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这是我第一次尝试使用oracle数据库。 我尝试为我的项目创建一些更新表单,但是当我单击更新按钮时它不会更新。但是如果它有效,在我们按下更新按钮后,它会显示在另一个页面上,但在另一个页面上它根本不会改变。我找不到这些信息。我希望有人能帮助我。无法更新php&oracle
对不起,我的英语不好。
<?php
include 'ora_connect.php';
if (isset($_POST['Update'])) {
$id = $_POST['ID'];
$nama = $_POST['NAMA'];
$departemen = $_POST['DEPARTEMEN'];
$username = $_POST['USERNAME'];
$password = $_POST['PASSWORD'];
$status = $_POST['STATUS'];
$sql = "UPDATE wlan_user SET USERNAME='$username',PASSWORD='$password',NAMA='$nama',STATUS='$status' WHERE ID=$id";
header("Location: index.php?page=monitoring");
}
$sqlparse =oci_parse($conn,$sql);
$result=oci_execute($sqlparse) or die(oci_error());
?>
<?php
$id = isset($_GET['ID']) ? $_GET['ID'] : '';
$query = "SELECT * FROM wlan_user WHERE ID=$id";
$statmen = oci_parse($conn, $query);
oci_execute($statmen, OCI_DEFAULT);
while ($res = oci_fetch_array($statmen, OCI_BOTH))
{
$nama = $res['NAMA'];
$departemen = $res['DEPARTEMEN'];
$username = $res['USERNAME'];
$password = $res['PASSWORD'];
$status = $res['STATUS'];
}
?>
<form name="form1" method="post">
<table border="0">
<tr>
<td>Nama</td>
<td><input type="text" name="nama" value=<?php echo "'$nama'"; ?>></td>
</tr>
<tr>
<td>Departemen</td>
<td><input type="text" name="departemen" value=<?php echo $departemen; ?>></td>
</tr>
<tr>
<td>Username</td>
<td><input type="text" name="username" value=<?php echo "'$username'"; ?>></td>
</tr>
<tr>
<td>Password</td>
<td><input type="password" name="password" value=<?php echo $password; ?>></td>
</tr>
<tr>
<td>Status</td>
<td><input type="radio" name="type" value="A" checked>A<br><input type="radio" name="type" value="B">B<br><input type="radio" name="type" value="C">C</td>
</tr>
<tr>
<td><input type="hidden" name="id" value=<?php echo $_GET['id']; ?>></td>
<td><input type="submit" name="Update" value="Update"></td>
</tr>
</table>
</form>
我不是那么好的php,但为什么你不添加检查,如果连接成功或没有,这是正确的'WHERE ID = $ id'?你应该在旁边添加引号吗? – Moudiz
你从哪里得到$ conn? –
@AchinthaGunasekara它来自我的ora_connect.php – Albanna