如何在extjs 4.2中包装表单的提交数据? [英] How to wrap submit data of form in extjs 4.2?

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本文介绍了如何在extjs 4.2中包装表单的提交数据?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我现在提交了:

method: save
name: Michael
birthday: 1983-02-01

但我需要:

method: save
data[name]: Michael
data[birthday]: 1983-02-01

和字段名称必须类似于 birthday ,而不是 data [birthday] .

and field name must be like birthday not data[birthday].

推荐答案

假定您正在处理表单提交,而您拥有一个代表该表单的控件:

Assuming that you're handling a form submission where you have a control representing the form:

var formData = form.getFieldValues();

来自Ext.form.Basic.getFieldValues

然后通过ajax提交:

And then submitting via ajax:

Ext.Ajax.request({url: "postlocation.php", method: "POST", data: formData});

来自Ext.Ajax.request

如果您不想提交表单,则可以覆盖表单上的按钮以调用模拟提交的过程.

If you don't want to submit your form, you can override a button on the form to invoke a process that simulates a submission.

// form def up here
buttons: [
    text: "Pseudo-Submit",
    id: "altsubmitbuttonthing"
]

在您的控制器中(或按钮的事件处理程序中:

In your controller (or an event handler for the button:

this.control({
     "button[id=altsubmitbuttonthing]": {
         click: function (control) {
             var form = control.up("form"), // <- now you have your form and you can do whatever you want with it's data.
                 formData = form.getFieldValues(),
                 preparedData = {};

             formData.theDateField = new Date(data.theDateField);
             formData.theIntField = parseInt(data.theIntField, 10);

             preparedData.data.birthday = formData.birthday;
             preparedData.data.name = formData.name;

             Ext.Ajax.request({
                 url: "/submissions",
                 method: "POST",
                 type: "json",
                 data: preparedData
             });
         }
     }
});

这篇关于如何在extjs 4.2中包装表单的提交数据?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!

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