileName = myFile.name;
    var myFileSize = myFile.size;
    var myFileType = myFile.type;

    //alert the information gathered and if it is right
    alert("FileName: " + myFileName + "- FileSize: " + myFileSize + " - FileType: " + myFileType);

    //check above by alert if file size is below a certain MB and of image type JPEG, PNG, GIF

    //Let's upload the complete file object
    uploadFile(myFile);
}

function uploadFile(myFileObject) {
    // Open Our formData Object
    var formData = new FormData();

    // Append our file to the formData object
    // Notice the first argument "file" and keep it in mind
    formData.append('my_uploaded_file', myFileObject);

    // Create our XMLHttpRequest Object
    xhr.onreadystatechange = addImage;

    // Open our connection using the POST method
    xhr.open("POST", 'upload.php');
    //Send the proper header information along with the request
    xhr.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');

    // Send the file
    xhr.send(formData);
}

function addImage() {
    if (xhr.readyState === 4 && xhr.status === 200) {
        alert("WIN");
        alert(xhr.responseText);
    } else {
        alert("ERROR ERROR: " + xhr.status);
    }
}

AJAX REQUEST CODE

function createRequest() {
  try {
    request = new XMLHttpRequest();
  } catch (tryMS) {
    try {
      request = new ActiveXObject("Msxml2.XMLHTTP");
    } catch (otherMS) {
      try {
        request = new ActiveXObject("Microsoft.XMLHTTP");
      } catch (failed) {
        request = null;
      }
    }
  } 
  return request;
}

PHP CODE

<?php
    echo "hello";
?>

<?php
/*
define("UPLOAD_DIR", "../images");

if (!empty($_FILES["myFile"])) {
    $myFile = $_FILES["myFile"];

    if ($myFile["error"] !== UPLOAD_ERR_OK) {
        echo "<p>An error occurred.</p>";
        exit;
    }

    // ensure a safe filename
    $name = preg_replace("/[^A-Z0-9._-]/i", "_", $myFile["name"]);

    // don't overwrite an existing file
    $i = 0;
    $parts = pathinfo($name);
    while (file_exists(UPLOAD_DIR . $name)) {
        $i++;
        $name = $parts["filename"] . "-" . $i . "." . $parts["extension"];
    }

    // preserve file from temporary directory
    $success = move_uploaded_file($myFile["tmp_name"],
        UPLOAD_DIR . $name);
    if (!$success) { 
        echo "<p>Unable to save file.</p>";
        exit;
    }

    // set proper permissions on the new file
    chmod(UPLOAD_DIR . $name, 0644);
}
*/
?>

I pasted all my PHP code although right now i'm just working with the echo PHP code to get a response, I am sure there are many things (besides security checking) with my upload php code. But if you guys can point me in the right direction or help me figure out if or why not the ajax call is working. I believe it is personally just no responseText. This is the first time i've worked with Ajax creating my own PHP code to be run.

Answer 1

In your form's action attribute you have scripts/upload.php but the ajax url is upload.php.

Do not set the content type header, it will be set automatically to multipart/form-data when you pass a formdata object to XMLHttpRequest.send .

The FormData object lets you compile a set of key/value pairs to send using XMLHttpRequest. Its primarily intended for use in sending form data, but can be used independently from forms in order to transmit keyed data. The transmitted data is in the same format that the form's submit() method would use to send the data if the form's encoding type were set to "multipart/form-data".