Ajax表單提交圖像時出錯

Question

當在同一頁面上提交表單並使更改出現時，我很難嘗試進行AJAX編輯更改，但圖像拋出錯誤： Undefined index: image in update.php on line 22 and 24 。 它拒絕傳遞值。

表格（editForm.php）：

<div class="modal-content editDisplay">
 <div class="modal-header">
  <button type="button" class="close" data-dismiss="modal"><span aria-hidden="true">×</span><span class="sr-only">Close</span></button>
  <h4 class="modal-title" id="editModalLabel">Edit Item</h4>
</div>
<form class="editForm" method="post" enctype="multipart/form-data">
  <div class="modal-body">
        <div class="form-group">
            <label for="inputName">Name</label>
            <input type="text" class="form-control" id="inputName" name="Product_Name" placeholder="Name" value="<?php echo $product ?>">
            <input type="hidden" name="oldProduct" value="<?php echo $oldProduct ?>">
        </div>
        <div class="form-group">
            <label for="inputDescription">Description</label>
            <textarea class="form-control" id="inputDescription" name="Description" placeholder="Description"><?php echo $description ?></textarea>
        </div>
        <div class="form-group">
            <label for="inputPrice">Price</label>
            <input type="text" class="form-control" id="inputPrice" name="Price" placeholder="Price" value="<?php echo $price ?>">
        </div>
        <div class="form-group">
            <label for="inputQuantity">Quantity</label>
            <input type="number" class="form-control" id="inputQuantity" name="Quantity" placeholder="Quantity" value="<?php echo $quantity ?>">
        </div>
        <div class="form-group">
            <label for="inputSalePrice">Sale Price</label>
            <input type="text" class="form-control" id="inputSalePrice" name="Sale_Price" placeholder="Sale Price" value="<?php echo $salePrice ?>">
        </div>
        <div class="form-group">
            <label for="inputImage">Image Upload</label><br>
            <fieldset class="file-fieldset">
                <span class="btn btn-default btn-file">
                    <span class="glyphicon glyphicon-upload"></span>&nbsp;&nbsp;Browse Browse <input name="image" type="file" id="inputImage"/><br>
                </span>
                <input type="hidden" name="prevPicture" value="<?php $image ?>"/>
                <span style="margin-left:8px;" value=""><?php echo $image ?></span>

            </fieldset>
        </div>
  </div>
  <div class="modal-footer">
    <button type="reset" class="btn btn-default">Reset</button>
    <button type="submit" class="btn btn-primary" id="saveButton" name="update">Save Changes</button>
  </div>
 </form>
</div>

PHP（update.php）：

<?php

include('connection.php');
include('LIB_project1.php');

$productName = $_POST['Product_Name'];
$productDescription = $_POST['Description'];
$price = $_POST['Price'];
$quantity = $_POST['Quantity'];
$salePrice = $_POST['Sale_Price'];
$oldImage = $_POST['prevPicture'];
$oldProduct = $_POST['oldProduct'];


//$productName = 'Jaime';
//$productDescription = 'This is crazy';
//$price = '0';
//$quantity = '12234';
//$salePrice = '0';
//$oldImage = $_POST['prevPicture'];
//$oldProduct = $_POST['oldProduct'];
$imageName= $_FILES['image']['name']; //TODO line 22
echo ' The image is '.$imageName;
$image_temp = $_FILES['image']['tmp_name']; // line 24
echo 'Product name is: '.$productName;

//$productName = 'Dodo Square';
//$productDescription = 'Flower on a Bee. Such Beauty!';
//$price = 9;
//$quantity = 8;
//$salePrice = 230;
//$newImage = '038.jpg';
//$oldProduct = 'Times Square';

//working under the assumption that the image already exist in the database
$targetDirectory = 'productImages'; 
$files = scandir($targetDirectory,1);
//code passed
for($i=0; $i<sizeof($files); $i++)
{
    if($oldImage==$files[$i])
    {
        unlink('productImages/'.$oldImage);
    }
}

$target = "productImages";

//add the image to the directory
$target = $target.'/'.$imageName;
move_uploaded_file($image_temp,$target);  

updateProduct($conn,'product',':productName', ':productDescription', ':price', ':quantity', ':imageName', ':salePrice',                          'Product_Name', 'Description', 'Price', 'Quantity', 'Image_Name', 'Sale_Price', $productName, $productDescription, $price, $quantity,$imageName, $salePrice, $oldProduct, ':oldProduct');
//header('location:admin.php');

?>

的UpdateProduct（...）

/*
 * This is a function to update Product
 * 
 */
 function updateProduct(PDO $connection,$table,$bindProductName, $bindProductDescription, $bindPrice, $bindQuantity, $bindImageName, $bindSalePrice,$productNameColumn, $productDescriptionColumn, $priceColumn, $quantityColumn, $imageNameColumn, $salePriceColumn,                                    $productName, $productDescription, $price, $quantity, $imageName, $salePrice, $oldProduct, $bindOldProduct)
 {
     $result = false;

     $sql = 'UPDATE ' . $table . ' SET ' . $productNameColumn . ' = ' . $bindProductName . ',' . $productDescriptionColumn . ' = ' .                            $bindProductDescription . ',' . $priceColumn . ' = ' . $bindPrice . ',' . $quantityColumn . ' = ' . 
       $bindQuantity . ',' . $salePriceColumn . ' = ' . $bindSalePrice . ',' . $imageNameColumn . ' = ' . $bindImageName . ' WHERE ' .                      $productNameColumn . ' = ' . $bindOldProduct;

    $smtp = $connection->prepare($sql);
    $smtp -> bindParam($bindProductName, $productName);
    $smtp -> bindParam($bindProductDescription, $productDescription);
    $smtp -> bindParam($bindPrice, $price);
    $smtp -> bindParam($bindQuantity, $quantity);
    $smtp -> bindParam($bindImageName, $imageName);
    $smtp -> bindParam($bindSalePrice, $salePrice);
    $smtp -> bindParam($bindOldProduct, $oldProduct);

    if($smtp->execute() )
    {
        $result = true;
    }

    return $result;
}

AJAX（顯示已編輯的更改）問題：需要提交這些已編輯的更改

 $(document).ready(function() 
 {
           //the user click save edit
   $(".edit").on("submit",function(e)
   { 
       e.preventDefault();
       $.ajax({
            type:"POST",
            url:'update.php', //I will put project id here as well
            data:$(".editForm").serialize(),
            success:function(smsg)
            {
                alert(smsg);
               //update the number of items the user has in their shopping cart
                $.get('admin.php',function(data){
                $('#refresh').load("admin.php #refresh");
                    //alert('success');
                });
            }
        });


   });


   });

Answer 1

在PHP中添加一系列測試，您將很快自己解決這個問題。

您已經在警告PHP ajax處理器文件發送的響應：

alert(smsg);

因此，使用它來解決/診斷出現問題的地方。

第一個測試可能是在PHP文件的頂部放置一個“我在這里”的消息 - 至少你知道ajax本身正在工作。 因此，將update.php的頂部修改為：

<?php
echo "Got to here";
die();

如果有警報，那么就去掉die()並在文件的不同位置回顯幾個這樣的測試。 使用此方法縮小錯誤的位置（在PHP文件中）。

回收收到的數據，以便了解其中的內容。修改update.php為：

$productName = $_POST['Product_Name'];
$productDescription = $_POST['Description'];
$price = $_POST['Price'];
$quantity = $_POST['Quantity'];
$salePrice = $_POST['Sale_Price'];
$oldImage = $_POST['prevPicture'];
$oldProduct = $_POST['oldProduct'];

echo "$productName: " .$productName. " -- $productDescription: " .$productDescription. " - etc etc etc";
die();

我相信你會很快發現錯誤 - 當然比編寫詳細的SO問題和等待答案更快......

祝好運！

Answer 2

var inputImage = $("#inputImage");
var fd = new FormData(document.getElementById("editform")); 

fd.append("image", inputImage);

$.ajax({
    url: "",
    type: "POST",
    data: fd,
    processData: false, 
    contentType: false, 
    success: function(response) {

    }
});

默認情況下，圖片未在帖子中添加到表單中，您需要獲取整個表單並在發送之前將圖像附加到表單。 我為asp.net做了這個，它也適用於PHP。

Answer 3

也許是因為圖像永遠不會通過ajax發送到服務器，所以$_FILES在'image'索引下沒有任何內容。 看看如何使用jquery序列化進行文件上傳

或者考慮使用FormData對象。

Answer 4

更改<input name="image" type="file" id="inputImage"/ to <input name="image" type="file" id="inputImage"/>