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我要使用 Javascript，并且正在寻找制作表格。我找到了一些我编辑的代码，但它不起作用。我正在尝试制作一个带有 2 个选择下拉菜单的表单。当访问者从第一个下拉列表中选择一项服务时，第二个下拉列表将自动更新为每个服务人员的姓名。我已将代码上传到 jsFiddle。网址是http://jsfiddle.net/mrtxR/。我认为这真的很简单，但找不到任何教程和指南。

// The data that the service should return
// JSFiddle will echo it back for us on that URL
var doctors = {
    success: true,
    doctors: [
        {
            id: 71,
            name: "George"
        },
        {
            id: 72,
            name: "James"
        }
    ]
}

// This is what your JSON from PHP should look like
var jsonDoctors = JSON.stringify(doctors);
console.log(jsonDoctors);

// Bind change function to the select
jQuery(document).ready(function() {
    jQuery("#services").change(onServiceChange);
});

function onServiceChange()
{
    var serviceId = jQuery(this).val();    

     $.ajax({
        url: '/echo/json/',
        type: 'post',
        data: {
            serviceId: serviceId,
            json: jsonDoctors // jsFiddle echos this back to us
        },
        success: onServicesRecieveSuccess,
        error: onServicesRecieveError
    });
}

function onServicesRecieveSuccess(data)
{
    // Target select that we add the states to
    var jTargetSelect = jQuery("#doctors");

    // Clear old states
    jTargetSelect.children().remove();

    // Add new states
    jQuery(data.doctors).each(function(){        
        jTargetSelect.append('<option value="'+this.id+'">'+this.name+'</option>');
    });
}

function onServicesRecieveError(data)
{
    alert("Could not get services. Please try again.");
}
4

1 回答 1

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您最后的评论是正确的，您应该将 serviceId 添加到每个医生。您的假 javascript 可能如下所示：

// The data that the service should return
// JSFiddle will echo it back for us on that URL
var doctors = {
    success: true,
    doctors: [
        {
            id: 71,
            serviceId : 1,
            name: "George"
        },
        {
            serviceId : 2,
            id: 72,
            name: "James"
        },
        {
            serviceId : 3,
            id: 73,
            name: "Ron"
        },
        { 
            serviceId : 1,
            id : 77,
            name : "Barak",

        }
    ]
}

function getJsonDoctors(serviceId) {
    var result = [];
    var l = doctors.doctors;
    for (var i = 0 ; i < l.length ; i++) {
        if (l[i].serviceId == serviceId) {
            result.push(l[i]);   
        }
    }
    return JSON.stringify({success : true,doctors : result});
}

// This is what your JSON from PHP should look like
var jsonDoctors = JSON.stringify(doctors);
console.log(jsonDoctors);

// Bind change function to the select
jQuery(document).ready(function() {
    jQuery("#services").change(onServiceChange);
});

function onServiceChange()
{
    var serviceId = jQuery(this).val();    

     $.ajax({
        url: '/echo/json/',
        type: 'post',
        data: {
            serviceId: serviceId,
            json: getJsonDoctors(serviceId) // jsFiddle echos this back to us
        },
        success: onServicesRecieveSuccess,
        error: onServicesRecieveError
    });
}

function onServicesRecieveSuccess(data)
{
    // Target select that we add the states to
    var jTargetSelect = jQuery("#doctors");

    // Clear old states
    jTargetSelect.children().remove();

    // Add new states
    jQuery(data.doctors).each(function(){        
        jTargetSelect.append('<option value="'+this.id+'">'+this.name+'</option>');
    });
}

function onServicesRecieveError(data)
{
    alert("Could not get services. Please try again.");
}
于 2013-04-01T18:22:21.990 回答