我刚刚学会了php在页面上显示信息,我收到了一个错误,请帮帮我..谢谢
此filepolitik.php
<?php
include 'koneksi.php';
$tabel1 = 'isiberita';
$tabel2 = 'kategori';
$id=1;
$hal = $_GET['hal']; //*Error : Notice: Undefined index: hal ???
//validasi halaman
if (!isset($_GET['hal'])) {
$page = 1;
} else {
$page = $_GET['hal'];
}
//data tampil perhalaman
$max_result = 2;
$from = (($page * $max_result) - $max_result);
//query tampil perhalaman
$sql=$conn->query("SELECT * FROM $tabel1,$tabel2 where $tabel1.id_kategori=$id and $tabel1.id_kategori=$tabel2.id_kategori
ORDER BY tanggal DESC LIMIT $from,$max_result");
while ($tampil = $sql->fetch(PDO::FETCH_ASSOC))
{
$data = substr($tampil['isi'],0,200);
echo "<div class='box'>";
echo "<p align='justify'>";
echo "<img class='gambar' src='$tampil[gambar]' width=150px height=150px align='left' />";
echo "<font valign='top'>";
echo "<strong>";
echo $tampil['judul'];
echo "</strong>";
echo $data;
echo "<a href='index.php?menu=detail_politik&id==$tampil[id_berita]'> baca lengkap>>>></a>";
echo "</font></p></div><br>";
}
//total
$total_result = mysql_result(mysql_query("SELECT count(*) as Num From $tabel1 where $tabel1.id_kategori=$id"),0);
//Fatal error: Uncaught Error: Call to undefined function mysql_result() ??
$total_pages = ceil($total_result / $max_result);
echo "<center> piliih Halaman <br />";
//proses link
if ($hal > 1) {
$prev = ($page - 1);
echo "<a href=$_SERVER[PHP_SELF]?menu=politik&hal$prev>
<-Sebelumnya </a>";
}
//link daftar urutan Halaman
for ($i=1; $i <= $total_pages; $i++) {
if (($hal)==$i) {
echo "$i";
} else {
echo "<a href=$_SERVER[PHP_SELF]?menu=politik&hal=$i>$i</a>";
}
}
//link halaman selanjutnya
if ($hal < $total_pages) {
$next = ($page + 1);
echo "<a href=$_SERVER[PHP_SELF]?menu=politik&hal=$next>Selanjutnya-></a>";
}
echo "</center>";
?>
此输出