Question

您能否看一下这段代码，让我知道为什么我无法上传文件中存储的所有图片。代码只会将one图像上传到服务器中。

HTML

        <form id="uploader" action="#" method="post">
            Package Name
            <input type="text" name="rental_id_box" id="rental_id_box"/>
            <br /> Package Images
            <input id="selected_imgs" name="selected_imgs" type="file" multiple>
        </form>

JS：

    $("#btn-upload").on("click", function() {

      var fd = new FormData($('form')[0]);
    //fd.append( 'file', input.files[0] );
      var request = $.ajax({
        type: "POST",
        url: "upload.php",
        data: fd,
        processData: false,
        contentType: false,
        cache: false,
        beforeSend: function() {
        console.log(fd);
        }
      });

      request.done(function(data) {
          $("#view").html(data)
      });

      request.fail(function(jqXHR, textStatus) {
        console.log("Request failed: " + textStatus);
      });
    });

和PHP

<?php
$title = $_POST['rental_id_box'];
foreach($_FILES as $index => $file)
    {
        $fileName     = $file['name'];
        $fileTempName = $file['tmp_name'];

        if(!empty($file['error'][$index]))
        {
            return false;
        }
        if(!empty($fileTempName) && is_uploaded_file($fileTempName))
        {
            move_uploaded_file($fileTempName, "uploads/". $fileName);
            echo '<p>Click <strong><a href="uploads/' . $fileName . '" target="_blank">' . $fileName . '</a></strong> to download it.</p>';
        }
    }

Answer 1

...&& is_uploaded_file($_FILES['name']['tmp_name'])

Answer 2

if(count($_FILES['selected_imgs']['name']) > 0){
for($i=0; $i<count($_FILES['selected_imgs']['name']); $i++) {
    $tmpFilePath = $_FILES['selected_imgs']['tmp_name'][$i];
    print_r($tmpFilePath);
    if(($tmpFilePath != "")&&(is_uploaded_file($tmpFilePath))){            
        $filePath = "upload/" . date('d-m-Y-H-i-s').'-'.$_FILES['selected_imgs']['name'][0];
        move_uploaded_file($tmpFilePath, $filePath);
      }
}

}

将多个映像上载到服务器的问题

2 个答案: