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如何对下面的成功回调函数进行编码，以便能够访问下面返回的 JSON 中的对象。显然，我将无法再访问返回的对象success: function(data) {if (data.returned === true) {了。我将如何做到这一点？

jQuery代码：

$("#projects").click(function() {
                jQuery.ajax({ type: "POST", dataType: "JSON",
                    url: "<?=base_url()?>index.php/home/projectsSlider",
                    json: {returned: true}, success: function(data) {
                        if (data.returned === true) {
                            $("#resultProjects").html(JSON.stringify(data.Projects));
                            $("#resultScreenshots").html(JSON.stringify(data.Screenshots));

                            $("#content").fadeOut(150, function() {
                                $(this).replaceWith(projectsSlider(data.projectId, data.projectName, data.startDate, data.finishedDate, data.projectDesc, data.createdFor, data.contributors, data.screenshotURI, data.websiteURL), function() {
                                    $(this).fadeIn(150);
                                });
                            });
                        }
                    }
                });
            });

返回的 JSON：

{
    "Projects": [
        {
            "projectId": "932713684f9073189ec7b",
            "projectName": "Cloud859Collective",
            "startDate": "April 19th, 2012",
            "finishedDate": "April 25th, 2012",
            "createdFor": "ClasskCreations",
            "contributors": "Mike Grigsby",
            "projectDesc": "This website was created with a friend in mind. His name is Kevin Johnson and he is a rapper. He needed a website that would allow him to host and share his music."
        },
        {
            "projectId": "10599012654f907093714e9",
            "projectName": "Nurbell Studio",
            "startDate": "April 15th, 2012",
            "finishedDate": "April 19th, 2012",
            "createdFor": "Nurbell LLC",
            "contributors": "Mike Grigsby",
            "projectDesc": "This is the page you are currently looking at. This is the official Nurbell homepage. Complete with a frontend and a backend."
        }
    ],
    "Screenshots": [
        {
            "screenshotURI": "http://nurbell.com/vd/1.0/images/project-data/kevo.png"
        },
        {
            "screenshotURI": "http://nurbell.com/vd/1.0/images/project-data/nurbell.png"
        }
    ]
}
4

2 回答 2

2

我不确定你在这里问什么。我认为你应该看看 javascript 命名空间。这样，您可以在对象（或命名空间）中创建一个属性并将 json 结果放入该属性中。

像这样的东西：

 var myProjects = {
     projects: null,

     getProjects: function() {
           // do the ajax thing with something like
           myProjects.projects = data.projects;
     },

     placeProjects: function() {
           if (myProjects.projects == null) myProjects.getProjects();
           $.each(myProjects.projects, function(i,e){
                //place project content
           }
     },
 }

 // define the click event
 $("#projects").click(myProjects.placeProjects());

数据将被存储，直到您将其删除或重新加载页面。您可以在 firebug 的 DOM 检查器中看到此对象。希望有帮助

编辑 ：

我已经在这个 jsFiddle http://jsfiddle.net/BTbJu/5中实现了这个想法 运行它，单击 Div 中的文本以加载第一个项目。继续点击旋转。

于 2012-07-14T20:36:34.523 回答
0

不确定我是否正确理解了这个问题，但您似乎担心在调用 JSON.stringify 时您的原始 json 可能会被永久修改。

不， JSON.Stringify 将返回一个新字符串。闭包参数“数据”引用的原始 json 将保持不变。您将通过数据引用很好地访问所有属性。

实际上，data.projectId 在闭包范围内会出错，应该评估为 undefined。

试试这个：

$("#projects").click(function () {
    jQuery.ajax({
        type: "POST",
        dataType: "JSON",
        url: "<?=base_url()?>index.php/home/projectsSlider",
        success: function (data) {
            var projects = data.Projects; //array
            var screenshots = data.Screenshots; //array

            //direct, one-off indexed
            console.log(projects[0].projectId);
            console.log(data.Projects[0].projectId);

            //looped with map
            projects.map(function(project, index) {
                console.log(project.projectId);
            });

            //traditional for
            for(var i1 = 0; i1 < projects.length; i1++) {
                console.log(projects[i1]);
            }

            //direct, one-off indexed
            console.log(screenshots[0].screenshotURI);
            console.log(data.Screenshots[0].screenshotURI);

            //looped
            screenshots.map(function(screenshot, index) {
                console.log(screenshot.screenshotURI);
            });

            if (data.returned === true) {
                //your stringify code etc.
            }
        }
    });
});
于 2012-07-14T20:33:40.663 回答