为什么AJAX成功调用中的代码不起作用? [英] Why is code in AJAX success call is not working?

问题描述

我有一个AJAX脚本，可将数据从表单插入MySQL数据库.这是AJAX.

I have an AJAX script to insert data from a form to MySQL database. This is the AJAX.

<!-- SUBMIT FORM VIA AJAX -->
$("#f_product").on('submit',function(event){ 
    event.preventDefault();

    data = $(this).serialize();

    $.ajax({
    type: "POST",
    url: "<?php echo site_url('con_product/ins_product'); ?>",
    data: data
    }).success(function() {
        alert("Products list is ready to be printed");
        window.open("<?php echo site_url('con_product/print_product'); ?>","_blank");
        window.open("<?php echo site_url('con_product/form_product'); ?>","_self");
    });
});
<!-- END SUBMIT FORM VIA AJAX -->

AJAX脚本成功将表单中的数据插入数据库.但是以某种方式，成功的脚本无法正常工作.为什么?

The AJAX script successfully insert data from the form to database. But somehow, the script on success is not working. Why?

此AJAX的行为是:

• 将数据插入数据库-成功
• 显示提醒
• 打开新页面以进行打印.
• 将当前页面刷新为新表单.

推荐答案

success是callback函数的名称，而不是promise.

The success is the name of the callback function and not the promise.

您应该使用:

$.ajax({
    type: "POST",
    url: "<?php echo site_url('con_product/ins_product'); ?>",
    data: data,
    success: function() {
        alert("Products list is ready to be printed");
        window.open("<?php echo site_url('con_product/print_product'); ?>","_blank");
        window.open("<?php echo site_url('con_product/form_product'); ?>","_self");
    }
});

或done承诺:

$.ajax({
    type: "POST",
    url: "<?php echo site_url('con_product/ins_product'); ?>",
    data: data
}).done(function() {
    alert("Products list is ready to be printed");
    window.open("<?php echo site_url('con_product/print_product'); ?>","_blank");
    window.open("<?php echo site_url('con_product/form_product'); ?>","_self");
});

这篇关于为什么AJAX成功调用中的代码不起作用?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！