PHP5对Mysql5的任意数据库表的管理代码示例（三）

续：点击编辑一个条目会跳转至edit.php
//edit.php

Edit an entry

$database = "sunsite";
$tablename = $_REQUEST['tablename'];
echo "

Data from $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)
{
echo "Found these entries in the database:

";
echo "";
while ($r = mysql_fetch_array($result))
{
echo "";
$colname[$column] = $r[0];
$column = $column + 1;
}
echo "";
mysql_free_result($result);

$query = "select * from $tablename";
$result = mysql_db_query($database, $query);
if ($result)
while ($r = mysql_fetch_array($result))
{
echo "

";
echo "";
for($col=1;$col<$column;$col++) echo "";
echo "";
}
echo "

$r[0]
$r[0]	$r[$col]

";
}
else echo "No data.";
mysql_free_result($result);
?>
">Finish


然后选择其中一个id，会跳出具体的内容并要求用户修改。
//editing.php

Editing an entry

$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);
}

$temp = $_GET[$colname[0]];
$query = "select * from $tablename where $colname[0]=$temp";
$result = mysql_db_query($database,$query);
$r = mysql_fetch_array($result);

?>

for ($col=1;$col<$column;$col++)
echo "";
?>
">
">

:
$colname[$col]:

最后是写入数据库
//editdb.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 = "update $tablename set $colname[1]='$para[1]'";
for($col=2;$col<$column;$col++)
$query = $query . ",$colname[$col]='$para[$col]'";
$query = $query . " where $colname[0]='$para[0]';";

$result = mysql_db_query($database, $query);
Header("Location: edit.php?tablename=$tablename");
}
else
{
echo "No name Entered. Please go back and reenter name";
}
?>

待续。