Postingan lainnya
MySQL Query Error
ini error kayak gini cara fixnya gan Fatal error: Uncaught Error: Call to undefined function mysql_query() in D:\xampp\htdocs\rifan\daftarbarang.php:23 Stack trace: #0 {main} thrown in D:\xampp\htdocs\rifan\daftarbarang.php on line 23
<head>
<title>Daftar Barang</title>
</head>
<body>
<?php
include "menubarang.php";
?>
<table width="519" border="1">
<caption align="top">Daftar Barang</caption>
<tr>
<th width="25" scope="col">No</th>
<th width="103" scope="col">Kode Barang </th>
<th width="128" scope="col">Nama Barang </th>
<th width="60" scope="col">Harga</th>
<th width="81" scope="col">Persediaan</th>
<th width="82" scope="col">Action Edit</th>
<th width="82" scope="col">Action Delete</th>
</tr>
<?php
include "db.php";
$sql = ("SELECT * FROM barang");
$query = mysql_query($sql);
$no = 1;
while($data = mysql_fetch_array ($query)){
echo "ada yang error: ".mysqli_error()
?>
<tr>
<td><?php echo $no?></td>
<td><?php echo $data['kodebarang']?></td>
<td><?php echo $data['namabarang']?></td>
<td><?php echo $data['harga'];?></td>
<td><?php echo $data['persediaan']?></td>
<td>
<a href="editbarang.php?kode=<?php echo $data['kodebarang']?>">Edit</a>
</td><td><a href="deletebarang.php?kode=<?php echo $data['kodebarang']?>">Delete</a></td>
</tr>
<?php
$no++;}
?>
</table>
</body>
</html>
dan juga ini gan Fatal error: Uncaught Error: Call to undefined function mysqli() in D:\xampp\htdocs\rifan\tambah.php:69 Stack trace: #0 {main} thrown in D:\xampp\htdocs\rifan\tambah.php on line 69
<head>
<title>Tambah Data Barang</title>
<script language="javascript">
function cekform(){
//ini untuk ngecek formnya (semua form tidak boleh kosong)
if(document.frmbarang.txtkode.value==""){
alert('Kode Barang Harus Diisi');
document.frmbarang.txtkode.focus();
return false;
} else if(document.frmbarang.txtnama.value==""){
alert('Nama Barang Harus Diisi');
document.frmbarang.txtnama.focus();
return false;
} else if(document.frmbarang.txtharga.value==""){
alert('Harga Barang Harus Diisi');
document.frmbarang.txtharga.focus();
return false;
} else if(document.frmbarang.txtpersediaan.value==""){
alert('Persediaan Barang Harus Diisi');
document.frmbarang.txtpersediaan.focus();
return false;
} else {
return true;
}
}
</script>
</head>
<body>
<?php
include "menubarang.php";
?>
<form action="" method="post" name="frmbarang" onsubmit="return cekform()">
<table width="500" border="1">
<tr>
<td width="163">Kode Barang </td>
<td width="321"><input name="txtkode" type="text" id="txtkode" size="5" maxlength="5" /></td>
</tr>
<tr>
<td>Nama Barang </td>
<td><input name="txtnama" type="text" id="txtnama" /></td>
</tr>
<tr>
<td>Harga</td>
<td><input name="txtharga" type="text" id="txtharga" /></td>
</tr>
<tr>
<td>Persediaan</td>
<td><input name="txtpersediaan" type="text" id="txtpersediaan" /></td>
</tr>
<tr>
<td> </td>
<td><input name="tblIsi" type="submit" id="tblIsi" value="Tambah Barang" /></td>
</tr>
</table>
</form>
</body>
</html>
<?php
include "db.php";
if(isset($_POST['tblIsi'])){
$kode = $_POST['txtkode'];
$nama = $_POST['txtnama'];
$harga = $_POST['txtharga'];
$persediaan = $_POST['txtpersediaan'];
$sql = "INSERT INTO barang VALUES('$kode','$nama','$harga','$persediaan')";
$query = mysql_query($sql);
if($query){
echo "<script>alert('Data barang berhasil dimasukkan ke database')</script>";
} else {
echo "<script>alert('Data barang gagal dimasukkan ke database')</script>";
echo mysql_error();
}
}
?>
tolong bantuanya gan
0
1 Jawaban:
coba jalanin query nya pake mysqli mas. mysql udah deprecated.
0
