AngularJS ng-repeat ID过滤器 [英] Angularjs ng-repeat filter by id

问题描述

我在mysql中有2个表，我用php(mysqli)调用数据，所以我将此数据放在$scope.student= <?php echo json_encode( $student) ?>;和$scope.ratings= <?php echo json_encode( $ratings) ?>;的javascript var中，2个表具有键id(ID_studen)

hi i have 2 tables in mysql, i call the data with php(mysqli), so i put this data in javascript var with $scope.student= <?php echo json_encode( $student) ?>; and $scope.ratings= <?php echo json_encode( $ratings) ?>; 2 tables have a key id(ID_studen)

第一个表只有一个个人数据，第二个表有数据，因此我需要过滤并通过此处的代码显示所有详细信息:

The first table have only a personal data, and the second have datails, so i need filter and show all details by ratings here code:

<div ng-app="" ng-controller="Ctrl">
  <ul ng-repeat="student in students">
   <li ng-repeat="rating in ratings (where student.tagid =ratings.tagid">  {{rating.note}}</li>

  </ul>
</div>

检查 jsfiddle.net

推荐答案

您可以使用$ filter，请参见下面的演示

You can use $filter please see demo below

app = angular.module("app", []);

app.controller("Ctrl", Ctrl);

function Ctrl($scope) {

  $scope.students = [{
    firstname: 'Buster',
    lastname: 'Bluth',
    tagid: '4134'
  }, {
    firstname: 'John',
    lastname: 'McClane',
    tagid: '9845'
  }, {
    firstname: 'Mister',
    lastname: 'Spock',
    tagid: '0905'
  }];

  $scope.ratings = [{
    matter: 'Mathematics',
    note: '12',
    tagid: '4134'
  }, {
    matter: 'Biology',
    note: '13',
    tagid: '9845'
  }, {
    matter: 'Lenguage:',
    note: '14',
    tagid: '0905'
  }];
}

<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.2.23/angular.min.js"></script>
<div ng-app="app" ng-controller="Ctrl">
  <ul ng-repeat="student in students">
    <li ng-repeat="rating in ratings | filter : {tagid: student.tagid}">{{student.firstname}} {{student.lastname}} | {{rating.matter}} {{rating.tagid}}</li>
  </ul>
</div>

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