cara menampilkan data lewat 2 combobox dengan tabel yang berbeda

adit · Dec 31, 2014
mohon pencerahannya pak..
dalam 1 form ada 2 combobox, combobox 1 menampilkan kartu kredit dan no.credit ke textboxt yang akn digunakan sedangkan combobox ke2 tempat dia malakukan transaksi gesek kartu untuk combo 1 sdah bsa tampil tpi untuk yg combo ke2 tdk bsa apakah scrip ini untuk 1 form sja pak mohon dngan sangat bntuannya..
 <?php    
$result = mysql_query("select * from edc");    
$jsArray = "var prdName = new Array();\n";    
echo '<select name="prdId" onchange="changeValue(this.value)">';    
echo '<option>--Silahkan Pilih --</option>';    
while ($row = mysql_fetch_array($result)) {    
    echo '<option value="' . $row['id_edc'] . '">' . $row['nama_edc'] . '</option>';    
    $jsArray .= "prdName['" . $row['id_edc'] . "'] = {name:'" . addslashes($row['nama_edc']) . "',desc:'".addslashes($row['bankRT_trans'])."'};\n";    
}    
echo '</select>';    
?>
                    <input type="hidden" name="nama_edc" id="nama_edc" />
                    <input type="hidden" latestvalue="0" onChange="menghitungTransaksi(&#39;trans&#39;);" id="bankRT_trans" name="bankRT_trans" maxlength="5" size="5" onblur="tCheckNumberPersen(this);">
                    <script type="text/javascript">    
<?php echo $jsArray; ?>  
function changeValue(id){  
document.getElementById('nama_edc').value = prdName[id].name;  
document.getElementById('bankRT_trans').value = prdName[id].desc;  
};  
                  </script>


<?php
$koneksi=mysql_connect("localhost","root","") or die (mysql_error());
$pilih_db=mysql_select_db("dbgestun") or die (mysql_error());
$perintah="SELECT * FROM krt_kredit";
$jalankan_perintah=mysql_query($perintah) or die (mysql_error());
 
$result = mysql_query("select * from krt_kredit");
$jsArray = "var card_type = new Array();\n";
echo '<select name="card_type" onchange="document.getElementById(\'no_credit\').value = card_type[this.value]">';
echo '<option>-------</option>';
while ($row = mysql_fetch_array($result)) {
    echo '<option value="' . $row['card_type'] . '">' . $row['card_type'] . '</option>';
    $jsArray .= "card_type['" . $row['card_type'] . "'] = '" . addslashes($row['no_credit']) . "';\n";
}
echo '</select>';
?>
Bankrate : <input type="text" name="no_credit" id="no_credit"/>  
<script type="text/javascript">
<?php echo $jsArray; ?>
</script>
Silahkan login untuk menjawab!
0
aaron49 · Jan 1, 2016 · 0 Suka · 0 Tidak Suka
mungkin bisa membantu sedikit pak, mohon maaf jikas salah. saya pun masih belajar pak :)
while ($row = mysql_fetch_array($result)) {
    echo '<option value="' . $row['card_type'] . '">' . $row['card_type'] . '</option>';
    $jsArray .= "card_type['" . $row['card_type'] . "'] = {card:'" . addslashes($row['no_credit']) . "'};\n";
}
echo '</select>';
?>
Bankrate : <input type="text" name="no_credit" id="no_credit"/>  
<script type="text/javascript">
document.getElementById('no_credit').value=card_type[id].card;
</script
0
Nia · Aug 9, 2016 · 0 Suka · 0 Tidak Suka
mohon bantuannya mas :)
ini saya ingin membuat paging dengan combobox tp ketika halamannya dipilih, datanya tdk muncul. kira2 kurangnya dmn ya?
<script></script>
<form>
<table id="tab_layout"width="100%">
	<tr height="10"></tr>
	<tr>
		<td style="background-color:white" align="justify">
				<table id="tab_data" width="80%">
					
				<?php 
				include "../config/koneksi.php";
				$tablename="log";
				// jumlah data perhalaman
				$rowsPerPage = 20;
				//nilai pertama
				$pageNum = 1;
				if(!empty($_GET['page']))
				{
					$pageNum = $_GET['page'];
				}
				$offset = ($pageNum - 1) * $rowsPerPage;
				// query database
				$query  = "SELECT * FROM $tablename ORDER by tanggal desc LIMIT $offset, $rowsPerPage";
				$result = mysql_query($query) or die('Error, query failed. ' . mysql_error());
				?>
					  <th>Pengguna</th>
					  <th>Aktifitas</th>
					  <th>Tanggal</th>
					</tr>
				<?php
				while($row = mysql_fetch_array($result))
				{
				?>
					<tr bgcolor="#fff">
					  <td><center><?php echo $row['pengguna']?></center></td>
					  <td><?php echo $row['aktifitas']?></td>
					  <td><center><?php echo $row['tanggal']?></center></td>
					</tr>
				<?php
				} //end of while
				?>
				  </table>
				<?php
				$query   = "SELECT COUNT(tanggal) AS numrows FROM $tablename";
				$result  = mysql_query($query) or die('Error, query failed. ' . mysql_error());
				$row     = mysql_fetch_array($result, MYSQL_ASSOC);
				$numrows = $row['numrows'];
				$maxPage  = ceil($numrows/$rowsPerPage);
				$nextLink = '&nbsp;';
				if($maxPage >1)
				{
					error_reporting(E_ALL ^ (E_NOTICE | E_WARNING));
					$nav .= "<form id=\"FNav\" name=\"FNav\" method=\"get\" action=\"\">";
					$nav .= "Halaman : <select name=\"page\" id=\"page\">";
					for($page = 1; $page <= $maxPage; $page++)
					{
						if($pageNum==$page)
						{
							$nav .= "<option selected>$page</option>";
						} else {
							$nav .= "<option>$page</option>";
						}
					}
					$nav .= "</select>";
					$nav .= "<input type=\"submit\" name=\"btn\" id=\"btn\" value=\"Pilih\" />";
					$nav .= "</form>";
				}
				echo '<p>'.$nav.'</p>';
					mysql_free_result($result);
				?>
		</td>
	</tr>
</table>
</form>