Question

我在将表单插入数据库时遇到问题。我无法弄清楚我错在哪里......

我的表格如下：

 <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
<script src="script.js"></script>

<form id="form" name="form">
    <div>
        <label>Name :</label>
        <input id="name" type="text">
        <label>Email :</label>
        <input id="submit" onclick="myFunction()" type="button" value="Submit">
    </div>
</form>

脚本script.js

function myFunction() {
    var name = document.getElementById("name").value;
    var dataString = 'name1=' + name;
    if (name == '') {
        alert("Please Fill All Fields");
    } else {
        $.ajax({
            type: "POST",
            url: "ajaxjs.php",
            data: dataString,
            cache: false,
            success: function(html) {
                alert(html);
            }
        });
    }
    return false;
}

和php脚本ajaxjs.php

<?php
include ('./includes/connection.php');
$name1 = $_POST['name1'];
if (isset($_POST['name1'])) {
    $query = mysql_query("INSERT INTO db VALUES('TEST','name1')");
}
?>

开发人员工具显示消息： script.js：5未捕获的ReferenceError：未定义dataStringmyFunction @ script.js：5onclick @ test.html：9

Answer 1

您的Ajax代码运行良好但是你的查询命令看起来错了请尝试

INSERT INTO table_name (column1,column2,column3,...)
VALUES (value1,value2,value3,...);

更新：源代码运作良好

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
<script type="text/javascript">
    function myFunction() {
        var name = document.getElementById("name").value;
        var dataString = 'name1=' + name;
        if (name == '') {
            alert("Please Fill All Fields");
        } else {
            $.ajax({
                type: "POST",
                url: "ajaxjs.php",
                data: dataString,
                cache: false,
                success: function(html) {
                alert(html);
                }
            });
        }
        return false;
    }
</script>
<form id="form" name="form">
<div>
<label>Name :</label>
<input id="name" type="text">
<label>Email :</label>
<input id="submit" onclick="myFunction()" type="button" value="Submit">
</div>
</form>

和ajaxjs.php

<?php
    // include ('./includes/connection.php');
    $name1 = $_POST['name1'];
    echo $name1;
    // if (isset($_POST['name1'])) {
    // $query = mysql_query("INSERT INTO db VALUES('TEST','name1')");
    // }
?>

表单提交使用Ajax无法正常工作

1 个答案: