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在 Oyeme 的建议下,我更改了代码,但仍然无法工作。它可以提醒“测试”,但不能提醒 DBuser_name 。

通过以下编码,我可以创建如下图所示的内容。我的问题是,如果用户单击列表中的一个人,我想将其提取 student.data.user_name到 localStorage。但是,在我下面的编码中,似乎我做错了?

function getStudentList() {
$.getJSON('http://mydomain.com/getStudents.php?jsoncallback=?', function(data) {
    $('#studentList li').remove();
    $('#load').hide();
    //students = data.user_name;
    $.each(data, function(index, student) {
        //$('#studentList').append('<li><a href="tutor_student_detail.html?user_name=' + student.data.user_name + '">' +
        $('#studentList').append('<li><a href="tutor_student_detail.html">' +
                '<h4>' + student.data.user_name + '</h4>' +
                '<p>' + student.data.role + '</p>' +
                '</a></li>');
                
        $("li a").click(function() {
       window.localStorage["view"] = $(this).data('user_name');
    });
    });
    $('#studentList').listview('refresh');
});

}

以下是tutor_student_detail.html(js部分)的编码

function start(){

var localUsername=window.localStorage.getItem("view");
    
$.getJSON('http://mydomain.com/getStudent.php?user_name='+localUsername+'&jsoncallback=?', displayStudent);
    
    alert("testing");
}

function displayStudent(data) {
    var DBuser_name=data[0].data.user_name;
    alert(DBuser_name);
    var employee = data.item;
    $('#username').text(student.data.user_name);
    $('#pw').text(student.data.password);
    $('#id').text(student.data.id);
    

    $('#actionList').listview('refresh');
    
}

json字符串的输出类似于

?([{"data":{"id":"4","user_name":"studentB","book":"4567","role":"Student"}}]);

在此处输入图像描述

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1 回答 1

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$('#studentList').append('<li><a href="tutor_student_detail.html"
                 data-name="'+student.data.user_name+'">' +
                '<h4>' + student.data.user_name + '</h4>' +
                '<p>' + student.data.role + '</p>' +
                '</a></li>');

$("li a").click(function() {
   window.localStorage["view"] = $(this).data('name');
});
于 2013-03-18T21:30:57.207 回答