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我有一个表命名AIRCONDITION(字段:ACName,空间,BTU,EnergyClass),我想编辑这些field.I使用下面的代码来选择要编辑的AIRCONDITION:编辑表值码不起作用
test1.php
<?php
$username = "george";
$password = "george123";
$hostname = "localhost";
$dbhandle = mysql_connect($hostname, $username, $password) or die("Could not connect to database");
$selected = mysql_select_db("my_db", $dbhandle);
$table = "aircondition";
$sql = "SELECT * FROM aircondition";
$result = mysql_query($sql, $dbhandle);
if(mysql_num_rows($result) > 0){
while($row = mysql_fetch_array($result)) {
echo $row['ACName']. "<a href='edit.php?edit=$row[ACName]'> Edit<br>
</a><br>";
}
}
?>
</body>
</html>
和我有这样一来更新特定字段(BTU例如)
edit.php
<?php
$username = "george";
$password = "george123";
$hostname = "localhost";
$dbhandle = mysql_connect($hostname, $username, $password) or die("Could not connect to database");
$selected = mysql_select_db("my_db", $dbhandle);
$id = intval($_GET['edit']);
if($id > 0) {
$res = mysql_query("SELECT * FROM aircondition WHERE ACNumber = '$id'");
$row= mysql_fetch_array($res);
$newbtu = mysql_real_escape_string($_POST['newbtu']);
$sql = "UPDATE aircondition SET BTU='$newbtu' WHERE ACNumber='$id'";
$res = mysql_query($sql) or die ("Error Updating".mysql_error());
echo "<meta http-equiv='refresh' content='0;url=edit.php?edit=$id'>";
}
?>
<form action="edit.php?edit=<?= $id; ?>" method="POST">
<input type="text" name="newbtu" placeholer="test" /><br>
<input type="submit" value="Update" />
</form>
</body>
</html>
然而,这一次似乎并没有工作。虽然我没有得到任何的错误,但是在这些领域没有更新。
您正在从ACName中获取并更新ACNumber。你确定这两个字段具有相同的值吗? – yajakass 2014-09-22 17:07:51
MySQL函数的折旧年限为PHP 5.5中,请使用MySQLi函数。 – Edward 2014-09-22 17:15:54
谢谢你,我的坏。我下次会更小心,以避免这样的错误。 – george123 2014-09-22 17:16:00