jQuery $ .ajax不显示任何内容 [英] jQuery $.ajax doesn't show anything

问题描述

我将使用AJAX和PHP文件处理HTML表单来管理数据，但是成功消息或错误消息均无效. 我试过萤火虫，但好像它在$ .ajax语句出现时等待一些东西，然后崩溃".

I'm going to work with a HTML form using AJAX and a PHP file to manage data but neither success message or error message work. I tried firebug but it's like it waits something when occur the $.ajax statement then it "crashes".

<form id="alcoholic-form" name="alcoholic-form" method="post">
    <fieldset>
        <input type="submit" id="submit" value="Invia">
    </fieldset>
</form><!-- form -->

<div id="risultato"></div>
<div id="errore"></div>

controller.js文件:

The controller.js file:

$(document).ready(function(){
    $("#submit").click(function(){

    //var formData = $("#alcoholic-form").serialize();
    var nome = "nico";
    var cognome = "basi";


    $.ajax({
        type: "POST",
        url: "prova.php", //data manager
        data: "nome=" + nome + "&cognome=" + cognome, //data to server
        dataType: "html", //return value format
        success:function(msg){
            $("#risultato").html(msg);
            alert(msg);
        },//success
        error: function(xhr, status, error) {
            var err = eval("(" + xhr.responseText + ")");
            alert(err.Message);
            }
    });//ajax
});//submit click
});//document ready 

然后是prova.php文件:

And then the prova.php file:

<?php /* questionario.php */ ?>

<?php

$nome = $_POST['nome'];
$cognome = $_POST['cognome'];

echo "nome = " . $nome . " cognome = " . $cognome;

?>

提前谢谢

推荐答案

您有一个带有提交"按钮的表单，当您单击提交"按钮时，默认操作是提交表单.

You have a form with a submit button, when you click on the submit button the default action is to submit the form.

由于您正在使用ajax请求，因此需要阻止默认表单提交.您可以调用event.preventDefault()来做到这一点.

Since you are using ajax request you need to prevent the default form submit. You can call event.preventDefault() to do that.

$(document).ready(function () {
    $("#submit").click(function (e) {
        //prevent the default form submit
        e.preventDefault();

        //var formData = $("#alcoholic-form").serialize();
        var nome = "nico";
        var cognome = "basi";


        $.ajax({
            type: "POST",
            url: "prova.php", //data manager
            data: "nome=" + nome + "&cognome=" + cognome, //data to server
            dataType: "html", //return value format
            success: function (msg) {
                $("#risultato").html(msg);
                alert(msg);
            }, //success
            error: function (xhr, status, error) {
                var err = eval("(" + xhr.responseText + ")");
                alert(err.Message);
            }
        }); //ajax
    }); //submit click
}); //document ready

---

但是由于您有一个表单和一个提交按钮，因此最好处理表单的提交事件，而不是按钮的点击事件

---

But since you have a form and a submit button, it will be better to handle the submit event of the form that the click event of the button

$(document).ready(function () {
    $("#alcoholic-form").submit(function (e) {
        //prevent the default form submit
        e.preventDefault();

        //var formData = $("#alcoholic-form").serialize();
        var nome = "nico";
        var cognome = "basi";


        $.ajax({
            type: "POST",
            url: "prova.php", //data manager
            data: "nome=" + nome + "&cognome=" + cognome, //data to server
            dataType: "html", //return value format
            success: function (msg) {
                $("#risultato").html(msg);
                alert(msg);
            }, //success
            error: function (xhr, status, error) {
                var err = eval("(" + xhr.responseText + ")");
                alert(err.Message);
            }
        }); //ajax
    }); //submit click
}); //document ready

这篇关于jQuery $ .ajax不显示任何内容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！