0

我已经阅读了这块板上的每条评论 4 小时，以解决此代码的问题。我希望有人可以在这里向菜鸟提供一些反馈。

<?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();
?>

我的 HTTP/mainpage 使用这种表单结构：

<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>

提前感谢您的所有帮助

4

5 回答 5

1

删除字符“ ` ”中

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

并再次测试

于 2013-10-04T07:21:27.567 回答
0

你的代码中有一些语法问题我已经修改了它检查它

     <?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'];

  echo $value;

 echo '<br></br>';
 $value2 = $_POST['email'];

  echo $value2;
  $value3 = $_POST['shirt'];


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

  $result = mysql_query($sql);

  mysql_close();

 ?>
于 2013-10-04T07:26:12.220 回答
0

嗨，我已经在我的电脑上尝试了你的脚本，并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，因为它已弃用。请使用 PDO 或 MYSQLi。

完整的工作示例： 

<?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>
于 2013-10-04T07:35:15.597 回答
0

你有额外的性格：

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

尝试这个：

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

此外，您很容易受到 SQL 注入的影响。使用real_escape_string在您的$_POST

于 2013-10-04T07:20:00.757 回答
-1

首先在最后一行关闭表单标签

</form>

现在删除查询末尾的`符号

以下是正确的替换它

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

在查询结束时

于 2013-10-04T07:20:02.140 回答