PHP5对Mysql5的任意数据库表的管理代码示例之二

来源：互联网作者： rocket 2008-03-17/10:36

正文

点击添加一个条目就会跳转到add.php
    //add.php
    <html>
<head><title>Add an entry to the database</title>
<meta http-equiv="Content-Type" content="text/html; charset=gb2312" />
</head>
<body bgcolor=#ffffff>
<h2>Add an entry</h2>
<?php
$database = "sunsite";
$tablename = $_GET['tablename'];
mysql_connect("localhost","root","") or die ("Problem connecting to DataBase");
$query = "show columns from $tablename";
$result = mysql_db_query($database,$query);
$column = 0;
if ($result)
{
while ($r = mysql_fetch_array($result))
{
$colname[$column] = $r[0];
$column = $column + 1;
}
mysql_free_result($result);
}
$id = 0;
do
{
$id++;
$query = "select * from $tablename where $colname[0]='$id'";
$result = mysql_db_query($database,$query);
}
while(mysql_fetch_array($result))
?>
<form method="post" action="addsoftware.php">
<table width=90% align=center>
<tr><td><?php echo "$colname[0]";?>:</td><td><?php echo "$id"; ?></td></tr>
<?
for ($col=1;$col<$column;$col++)
echo "<tr><td>$colname[$col]:</td><td><input type=text name=$colname[$col] size="100%"></td></tr>";
?>
<input type=hidden name="<?php echo "$colname[0]"; ?>" value="<?php echo "$id"; ?>">
<input type=hidden name=tablename value="<?php echo "$tablename"; ?>">
<tr><td></td><td><input type=submit value=" Add "></td></tr>
</table>
</form>
<a href="tables.php?tablename=<? echo "$tablename"?>">Finish</a>
</body>
</html> 其中id是自动寻找并生成的，而且保证在数据库中id是唯一的。点击add之后会将内容写入数据库，然后会继续让你添加内容，直至添加完毕点击finish回到上一层，显示出添加条目後的数据库内容。
下面是添加过程中对数据库进行操作的addsoftware.php
    //addsoftware.php
    <?php
$database = "sunsite";
$tablename = $_POST['tablename'];
mysql_connect("localhost","root","") or die ("Problem connecting to DataBase");
$query = "show columns from $tablename";
$result = mysql_db_query($database,$query);
$column = 0;
if ($result)
{
while ($r = mysql_fetch_array($result))
{
$colname[$column] = $r[0];
$column = $column + 1;
}
mysql_free_result($result);
}

for($col=0;$col<$column;$col++)
$para[$col] = $_POST[$colname[$col]];

if ($_POST['name'])
{
mysql_connect("localhost","root","") or die ("Problem connecting to DataBase");
$query = "insert into $tablename values ('$para[0]' ";
for($col=1;$col<$column;$col++)
$query = $query . ",'" . $para[$col] . "'";
$query = $query . ");";

$result = mysql_db_query($database, $query);
Header("Location: add.php?tablename=$tablename");
}
else
{
echo "No name Entered. Please go back and reenter name";
}
?>
<meta http-equiv="Content-Type" content="text/html; charset=gb2312" />