100分求一段代码.急......

Question

phpmyadmin吧,有现成为什么不用呢？

Answer 1

<form action=post>
<input type=text name=sql>
<input type=submit>
</form>
<?php
if($_SERVER['REQUEST_METHOD'] != 'POST') exit();
mysql_connect();
mysql_select_db("test");
mysql_query($_POST['sql']) or die('操作失败:'.mysql_error());
echo $_POST['sql'].' 成功';
?>

Answer 2

又被絮叨抢了,还好有点错误,改正一下,
<form action=post> 应该是 <form action="runsql.php" method="post">mysql_connect(); 应有参数 mysql_connect("localhost","username","password");
//runsql.php
<? 
<form action="runsql.php" method="post">
<input type=text name=sql>
<input type=submit>
</form>
<?php
if($_SERVER['REQUEST_METHOD'] != 'POST') exit();
mysql_connect("localhost","username","password");//根据情况修改mysql_select_db("test");
mysql_query($_POST['sql']) or die('操作失败:'.mysql_error());
echo $_POST['sql'].' 成功';
?>

Answer 3

恩，
phpmyadmin的代码也可以参考一下先