在提交之前添加POST参数

时间:2009-06-14 21:46:02

标签: javascript jquery forms

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。

7 个答案:

答案 0 :(得分:89)

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;
}); 

答案 1 :(得分:18)

以前的答案可以缩短,并且可以可读

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

答案 2 :(得分:9)

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()$.post()文档。

答案 3 :(得分:2)

您可以执行form.serializeArray(),然后在发布之前添加名称 - 值对:

var form = $(this).closest('form');

form = form.serializeArray();

form = form.concat([
    {name: "customer_id", value: window.username},
    {name: "post_action", value: "Update Information"}
]);

$.post('/change-user-details', form, function(d) {
    if (d.error) {
        alert("There was a problem updating your user details")
    } 
});

答案 4 :(得分:1)

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

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

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

答案 5 :(得分:0)

PURE JavaScript:

  

创建XMLHttpRequest:

function getHTTPObject() {
    /* Crea el objeto AJAX. Esta funcion es generica para cualquier utilidad de este tipo, 
       por lo que se puede copiar tal como esta aqui */
    var xmlhttp = false;
    /* No mas soporte para Internet Explorer
    try { // Creacion del objeto AJAX para navegadores no IE
        xmlhttp = new ActiveXObject("Msxml2.XMLHTTP");
    } catch(nIE) {
        try { // Creacion del objet AJAX para IE
            xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
        } catch(IE) {
            if (!xmlhttp && typeof XMLHttpRequest!='undefined') 
                xmlhttp = new XMLHttpRequest();
        }
    }
    */
    xmlhttp = new XMLHttpRequest();
    return xmlhttp; 
}
  

通过POST发送信息的JavaScript函数:

function sendInfo() {
    var URL = "somepage.html"; //depends on you
    var Params = encodeURI("var1="+val1+"var2="+val2+"var3="+val3);
    console.log(Params);
    var ajax = getHTTPObject();     
    ajax.open("POST", URL, true); //True:Sync - False:ASync
    ajax.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');
    ajax.setRequestHeader("Content-length", Params.length);
    ajax.setRequestHeader("Connection", "close");
    ajax.onreadystatechange = function() { 
        if (ajax.readyState == 4 && ajax.status == 200) {
            alert(ajax.responseText);
        } 
    }
    ajax.send(Params);
}

答案 6 :(得分:-1)

你可以做一个ajax电话。

这样,您就可以通过ajax&#39;数据自己填充POST数组:&#39;参数

var params = {
  url: window.location.pathname,
  time: new Date().getTime(), 
};


$.ajax({
  method: "POST",
  url: "your/script.php",
  data: params
});