Question

我正在使用Krajee的Bootstrap文件输入插件通过AJAX调用执行上传。

以下是Krajee插件AJAX部分的链接：Krajee plugin AJAX

我正在使用的JS和PHP（codeigniter）代码如下：

JS：

<script>        
    $("#file-upload").fileinput({ 
        'allowedFileExtensions' : ['csv'],
        'maxFileSize': 5120,
        'maxFileCount': 1,
        'uploadUrl': 'dashboard/uploader',
        'elErrorContainer': '#errorBlock',
        'uploadAsync': true,
        'msgInvalidFileExtension': 'Invalid extension for file "{name}". Only "{extensions}" files are supported.',
        'uploadExtraData': {csrf_token_name: $("input[name=csrf_token_name]").val()}
    });       
</script>

PHP：

public function uploader(){
    $config['upload_path'] = './csv_uploads/';
    $config['allowed_types'] = 'csv';
    $config['max_size'] = '5120';

    $this->upload->initialize($config);
    if (!$this->upload->do_upload("file-upload")){
        $data['error'] = 'The following error occured : '.$this->upload->display_errors().'Click on "Remove" and try again!';
        echo json_encode($data); 
    } else {
        echo json_encode("success"); 
    }            
}

现在我收到来自PHP的回复，不管它是错误还是成功的JSON，我已经浏览了插件文档，我仍然无法找到如何捕获AJAX响应并根据响应行事，因为我们使用ajax成功函数在jQuery中执行：

success: function (response) {
            //Deal with the server side "response" data.
         },

我该怎么做？

Answer 1

您可以在live demo

查看演示

如果您希望成功事件触发

，请务必设置 uploadAsync false

示例代码：

JS

$("#input-id").fileinput({
    showRemove:false,
    showPreview: false,
    uploadUrl: "../xxxx/xxxx/XXXXXX.php", // server upload action
    uploadAsync: false,
    uploadExtraData: function() {
        return {
            bdInteli: xxxx
        };
    }
});

// CATCH RESPONSE
$('#input-id').on('filebatchuploaderror', function(event, data, previewId, index) {
var form = data.form, files = data.files, extra = data.extra, 
    response = data.response, reader = data.reader;
});

$('#input-id').on('filebatchuploadsuccess', function(event, data, previewId, index) {
    var form = data.form, files = data.files, extra = data.extra, 
    response = data.response, reader = data.reader;
    alert (extra.bdInteli + " " +  response.uploaded);
});

PHP

$nombre = $_FILES["ficheroExcel"]["name"];
$bdInteli = $_POST['bdInteli'];
if (move_uploaded_file($_FILES["ficheroExcel"]["tmp_name"], $nombre) ){
    $output = array('uploaded' => 'OK' );
} else {
   $output = array('uploaded' => 'ERROR' );
}
echo json_encode($output);

Answer 2

您可以阅读events section on the plugin documentation page以了解插件提供的各种事件。

这取决于您如何在插件中设置ajax上传。该插件提供了两个ajax upload modes - synchronous and asynchronous，如文档中所述。如果您将uploadAsync属性设置为true，则它是异步的。

FOR AJAX SUCCESS TRAP：

您可以使用filebatchuploadsuccess事件进行同步上传
您可以使用fileuploaded事件进行异步上传

FOR AJAX ERROR TRAP：

您可以使用filebatchuploaderror事件进行同步上传
您可以使用fileuploaderror事件进行异步上传

在您的情况下，您已将uploadAsync设置为true - 因此请使用异步设置/事件。

Answer 3

您可以在测试中使用此示例代码。在我的测试中，我的响应数据如下：

response data:
{
"ver":"1.0",
"ret":true,
"errmsg":null,
"errcode":0,
"data":{
    "status":"upload success",
    "originalFilename":"testFileName.txt",
    "fileName":"excelFile",
    "fileType":"text/plain",
    "fileSize":1733
}

 javascript code:
 $('#input-id').on('fileuploaded', function(event, data, previewId, index) {
    var response = data.response;
    if(response.ret ) {
        alert("upload success!"+data.response.data);
    }else{
        alert("upload failed!"+response.errmsg)
    }
    alert('File uploaded triggered'+form+"response:"+response);
    console.info(response.data);
});

Answer 4

refer this answer，我已经这样做了

javascript ：

$('#input-id').on('fileuploaded', function(event, data, previewId, index) {
    var form = data.form, files = data.files, extra = data.extra,
        response = data.response, reader = data.reader;
    console.log('File uploaded successfully : ID '+ data.response.d);
});

在 ASHX 文件中添加对上下文的响应：

context.Response.ContentType = "application/json";
string myId = "NewwId 1";
var wrapper = new { d = myId };
context.Response.Write(Newtonsoft.Json.JsonConvert.SerializeObject(wrapper));

Krajee Bootstrap文件输入，捕获AJAX成功响应

5 个答案: