使用Angular js和php上传多文件

Question

我们正在接管一个使用Angular js进行数据和多文件上传的表单的网站（最多4个文件）。

我们正在发布到.php文件。 我已经能够分离和捕获现场数据-但似乎只能“看到”一个文件。 如果我上传了多个文件，我只会看到添加到表单中的最后一个文件。

有人可以帮我使用PHP获取文件吗？

表单HTML（用于文件）：

<ul class="attachments">
<li ng-repeat="attachment in attachments" class="attachment">
    {{attachment.name}}
<a href="" ng-click="removeAttachment(attachment)"><i class="btn-submit-att-x-hover-off"></i></a>
</li>
</ul>
    <button type="button" class="attachments-button btn btn-default" ngf-select ngf-change="onFileSelect($files)" ngf-multiple="true" ng-if="attachments.length < 4">
    <i class="btn-submit-attch"></i>Upload Attachments
    </button>

Angular JS代码-对于我所看到的文件：

$scope.attachments = [];

//onFileSelect adds file attachments to the $scope up to the
//configured maximum value.
    $scope.onFileSelect = function ($files) {
    var MAX_ATTACHMENTS = 4;
    for(var i=0; i < $files.length; i++){
        if($scope.attachments.length >= MAX_ATTACHMENTS){
        break;
        }
    $scope.attachments.push($files[i]);
        }            
    };

    $scope.removeAttachment = function (attachment) {
        var index = $scope.attachments.indexOf(attachment);
        if (index > -1) {
        $scope.attachments.splice(index, 1);
        }
    };

//POST form data
    $scope.upload = Upload.upload({
        url: 'forms/pressrelease.html',
        method: 'POST',
        data: $scope.model,
        file: $scope.attachments,
        fileFormDataName: "files"
    })

    .success(function (data, status, headers, config) {
        alertSuccess("Your press release has been submitted successfully!");
        $scope.showSuccess = true;
    }

我们使用过的仅有的PHP代码可以使我们得到任何回报-但同样-仅1个，或者最后一个文件名是：

    if(isset($_FILES['files'])){ 
$files = $_FILES['files'];

$sentfilename = $_FILES['files']['name'];
$email_body .= "I see files: $sentfilename<br />";

$tmpsentfile = $_FILES['files']['tmp_name'];

$i=0;

$getfile = $files[0];
$sentfilename = $getfile;
$sentfilesize = $_FILES['files']['size'];
$errormessage = $_FILES['files']['error'];

$email_body .= "I see files: $sentfilename<br />";
}

如果有人可以向我们展示PHP代码，则需要使用它们捕获这些上传的文件。

谢谢

Answer 1

创建一个名称数组并发布动态文件名：

var names = []
foreach file:
    names.push(files[i].name)

在上传中：

  method: 'POST',
    data: $scope.model,
    file: $scope.attachments,
    fileFormDataName: names