Warning: file_get_contents(/data/phpspider/zhask/data//catemap/1/angularjs/20.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
Javascript 如何在AngularJS中格式化作用域?_Javascript_Angularjs_Angularjs Scope - Fatal编程技术网

Javascript 如何在AngularJS中格式化作用域?

Javascript 如何在AngularJS中格式化作用域?,javascript,angularjs,angularjs-scope,Javascript,Angularjs,Angularjs Scope,我有几个$scope的内容 HTML: <td>{{ project.captation }}</td> <td>{{ project.perso }}</td> $http({ url: "php/random.php", method: "GET" }).success(function(data) { $scope.project = data; }); 内容2 - Lorem ipsum- quia dolor

我有几个
$scope
的内容

HTML

<td>{{ project.captation }}</td>
<td>{{ project.perso }}</td>
$http({
    url: "php/random.php",
    method: "GET"
}).success(function(data) {
    $scope.project = data;
}); 
内容2

- Lorem ipsum- quia dolor sit ame- consectetur adipiscing elit
JS(控制器)

<td>{{ project.captation }}</td>
<td>{{ project.perso }}</td>
$http({
    url: "php/random.php",
    method: "GET"
}).success(function(data) {
    $scope.project = data;
}); 
仅在存在
-
的情况下,我想用
  • 替换每个
    -

  • 如何做到这一点

    您需要创建一个自定义过滤器,用
  • 替换
    -
  • 。请参阅此以开始使用过滤器。然后在HTML模板中使用以下筛选器:

    <td>{{ project.captation | filterNameHere }}</td>
    <td>{{ project.perso | filterNameHere }}</td>
    
    {{project.captation | filterNameHere}
    {{project.perso | filterNameHere}
    
    可能是这样的:

    var items = $scope.project.perso.split('-');
    if (items.length > 1) {
        var list = "";
        for (var i = 0; i < items.length; i++) {
            list += "<li>" + items[i] + "</li>";
        }
        $scope.project.perso = list;
    }
    
    var items=$scope.project.perso.split('-');
    如果(items.length>1){
    var list=“”;
    对于(变量i=0;i”+项目[i]+“”;
    }
    $scope.project.perso=列表;
    }
    
    谢谢!我把它和@Sandeep Panda的anwser一起使用!