用于从 POST 表单向 mySQL 添加数据的 php [英] php for adding data to mySQL from POST form

查看:48
本文介绍了用于从 POST 表单向 mySQL 添加数据的 php的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我已经阅读了关于此代码故障排除的每条评论 4 小时.我希望有人可以向这里的菜鸟提供一些友好的反馈.

我的 HTTP/主页使用这种表单结构:

<字段集><div class="row-fluid"><div class="span12"><label class="no">Name</label><input name="name" placeholder="Name" type="text">

<div class="row-fluid"><div class="span12"><label class="no">电子邮件</label><input name="email" placeholder="Email" type="text">

<div class="row-fluid"><div class="span12"><div class="row-fluid"><选择样式=显示:无;"name="shirt" class="selectpicker span12"><option selected="selected">选择首选衬衫</option><option>男式衬衫</option><option>女式衬衫</option><option>青年/儿童衬衫</option></选择>

<div class="formFoot"><button type="submit" class="btn">提交</button>

提前感谢您的帮助

解决方案

我已经在我的 PC 中尝试了您的脚本,但在查询中发现了一个额外的单引号.此外,您的选择输入隐藏在 HTML 文件中.

不要在实际项目中使用 Mysql api,因为它已被弃用.请使用 PDO 或 MYSQLi.

完整工作示例:

HTML:

<头><title>项目</title><身体><form id="joinForm" action="join.php" method="post" accept-charset="utf-8"><字段集><div class="row-fluid"><div class="span12"><label class="no">Name</label><input name="name" placeholder="Name" type="text">

<div class="row-fluid"><div class="span12"><label class="no">电子邮件</label><input name="email" placeholder="Email" type="text">

<div class="row-fluid"><div class="span12"><div class="row-fluid"><select style="" name="shirt" class="selectpicker span12"><option selected="selected">选择首选衬衫</option><option>男式衬衫</option><option>女式衬衫</option><option>青年/儿童衬衫</option></选择>

<div class="formFoot"><button type="submit" class="btn">提交</button>

</fieldset></表单></html>

I have been reading every comment on this board for 4 hours in relation to troubleshooting this code. I was hoping someone can provide some kind feedback to a noob here.

<?php
//Join.php
define('DB_NAME', 'biblaunch');
define('DB_USER', 'xxxxxxx');
define('DB_PASSWORD', 'xxxxxxxxx');
define('DB_HOST', 'xxxxxxxxx.hostedresource.com');

$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);

if (!$link) {
    die('Could not connect: ' . mysql_error());
}

$db_selected = mysql_select_db(DB_NAME, $link);

if (!$db_selected) {
    die('Can\'t use ' . DB_NAME . ': ' . mysql_error());
}

$value = $_POST['name'];
$value2 = $_POST['email'];
$value3 = $_POST['shirt'];


$sql = "INSERT INTO customer (name, email, shirt) VALUES ('$value', 
'$value2','$value3')";`

$result = mysql_query($sql);

mysql_close();
?>

my HTTP/mainpage uses this form structure:

<form id="joinForm" action="join.php" method="post" accept-charset="utf-8">
                       <fieldset>                       
                        <div class="row-fluid">
                        <div class="span12">
             <label class="no">Name</label>
                 <input name="name" placeholder="Name" type="text">
                        </div>
                        </div>
                                     <div class="row-fluid">
                         <div class="span12">
             <label class="no">Email</label>
             <input name="email" placeholder="Email" type="text">
                                    </div>
                                    </div>
                                 <div class="row-fluid">
                 <div class="span12">   
                     <div class="row-fluid">
    <select style="display: none;" name="shirt" class="selectpicker span12">

<option selected="selected">Select Preferred Shirt</option>

<option>Mens Style Shirt</option>       

<option>Ladies Style Shirt</option>                                            
<option>Youth/Child Shirt</option>
    </select>
                        </div>
                        </div>
                        </div>
<div class="formFoot">                     
<button type="submit" class="btn">Submit</button>

Thank you for all your help in advance

解决方案

Hi I have tried your script in my PC and found an extra single quote in query. In addition your select input was hidden in HTML file.

Do not use Mysql api in real project as its deprecated. Please use PDO or MYSQLi.

Full working Example:

<?php

//join.php
define('DB_NAME', 'test');
define('DB_USER', 'root');
define('DB_PASSWORD', 'root');
define('DB_HOST', 'localhost');


$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);

if (!$link) {
    die('Could not connect: ' . mysql_error());
}

$db_selected = mysql_select_db(DB_NAME, $link);

if (!$db_selected) {
    die('Can\'t use ' . DB_NAME . ': ' . mysql_error());
}

$value      = $_POST['name'];
$value2     = $_POST['email'];
$value3     = $_POST['shirt'];


$sql = "INSERT INTO customer (name, email, shirt) VALUES ('$value', '$value2','$value3')";

$result = mysql_query($sql) or die(mysql_error());

mysql_close();


?>

HTML:

<!doctype html>
<head>
    <title>Items</title>
</head>
<body>
    <form id="joinForm" action="join.php" method="post" accept-charset="utf-8">
        <fieldset>                       
            <div class="row-fluid">
                <div class="span12">
                    <label class="no">Name</label>
                    <input name="name" placeholder="Name" type="text">
                </div>
            </div>
            <div class="row-fluid">
                <div class="span12">
                    <label class="no">Email</label>
                    <input name="email" placeholder="Email" type="text">
                </div>
            </div>
            <div class="row-fluid">
                <div class="span12">   
                    <div class="row-fluid">
                        <select style="" name="shirt" class="selectpicker span12">

                            <option selected="selected">Select Preferred Shirt</option>
                            <option>Mens Style Shirt</option>       
                            <option>Ladies Style Shirt</option>                                            
                            <option>Youth/Child Shirt</option>
                        </select>
                    </div>
                </div>
            </div>

             <div class="formFoot">                     
                <button type="submit" class="btn">Submit</button>
            </div>
        </fieldset>

    </form>

</body>
</html>

这篇关于用于从 POST 表单向 mySQL 添加数据的 php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!

查看全文
相关文章
PHP最新文章
热门教程
热门工具
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆