ajax异步请求的一个jscript

52007

<form action="" method="post" enctype="multipart/form-data">
        <input name="q" id="txt" class="txt" type="text" />   
        <input type="submit" class="btn" name="submit" value="Upload" /><br />
</form>
异步请求的表单，怎么让他点击按钮把name="q"的值传到test.php，然后回传请求数据<divid="txtHint" class="file-box">ajax所请求的数据显示位置</div>

没事玩玩破解

[JavaScript] 纯文本查看 复制代码

    var submit = document.querySelector('input[type="submit"]');
    submit.onclick = function () {
        var txt = document.getElementById('txt');
        var txtHint = document.getElementById('txtHint');
        var xhr = new XMLHttpRequest();
        xhr.open('post','test.php',true);
        xhr.onreadystatechange = function () {
            if (xhr.readyState === 4 && xhr.status === 200){
                var response = JSON.parse(xhr.responseText);
                txtHint.innerText = response.name;
            }
        };
        xhr.send('name='+ txt.value);
    };