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codingan error terus
gimana cara mbenerin kodingan saya yang kayak gini kasusnya Warning: mysqli_query() expects at least 2 parameters, 1 given in C:\xampp\htdocs\forminput\index.php on line 13, saya udak otak atik pake cara ini itu tetep aja hasilnya sama. berikut kodenya :
<?php
include('koneksi.php');
?>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Data Radio</title>
</head>
<body bgcolor="white cyan">
<?php
$sql = 'select * from data_radio';
$query = mysqli_query('$sql, $database');
?>
<h2><strong><p align="center">DATA RADIO</p></strong></h2>
<table width="1400" border="1" cellpadding="0" cellspacing="0" align="center">
<!--DWLayoutTable-->
<tr>
<td width="60" height="50" align="center" valign="middle" bgcolor="#00FFFF"><b>No</b></td>
<td width="460" align="center" valign="middle" bgcolor="#00FFFF"><b>Nama Radio</b></td>
<td width="460" align="center" valign="middle" bgcolor="#00FFFF"><b>Nama Udara</b></td>
<td width="100" align="center" valign="middle" bgcolor="#00FFFF"><b>Frekuensi</b></td>
<td width="390" align="center" valign="middle" bgcolor="#00FFFF"><b>Wilayah Layanan</b></td>
<td width="133" align="center" valign="middle" bgcolor="#00FFFF"><a href="form.php"><b>TAMBAH</b></a></td></tr>
<?php
while($data_radio = mysqli_fetch_array($query)){
?>
<tr>
<td p align="center" bgcolor="#FFFFFF"><?php echo $data['nama_radio']; ?></td>
<td p align="center" bgcolor="#FFFFFF"><?php echo $data['nama_udara']; ?></td>
<td p align="center" bgcolor="#FFFFFF"><?php echo $data['frekuensi']; ?></td>
<td p align="center" bgcolor="#FFFFFF"><?php echo $data['wilayah_layanan']; ?></td>
<td p align="center" bgcolor="#FFFFFF">
Edit | Hapus
</td>
</tr>
<?php
}
?>
</table>
</body>
</html>
7 Jawaban:
Kutip pada mysqli_query dihapus. Sperti ini. Kalo masih gak bisa. $sql ditukar posisi dengan $database. $query = mysqli_query($sql, $database)
keluarnya malah peringatan seperti ini Warning: mysqli_query() expects parameter 1 to be mysqli, string given in C:\xampp\htdocs\forminput\index.php on line 13
coba kayak gini gan
<?php
$sql = "SELECT * FROM data_radio";
$query = mysqli_query($database, $sql);
while($data = mysqli_fetch_array($query)):
?>
<tr>
<td p align="center" bgcolor="#FFFFFF"><?php echo $data['nama_radio']; ?></td>
<td p align="center" bgcolor="#FFFFFF"><?php echo $data['nama_udara']; ?></td>
<td p align="center" bgcolor="#FFFFFF"><?php echo $data['frekuensi']; ?></td>
<td p align="center" bgcolor="#FFFFFF"><?php echo $data['wilayah_layanan']; ?></td>
<td p align="center" bgcolor="#FFFFFF">
Edit | Hapus
</td>
</tr>
<?php endwhile; ?>
malah muncul dua peringatan, mohon bantuannya saya baru belajar
Warning: mysqli_query() expects parameter 1 to be mysqli, string given in C:\xampp\htdocs\forminput\index.php on line 13
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in C:\xampp\htdocs\forminput\index.php on line 14
Oh iya bagian
while( $data = mysqli_fetch_array($query))
ganti jadi
while( $data = mysqli_fetch_assoc($query))
Itu msyqli nya kan butuh 2 parameter. Parameter pertama database,parameter kedua querynya
// pada file koneksi.php pastikan variabel ($database) yg menampung koneksinya
$database = mysqli_connect('localhost','user','password','database');
$sql = 'select * from data_radio';
$query = mysqli_query('$sql, $database');
// ubah menjadi seperti ini, karena aturan fungsi mysqli_query(Koneksi , query string)
$query = mysqli_query($database, $sql);
Semoga Bermanfaat, Selamat Belajar... Happy Sharing (^_^)