javascript - 无法使用 QUnit Test 中声明的变量

Question

我对 Javascript 比较陌生，也是第一次使用 QUnit。我在一个文件中创建了一些全局函数，并使用“script”标签将该文件包含到我的 javaScriptTests.html 文件中。

在 javaScriptTests.html 中，我声明了一个对象来传递到声明的函数并返回结果。然而，测试失败了。乍一看，它似乎是 QUnit，但我认为实际问题可能是我糟糕的 javascript 技能，无法以一种万无一失的方式定义函数。

QUnit 错误是:

我在 bradri.js 中声明的函数如下所示:

var findNodeIndex = function(id, nodes){
    // return node index with a given id
    var length = nodes.length || 0; 
    for (var index=0; index < length; index++){
        if (nodes[index].name === id){
            return index;
        };      
    };
};

var buildLinks = function (nodes){
    // build the links table for d3
    var links = [],
        length = nodes.length;
    for (var i=0; i < length; i++){
        if (nodes[i].mother){
            //console.log(i, nodes[i].mother);
            links.push({source: i, target: findNodeIndex(nodes[i].mother)});
        };
        if (nodes[i].father){
            links.push({source: i, target: findNodeIndex(nodes[i].farther)});
        };
    }
    return links;
};

节点未在此文件中定义。该错误似乎表明 QUnit 希望在 bradri.js 中定义 var 'nodes'，即使它仅在内部使用，并且 'nodes' 是从 javaScriptTests.html 传入的。

这就是我的测试:

QUnit.module("unrelated test", {
  setup: function() {
    // add it to current context 'this'
    this.testNodes = [
    {name: 'a', mother: '', farther: ''},
    {name: 'b', mother: '', farther: ''},
    {name: 'c', mother: '', farther: ''},
    {name: 'd', mother: '', farther: ''},
    {name: 'e', mother: '', farther: ''},
    {name: 'f', mother: 'a', farther: 'b'},
    {name: 'g', mother: 'a', farther: 'b'},
    {name: 'h', mother: 'a', farther: 'b'},
    {name: 'i', mother: 'c', farther: 'd'},
    {name: 'j', mother: 'c', farther: 'd'},
    {name: 'k', mother: '', farther: ''},
    {name: 'l', mother: 'e', farther: 'f'},
    {name: 'm', mother: 'j', farther: 'k'},
    {name: 'n', mother: 'l', farther: 'm'}
    ];
  }
});

QUnit.test( "Unit testing of custom D3 code", function( assert ) {

var result = '[{"source":5,"target":0},{"source":6,"target":0},{"source":7,"target":0},{"source":8,"target":2},{"source":9,"target":2},{"source":11,"target":4},{"source":12,"target":9},{"source":13,"target":11}]';

var temp = buildLinks(this.testNodes); // IT FAILS HERE
//JSON.stringify(temp)
//assert.equal(result, result, "We expect value to be hello" );
});

Answer 1

调用 findNodeIndex(nodes[i].mother); 时缺少第二个 nodes 参数； 将这些调用更改为 findNodeIndex(nodes[i] .mother，节点)；