Question

我编写了一个JavaScript来显示来自JSON对象的数据。会员 employeeType 有两位员工成员;一个包含一个包含3个员工对象的数组，每个对象包含三个成员，另一个包含3个成员，即“employee”，“id”和“email”。如果成员有多个对象，我可以显示成员但如果对象在员工中是单独的，则无法显示对象的成员。我可能不清楚解释它，但你可以看看这个fiddle here。

我的JSON看起来像这样;

var data = {"employeeSearch":{"employeeType":[{"employees":[{"employee":"name1", "id":"1", "email":"name1@company.com"},{"employee":"name2", "id":"2", "email":"name2@company.com"},{"employee":"name3", "id":"3", "email":"name3@company.com"}]},{"employees":{"employee":"name4", "id":"4", "email":"name4@company.com"}}]}};

此脚本显示数据

for(var i = 0; i < data.employeeSearch.employeeType.length; i++) {
    for (var j = 0; j < data.employeeSearch.employeeType[i].employees.length; j++) {
        countItem++;
        output += "<tr><td>"+ countItem +"</td>" + 
                "<td>"+ data.employeeSearch.employeeType[i].employees[j].employee + "</td>" + 
                "<td>"+ data.employeeSearch.employeeType[i].employees[j].id +"</td>" + 
                "<td>"+ data.employeeSearch.employeeType[i].employees[j].email +"</td></tr>";
    }
}
return output;

这是它在JsonViews中的外观

enter image description here

现在如何显示最后一个员工对象？请指出正确的方向，并解释我在做什么，我应该怎么做。

JSFiddle

Answer 1

请注意，第二个员工不是数组而是一个对象，它应该是一个数组。

jsfiddle：http://jsfiddle.net/fish_ball/HASuY/1/

var data = {
  "employeeSearch": {
    "employeeType": [{
       "employees":[
          {"employee":"name1", "id":"1", "email":"name1@company.com"},
          {"employee":"name2", "id":"2", "email":"name2@company.com"},
          {"employee":"name3", "id":"3", "email":"name3@company.com"}
        ]},{
        // Notice the differences HERE!
        "employees":[
          {"employee":"name4", "id":"4", "email":"name4@company.com"}
        ]
    }]
  }
};

Answer 2

如前所述，如果可以，请更改数据结构，以便在两种情况下都有一个数组。

如果您不能这样做并假设某个员工从未拥有length属性，您可以对其进行简单测试，然后在运行时将其转换为数组：

for(var i = 0; i < data.employeeSearch.employeeType.length; i++) {
    var employees = data.employeeSearch.employeeType[i].employees;
    if (typeof employees.length !== 'number') { // not an array? Make it one.
        employees = [employees];
    }
    for (var j = 0; j < employees.length; j++) {
        // ...
    }
}

Answer 3

现在如何显示最后一个员工对象？请指出我正确的方向，并解释我在做什么，我该怎么做。

就像fish_ball所说，你的json格式并不普通。

您的data.employeeSearch.employeeType[0]包含一系列对象empployees，而data.employeeSearch.employeeType[1]包含一个对象employees

如何，您可以访问最后一个员工对象，如下所示：

getEmployees = function() {
    var countItem = 0;

            for (var j = 0; j < data.employeeSearch.employeeType[0].employees.length; j++) {
                countItem++;
                output += "<tr><td>"+ countItem +"</td>" + 
                            "<td>"+ data.employeeSearch.employeeType[0].employees[j].employee + "</td>" + 
                            "<td>"+ data.employeeSearch.employeeType[0].employees[j].id +"</td>" + 
                            "<td>"+ data.employeeSearch.employeeType[0].employees[j].email +"</td></tr>";
            }

                            output += "<tr><td>"+ (++countItem) +"</td>" + 
                            "<td>"+ data.employeeSearch.employeeType[1].employees.employee + "</td>" + 
                            "<td>"+ data.employeeSearch.employeeType[1].employees.id +"</td>" + 
                            "<td>"+ data.employeeSearch.employeeType[1].employees.email +"</td></tr>";


        return output;
    }

JSFiddle Demo

就像Felix King指出的那样，empolyeeType可能并不总是只有2.如果是这种情况，那么你需要检查empolyees.length或typeof并迭代数组如果empolyees.length为>0或typeof不是number。请参阅以下示例：

getEmployees = function() {
var countItem = 0;

for (var i = 0; i < data.employeeSearch.employeeType.length; i++){

    if (data.employeeSearch.employeeType[i].employees.length > 0){

        for (var j = 0; j < data.employeeSearch.employeeType[i].employees.length;j++){
            countItem++;            
            output += "<tr><td>"+ countItem +"</td>" + 
                        "<td>"+ data.employeeSearch.employeeType[i].employees[j].employee + "</td>" + 
                        "<td>"+ data.employeeSearch.employeeType[i].employees[j].id +"</td>" + 
                        "<td>"+ data.employeeSearch.employeeType[i].employees[j].email +"</td></tr>";
        }

    }else{
                        output += "<tr><td>"+ (countItem+1) +"</td>" + 
                        "<td>"+ data.employeeSearch.employeeType[i].employees.employee + "</td>" + 
                        "<td>"+ data.employeeSearch.employeeType[i].employees.id +"</td>" + 
                        "<td>"+ data.employeeSearch.employeeType[i].employees.email +"</td></tr>";
    }

}


    return output;
}

Another JSFiddle Demo

显示具有一个或多个对象的JSON数据成员

3 个答案: