PHP5对Mysql5的任意数据库表的管理代码示例(二)

王朝mysql·作者佚名  2008-05-31
窄屏简体版  字體: |||超大  

续: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" />

待续。

 
 
 
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