警告:mysqli_query()期望至少2个参数,第3行的C:\ xampp \ htdocs \ penjualan \ proses.php中给定1个警告:mysqli_fetch_array()期望参数1为mysqli_result,在C:\中给定null第25行上的xampp \ htdocs \ penjualan \ proses.php。
ini adalah kode nya,
<?php
include "db/koneksi.php";
$data=mysqli_query("select * from tblbarang"); ...................... line3
$op=isset($_GET['op'])?$_GET['op']:null;
if($op=='kode'){
echo"<option>Kode Barang</option>";
while($r=mysqli_fetch_array($data)){
echo "<option value='$r[kode]'>$r[kode]</option>";
}
}elseif($op=='barang'){
echo'<table id="barang" class="table table-hover">
<thead>
<tr>
<Td colspan="5"><a href="?page=barang&act=tambah" class="btn btn-primary">Tambah Barang</a></td>
</tr>
<tr>
<td>Kode Barang</td>
<td>Nama Barang</td>
<td>Harga Beli</td>
<td>Harga Jual</td>
<td>Stok</td>
</tr>
</thead>';
while ($b=mysqli_fetch_array($data)){ ................. line 25
echo"<tr>
tolonng