从PHP插入MySQL(jQuery/AJAX)

9 php mysql ajax jquery

我已经看过很多教程,但是它们很混乱,并且做我想做的事情,我只是不知道如何使用这些教程中的现有东西并让它们以我想要的方式工作.

我有一个非常简单的表单,包含文本框,标签和提交按钮.当用户在表单中输入内容,然后单击提交,我想使用php和ajax(使用jquery)将表单的结果插入到mysql数据库中.

有人可以告诉我这是如何实现的吗?只有一些非常基本的东西才是让我开始的.任何帮助表示赞赏.

谢谢

Sim*_*n H 20

嗨,这里只是一个如何做到这一点的简单例子:

HTML:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
    <head>
        <title>Quick JQuery Ajax Request</title>
        <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">

        <!-- include the jquery lib -->
        <script type="text/javascript" src="jquery.js"></script>
        <script type="text/javascript">
            var ajaxSubmit = function(formEl) {
                // fetch where we want to submit the form to
                var url = $(formEl).attr('action');

                // fetch the data for the form
                var data = $(formEl).serializeArray();

                // setup the ajax request
                $.ajax({
                    url: url,
                    data: data,
                    dataType: 'json',
                    success: function() {
                        if(rsp.success) {
                            alert('form has been posted successfully');
                        }
                    }
                });

                // return false so the form does not actually
                // submit to the page
                return false;
            }
        </script>

    </head>
    <body>

        <form method="post" action="process.php"
              onSubmit="return ajaxSubmit(this);">
            Value: <input type="text" name="my_value" />
            <input type="submit" name="form_submit" value="Go" />
        </form>

    </body>
</html>
Run Code Online (Sandbox Code Playgroud)

process.php脚本:

<?php

function post($key) {
    if (isset($_POST[$key]))
        return $_POST[$key];
    return false;
}

// setup the database connect
$cxn = mysql_connect('localhost', 'username_goes_here', 'password_goes_here');
if (!$cxn)
    exit;
mysql_select_db('your_database_name', $cxn);

// check if we can get hold of the form field
if (!post('my_value'))
    exit;

// let make sure we escape the data
$val = mysql_real_escape_string(post('my_value'), $cxn);

// lets setup our insert query
$sql = sprintf("INSERT INTO %s (column_name_goes_here) VALUES '%s';",
                'table_name_goes_here',
                $val
);

// lets run our query
$result = mysql_query($sql, $cxn);

// setup our response "object"
$resp = new stdClass();
$resp->success = false;
if($result) {
    $resp->success = true;
}

print json_encode($resp);
?>
Run Code Online (Sandbox Code Playgroud)

请注意,这些都没有经过测试.我希望你能帮助你.

  • OOW!谢谢堆,@ ZeSimon.:) (2认同)