刷新或打开页面将空数据插入我的数据库!PHP,来自

Cri*_*pto 0 html php mysql forms

我有一个有几个输入的表单.但是,当我只是加载页面时,一个空白项目被添加到数据库中. 即使我第一次进入网站,清除了所有的cookie,它仍然会向数据库添加空数据 为什么会这样?

<body>





<hr></hr>
<div
style="

margin-left:auto;
margin-right:auto;

width:600px;
"
>
<form method="post" action="admin.php" name="main" id="main">
Post to:
<select name="wheretopost" onchange="testValue(this);" name="select" id="select">
<option value="blog">Blog</option>
<option name='links' value="links">Links</option>
<option value="apparel">Apparel</option>
<option value="goods">Goods</option>
</select>
<div class="productKind" style="padding:10px;">
Mens<input type="radio" name="productKind" id="productKind"  value="Mens">
Womens<input type="radio" name="productKind" id="productKind"  value="Womens">
Kids<input type="radio" name="productKind" id="productKind"  value="Kids">
</div>
<div class="goodsKind" style="padding:10px;">
Stickers<input type="radio" name="goodsKind" id="goodsKind"  value="Stickers">
Incense<input type="radio" name="goodsKind" id="goodsKind"  value="Incense">
Patches<input type="radio" name="goodsKind" id="goodsKind"  value="Patches">
</div>
<br/>
Subject:<br/>
<input type="text" name="title" style="width:100%;" />
<br/>
<br/>
TextArea:<br/>
 <textarea name="txtarea" style="width:100%;" rows="30">
 </textarea>

<center> <input type="submit" style="width:300px;" />
</center>
</form>

</div>






<?php
$type = $_POST["wheretopost"];
$title = $_POST["title"];
$body = $_POST["txtarea"];
$date = date("F j, Y");
?>


<?
$sql = "INSERT INTO `yoyo`.`posts` (`id`, `type`, `title`, `body`, `date`) VALUES (NULL, '$type', '$title', '$body', '$date');";
mysql_query($sql);

?>


 </body>
</html>
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请注意,它发布的页面是自己的.(这可能是问题吗?)

ok3*_*k32 6

在插入内容之前检查请求是否已发布:

if ($_SERVER['REQUEST_METHOD'] == 'POST') {
    $type = $_POST["wheretopost"];
    $title = $_POST["title"];
    $body = $_POST["txtarea"];
    $date = date("F j, Y");

    $sql = "INSERT INTO `yoyo`.`posts` (`id`, `type`, `title`, `body`, `date`) VALUES (NULL, '$type', '$title', '$body', '$date');";
    mysql_query($sql);
}
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