Question

Ajax函数（带嵌套）

function open_appointment(id)
{
save_method = 'open_appointment';
$('#form_open_appointment')[0].reset(); // reset form on modals
$('.form-group').removeClass('has-error'); // clear error class
$('.help-block').empty(); // clear error string

//Ajax Load data from ajax
$.ajax({
    url : "<?php echo site_url('ReceptionistController/ajax_edit_appointment')?>/" + id,
    type: "GET",
    dataType: "JSON",
    success: function(data)
    {

        $.ajax({
            url : "<?php echo site_url('DoctorController/ajax_edit_patient')?>/" +data.ap_patient, 
            type: "GET",
            dataType: "JSON",
            success: function(data)
            {
                $('[name="pt_name"]').val(data.pt_name);

            },
            error: function (jqXHR, textStatus, errorThrown)
            {
                alert('Failed');
            }
        });

        $('#modal_open_appointment').modal('show'); // show bootstrap modal when complete loaded
        $('.modal-title').text('Open Appointment'); // Set title to Bootstrap modal title

    },
    error: function (jqXHR, textStatus, errorThrown)
    {
        alert('Error get data from ajax');
    }
});
 }

我想在另一个中使用一个ajax调用。我已经编写了上面的代码，但它不起作用（嵌套时），并给出show＆＃34; Failed＆＃34; 。嵌入ajax调用函数存在问题

Answer 1

你可以使用下面的承诺

代码未经测试......

供参考： https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/Promise

var getAppointment = function() {

    return new Promise(function(resolve, reject) {

        $.ajax({
            url : "<?php echo site_url('ReceptionistController/ajax_edit_appointment')?>/" + id,
            type: "GET",
            dataType: "JSON",
            success: function(data)
            {
                resolve(data);
            },
            error: function (jqXHR, textStatus, errorThrown)
            {
                reject('Error get data from ajax');
            }
        });

    })
}

var editPatient = function(data) {
    return new Promise(function(resolve, reject) {

        $.ajax({
            url : "<?php echo site_url('DoctorController/ajax_edit_patient')?>/" +data.ap_patient, 
            type: "GET",
            dataType: "JSON",
            success: function(data)
            {
                resolve(data);

            },
            error: function (jqXHR, textStatus, errorThrown)
            {
                reject('Failed');
            }
        });
    }
}

getAppointment()
    .then(function(data) {
        return editPatient(data)
    })
    .then(function(data) {
        $('[name="pt_name"]').val(data.pt_name);
    })
    .catch(function(error) {
        console.log(error)
    ));

Answer 2

语法看起来很好，但首先，检查外部ajax请求它是否正常工作并正确返回data.ap_patient。之后调用内部ajax请求。并检查请求是否要更正操作方法的检查元素。

嵌套ajax调用函数

2 个答案: