在提交之前添加POST参数

Question

我有这个简单的形式:

<form id="commentForm" method="POST" action="api/comment">
    <input type="text" name="name" title="Your name"/>
    <textarea  cols="40" rows="10" name="comment" title="Enter a comment">
    </textarea>
    <input type="submit" value="Post"/>
    <input type="reset" value="Reset"/>
</form>

我需要在发送到服务器之前添加两个POST参数:

var params = [
               {
                 name: "url",
                 value: window.location.pathname
               },
               {
                  name: "time",
                  value: new Date().getTime()
               }
             ];

请不要修改表格.

Answer 1

要使用Jquery添加它:

$('#commentForm').submit(function(){ //listen for submit event
    $.each(params, function(i,param){
        $('<input />').attr('type', 'hidden')
            .attr('name', param.name)
            .attr('value', param.value)
            .appendTo('#commentForm');
    });

    return true;
}); 

Answer 2

以前的答案可以缩短,更具可读性.

$('#commentForm').submit(function () {
    $(this).append($.map(params, function (param) {
        return   $('<input>', {
            type: 'hidden',
            name: param.name,
            value: param.value
        })
    }))
});

Answer 3

如果要在不修改表单的情况下添加参数,则必须序列化表单,添加参数并使用AJAX发送:

var formData = $("#commentForm").serializeArray();
formData.push({name: "url", value: window.location.pathname});
formData.push({name: "time", value: new Date().getTime()});

$.post("api/comment", formData, function(data) {
  // request has finished, check for errors
  // and then for example redirect to another page
});

请参阅.serializeArray()和$.post()文档.

Answer 4

您可以在没有jQuery的情况下执行此操作：

    var form=document.getElementById('form-id');//retrieve the form as a DOM element

    var input = document.createElement('input');//prepare a new input DOM element
    input.setAttribute('name', inputName);//set the param name
    input.setAttribute('value', inputValue);//set the value
    input.setAttribute('type', inputType)//set the type, like "hidden" or other

    form.appendChild(input);//append the input to the form

    form.submit();//send with added input