hal1/view.php
<?php
include "koneksi.php";
include "cek.php";
$result = $koneksi->query("select * from pengunjung");
$baris = $result->fetch_assoc();
echo "Halaman Admin<br>";
echo "Nama Anda adalah : ".$_SESSION['namauser']. "<br><br>";
echo "<li><a href = edit.php?userid=$baris[id]>[Edit Data Diri]</a></li><br>
<li><a href=logout.php> Logout</a></li>";
?>
edit.php
<?php
include("koneksi.php");
include("cek.php");
$id=$_GET['userid'];
$query = "select * from pengunjung where id=$id";
$result = mysqli_query($koneksi,$query);
while ($baris = mysqli_fetch_array($result))
{
echo "<form method=post action=update.php>";
echo "Nama : <input type=text name=nama value=$baris[1]>";
echo "<br>";
echo "Email : <input type=text name=email value=$baris[2]>";
echo "<br>";
echo "Situs : <input type=text name=situs value=$baris[3]>";
echo "<br>";
echo "<input type=submit name=submit value=update>";
echo "<input type=hidden name=id value=$baris[0]>";
echo "</form>";
}
?>
update.php
<?php
include("koneksi.php");
$id=$_POST['id'];
$nama=$_POST['nama'];
$email=$_POST['email'];
$situs=$_POST['situs'];
$query = "update pengunjung set id='$id', nama='$nama', email='$email',
situs='$situs' where id='$id'";
$result = mysqli_query($koneksi,$query);
echo '<script language="JavaScript">alert("Data telah di update");
document.location="edit.php"</script>';
?>
我使用登录表单,并以具有特定ID的用户登录。我尝试编辑登录的用户的数据(NAMA,电子邮件,Situs)。问题是为什么它没有选择登录的用户的正确数据(NAMA,电子邮件,Situs)?每当我登录不同的用户时,然后单击编辑时,它总是从数据库上的第一个ID/用户显示数据(NAMA,电子邮件,SITUS)。helpp !!
使用ID变量
<?php
include("koneksi.php");
include("cek.php");
$id=$_GET['userid'];
$query = "select * from pengunjung where id=$id";
$result = mysqli_query($koneksi,$query);
while ($baris = mysqli_fetch_assoc($result))
{
?>
<form method=post action=update.php>
Nama : <input type=text name="nama" value="<?php echo $baris['column1']?>">
Email : <input type=text name="email" value="<?php echo $baris['column2']?>">
Situs : <input type=text name=situs value="<?php echo $baris['column3']?>">
<input type=hidden name=id value="<?php echo $id?>"d>
<input type=submit name=submit value=update>
</form>
<?php
}
?>
hal1/view.php
<?php
include"koneksi.php";
include "cek.php";
echo "Halaman Admin<br>";
echo "Nama Anda adalah : ".$_SESSION['namauser']. "<br><br>";
echo " <li><a href = edit.php?userid=$_SESSION[id]>[Edit Data Diri]</a></li><br>
<li><a href=logout.php> Logout</a></li>";
?>
好吧..所以我认为我通过使用会话解决了该问题,在这种情况下,登录的用户的$ _session [id]。我将会话代码放在check_user_login.php文件上,所以我不这样需要在hal1/view.php中再次进行查询。
check_user_login.php
<?php
include("koneksi.php");
$username = $_POST['username'];
$password = $_POST['password'];
$query = "select * from pengunjung where username='$username' and password='$password'";
$data = mysqli_query($koneksi,$query) or die ("Couldn't execute query");
$hasil = mysqli_fetch_assoc($data);
if ($username=="" || $password=="")
{
echo '<script language="JavaScript">alert("Username atau Password Kosong");
document.location="index.php"</script>';
}
else
if($username==$hasil['username'] and $password==$hasil['password'])
{
session_start();
$_SESSION['id'] = $hasil['id'];
$_SESSION['namauser'] = $username;
header("location:hal1/view.php");
else
{
echo '<script language="JavaScript">alert("Username atau Password Salah");
document.location="index.php"</script>';
}