PHP问题在这里，我制作了一个登录/注销类型的代码，具有insert和delete功能，支持login和logout。
所以问题是插入文本后我根本无法删除它，因为删除按钮就像一个简单的回头波顿，不做他的工作，什么都没有在if(isset($_POST['delete']))条件下似乎有效。
问题可能是我正在使用两个无效操作引用同一页面？导致第一个按钮工作和第二个不。
任何人都能理解为什么？
<html>
<header></header>
<body>
<!-- START PHP -->
<?php
//If not submit i put the submit form
if(!isset($_POST['send'])){
echo "<form name='send' action='' method='POST'>
<input type='text' name='text' value=''/>
<input type='submit' name='send' value='send' />
</form>";
}<!-- IF END -->
//If submit was set I insert $text into the db and I render
//the delete button
else {
$conn= mysql_connect('localhost','root','');
mysql_select_db('db_try',$conn ) or die(mysql_error());
$dato=$_POST['dato'];
mysql_query(" INSERT INTO test (value) VALUES ('$text') ") or die(mysql_error());
echo "Operation complete";
//Now i render the delete submit button...
echo "<form name='delete' action='' method='POST'>
<input type='submit' name='delete' value='delete' />
</form>";
//...and if i push it NOTHING, like it's only
//a return to the first form button
if(isset($_POST['delete'])){
mysql_query(" DELETE FROM test WHERE value='$text' ") or die(mysql_error());
echo "<br>Text'".$text."' deleted";
}
}<!-- ELSE END-->
?><!-- END PHP -->
</body>
</html>您的代码存在逻辑问题。单击删除按钮时，脚本将再次运行。您拥有的第一个条件 - if(!isset($_POST['send']))现在将通过，因为不再设置发送按钮，因此它会进入 if 语句并且永远不会运行您的删除代码。
您的脚本似乎也容易受到 SQL 注入的影响。
这是执行此操作的正确方法，这是一个快速提示，您需要在mysql插入安全性等方面做更多的工作。
<html>
<header>
<body>
<?php
$conn= mysql_connect('localhost','root','');
mysql_select_db('db_try',$conn ) or die(mysql_error());
if(isset($_POST['send'])){
$text = $_REQUEST['text'];
mysql_query(" INSERT INTO test (value) VALUES ('$text') ") or die(mysql_error());
$answer = "Operation complete";
$form = "<form name='delete' action='' method='POST'>
<input type='submit' name='delete' value='delete' />
</form>";
}
else if(isset($_POST['delete'])){
mysql_query(" DELETE FROM test WHERE value='$text' ") or die(mysql_error());
$answer = "Text'".$text."' deleted";
}
else {
$form = "<form name='send' action='' method='POST'>
<input type='text' name='text' value=''/>
<input type='submit' name='send' value='send' />
</form>";
}
print "<h1>" . $answer . "</h1>";
print $form;
?>
</body>
</header>
</html>
我认为
它也可能有效...
if (!isset($_POST['submit']) || isa($_POST['submit'] != 'login'))