Jquery - 在提交时动态构建表单操作？

Jquery - Dynamically build form action on submit?

提问by Probocop

I'm trying to have the action of a HTML form built when the user clicks the submit button.

我正在尝试在用户单击提交按钮时构建 HTML 表单的操作。

So the user fills in a form, clicks submit, then the action gets built thenit actually gets submitted. The reason being is because the form has a load of options on it which will be passed to a script.

因此，用户填写表单，单击提交，然后构建操作，然后实际提交。原因是因为表单上有大量选项，这些选项将传递给脚本。

How would I go about doing this with Jquery?

我将如何使用 Jquery 执行此操作？

回答by Matt Ball

Untested, but this should at least get you on your way:

未经测试，但这至少应该让你上路：

$('#myForm').submit(function (event)
{
    var action = '';
    // compute action here...
    $(this).attr('action', action);
});

回答by Jarek

Use jQuery submitevent:

使用 jQuery提交事件：

$(document).ready(function() {
  $('#yourFormId').submit(function() {
    $(this).attr('action', 'dynamicallyBuildAction'); 
    return false;
  });
});

回答by Guffa

The plain Javascript solution would have a function:

普通的 Javascript 解决方案将有一个功能：

function changeAction() {
  this.action = 'the dynamic action';
  return true;
}

In the form you would set the onsubmit event:

在表单中，您将设置 onsubmit 事件：

<form ... onsubmit="return changeAction();">

To do the same using jQuery would be:

使用 jQuery 做同样的事情是：

$(function(){
  $('IdOfTheForm').submit(function(){
    this.action = 'the dynamic action';
    return true;
  });
});