Ajax文件上传“尝试加载资源时发生错误”,并且Safari中的网络连接丢失

时间:2018-12-19 11:55:30

标签: php jquery ajax file-upload safari

我正在尝试上传文件和该文件的缩略图(使用cropper v2.3.0)。 该代码可在所有其他浏览器上使用,但在野生动物园中会出现错误。

问题描述如下:

  1. 在桌面上的safari浏览器上上传文件时,发生以下错误,并使用了其他错误,然后safari浏览器没有错误并获得成功消息。

  2. 我测试了两种方法,它们首先仅上传base64中的裁剪图像 编码,或者也可以将blob作为文件添加到formData中,但是在两种方式下均无法解决错误。

  3. 我也只尝试上传图像,然后有时会或有时不会发生错误。

  4. 如果使用裁纸器调整,则会发生错误(这是我的假设)

我的js代码提交表单

function addFile() {
 $("#result").html("");
 var myForm = $('#mainForm');
 var formData = new FormData(myForm[0]);

 $.ajax({
    url: "action.php", // Url to which the request is send
    type: "POST", // Type of request to be send, called as method
    data: formData, // Data sent to server, a set of key/value pairs (i.e. form fields and values)
    contentType: false, // The content type used when sending data to the server.
    cache: false, // To unable request pages to be cached
    processData: false, // To send DOMDocument or non processed data file it is set to false
    dataType: "json",
    success: function (data)            // A function to be called if request succeeds
    {
        $("#result").html(data.response + " : " + data.message);
    },
    error: function (res) {
        $("#result").html(res.responseText);
    }
 });
 return false;
}

我的动作php代码

    <?php

    $uploadThumbnailPath    = "dir";
    $thumbImgData           = $_POST['thumbImg'];

    $numberOfImages = 1;

    $isImageUploaded = 0;
    if ($thumbImgData != "") {
        //thumbnail image uploading code
        list($type, $thumbImgData) = explode(';', $thumbImgData);
        list(, $thumbImgData) = explode(',', $thumbImgData);
        $thumbImgData = base64_decode($thumbImgData);

        $myTimeStamp      = "thumbImg_" . time() . uniqid();
        $displayImageName = $myTimeStamp . ".png";
        $dir              = $uploadThumbnailPath;
        if (file_put_contents("$dir/$displayImageName", $thumbImgData)) {
            $jpgFormatImageName = $myTimeStamp . ".jpg";
            convertPNGtoJPG("$dir/$displayImageName", "$dir/$jpgFormatImageName", 238, 238, 238, 1);
            if (file_exists("$dir/$displayImageName")) {
                unlink("$dir/$displayImageName");
            }
            $isImageUploaded = 1;
        }
    } else {
        $arrayResponse = array("response" => "thumbImg_BLANK", "message" => 'thumbImg_BLANK');
        echo json_encode($arrayResponse);
        exit;
    }


    for ($i = 1; $i <= $numberOfImages; $i++) {

        if (isset($_POST["imgName$i"])) {
            $itemImagesName = "";
        } else {
            $itemImagesName = $_FILES["imgName$i"]['name'];
        }
        if ($itemImagesName != "") {
            $extension                   = pathinfo($itemImagesName, PATHINFO_EXTENSION);
            $uploadNewFileNameWithoutExt = "image_" . md5($i . time());
            $uploadDirPath               = "dir/p/";

            $uploadNewFileName[$i]     = $uploadNewFileNameWithoutExt . '.' . $extension;
            $uploadNewFileWithPathName = $uploadDirPath . $uploadNewFileName[$i];
            $mesUpload                 = uploadImageFileOnServer("imgName$i", $allowedExts, $maxFileSize, $uploadNewFileWithPathName);
        }
    }

    $itemImages = implode("#:#", $uploadNewFileName);

    $thumbnailImageName = "default_thumbnail.png";
    if ($isImageUploaded == 1) {
        $thumbnailImageName = $jpgFormatImageName;
    }

    if ($mesUpload == "FILE_UPLOADED") {
        $arrayResponse = array("response" => "OK", "message" => "OK UPLOAD SUCCESS");
        echo json_encode($arrayResponse);
        exit;
    } else {
        /* $mesUpload */
        $arrayResponse = array("response" => "FILE_FAILED", "message" => "FAIL TO UPLOAD");
        echo json_encode($arrayResponse);
        exit;
    }
    ?>

以下是该错误的屏幕截图 这是

screen shot 1

screen shot 2

请帮助我解决此问题。我为这个错误感到困惑,我还没有任何想法来解决这个问题。 如果有人要使用,我在网上上传了示例代码,请点击以下链接 https://tamapev.000webhostapp.com/upload-img/

1 个答案:

答案 0 :(得分:0)

问题可能是您将表单数据发送到错误的页面,首先确定action.php是在根目录中还是在名为“ upload-img”的目录中。然后将请求发送到给定的页面。

接下来显示“尝试加载资源时发生错误”的错误,以查看实际错误,在第一个屏幕截图中,面板上有一个小按钮,显示“ Response”,单击并更改改为“ JSON”

如果“ action.php”位于“ upload-img”中,则需要更改

url: "action.php",url: "/upload-img/action.php",