Question

我使用Angular-Upload提交文件。我需要提取文件名的2个部分，以用作返回api控制器的formData。这些文件目前名为mm / yyyy PipeName PipelineLocation.pdf 我需要实际删除日期，然后将PipeName添加到＆＃39; pipeName＆＃39;＆＃39;＆＃39;＆＃39;＆＃39;然后将PipelineLocation添加到＆＃39; locationName＆＃39;：＆＃39;＆＃39; 我使用的示例文件名为＆＃39; 02-2011 P3D4LB38A2 DDEC33D.pdf＆＃39;和＆＃39; 11-2008 ED34PL89G5 23FFWC580.pdf＆＃39;

我可以看到具有name属性的数组，但我不知道如何访问它。 Array from the view Plunker

app.controller('MainCtrl', function($scope, $upload) {
  $scope.$watch('files', function () {
    $scope.upload($scope.files);
  });



$scope.result={};
   $scope.upload = function (files) {
    if (files && files.length) {
        for (var i = 0; i < files.length; i++) {
            var file = files[i];
            $upload.upload({
                url: '/api/apiBatchPipeLine',
                fields: {

                    'typeId': 1,
                    'companyId': $scope.companyId.CompanyId,
                    'documentDate': $scope.model.documentDate,
                    'companyName': $scope.CompanyName,
                    'pipeName': ,
                    'locationName': ,
                    'typeName': 'Pipeline Reports'
                },
                file: file
            }).progress(function (evt) {
                var progressPercentage = parseInt(100.0 * evt.loaded / evt.total);
                  console.log('progress: ' + progressPercentage + '% ' + evt.config.file.name);
            }).success(function (data, status, headers, config) {
                console.log('file ' + config.file.name + 'uploaded. Response: ' +
                            JSON.stringify(data));
            });

        }
    }
  };
});

UPDATE 错误

TypeError: undefined is not a function
at l.$scope.upload (https://localhost:44300/MyScripts/Controllers/BatchSubmit/batchSubmitPipesController.js:68:31)
at Object.fn (https://localhost:44300/MyScripts/Controllers/BatchSubmit/batchSubmitPipesController.js:55:16)
at l.$get.l.$digest (https://localhost:44300/Scripts/angular.min.js:123:445)
at l.$get.l.$apply (https://localhost:44300/Scripts/angular.min.js:126:362)
at bg.$$debounceViewValueCommit (https://localhost:44300/Scripts/angular.min.js:219:34)
at bg.$setViewValue (https://localhost:44300/Scripts/angular.min.js:218:263)
at https://localhost:44300/Scripts/angular-file-upload-all.min.js:2:1349
at https://localhost:44300/Scripts/angular.min.js:138:513
at e (https://localhost:44300/Scripts/angular.min.js:40:339)
at https://localhost:44300/Scripts/angular.min.js:44:375

第68行

 if (file.name.test(regex)) { // used to validate the filename

第55行

 $scope.$watch('files', function (files) { // this can be simplified like so.
    $scope.upload(files);

更新PICS

Answer 1

您可以使用正则表达式提取文件名的所需部分：

实现此目的的正则表达式可能类似于：

^\d+\D\d+\s*(\S*)\s*(\S*)\..*$

这是寻找任何数字序列，后跟非数字，后跟任何数字序列。然后是所有空格，然后是一系列非空格（你的pipename），然后是空格和另一个非空格序列（你的位置名）。最后的最后一点是确保它与文件扩展名匹配。

app.controller('MainCtrl', function($scope, $upload) {

  var regex = /^\d+\D\d+\s*?(\S*)\s*(\S*)\..*$/i;

  $scope.$watch('files', function (files) { // this can be simplified like so.
    $scope.upload(files);
  });

$scope.result={};
   $scope.upload = function (files) {
    if (files && files.length) {
        for (var i = 0; i < files.length; i++) {
            var file = files[i];

            if(regex.test(file.name)) { // used to validate the filename
                var matches = file.name.match(regex);

                $upload.upload({
                    url: '/api/apiBatchPipeLine',
                    fields: {

                        'typeId': 1,
                        'companyId': $scope.companyId.CompanyId,
                        'documentDate': $scope.model.documentDate,
                        'companyName': $scope.CompanyName,
                        'pipeName': matches[1],
                        'locationName': matches[2],
                        'typeName': 'Pipeline Reports'
                    },
                    file: file
                }).progress(function (evt) {
                    var progressPercentage = parseInt(100.0 * evt.loaded / evt.total);
                      console.log('progress: ' + progressPercentage + '% ' + evt.config.file.name);
                }).success(function (data, status, headers, config) {
                    console.log('file ' + config.file.name + 'uploaded. Response: ' +
                                JSON.stringify(data));
                });
            }
        }
    }
  };
});

如何使用angular解析和修剪多个文件名

1 个答案: