|
|
Posted by Mark Sargent on 05/12/05 08:19
Kristen G. Thorson wrote:
> I don't understand what you're trying to do. Do you want two separate
> list boxes, one for product types and one for makers? If so, use two
> queries. Use the same block of code you have for product types to
> populate the makers box. If that's not what you're looking for, you
> need to describe what your goal is. You might also give more of your
> table structure since you're asking about joins.
>
>
> kgt
>
>
> Mark Sargent wrote:
>
>> Hi All,
>>
>> with wanting to show both product types(Switch, Router etc) and
>> Makers(Cisco, Avaya, etc) on the one page in select boxes, I was
>> wondering, do you use 2 seperate queries to the database or do you
>> inner join to get all in 1..? I have set up different tables with
>> related id's etc. So, to get Products.product_name and
>> Products.product_id along with what the below query pulls, what wold
>> be best..? I believe if I do a 2nd query, than some variables need to
>> be identified with the particular query, yes..? Cheers.
>>
>> <html>
>> <body>
>> <hr>
>> <h1 align=center>JUMBO STATUS</h1><p>
>> <center>Used Hardware Specialist</center>
>> <hr>
>> <h2 align=center>Admin</h2>
>> <table align="center" border="2">
>> <tr>
>> <td><h3 align=center>PRODUCT TYPE</h3></td>
>> </tr>
>> <tr><td><center>
>> <select name="poduct_type">
>> <?php
>> $db = mysql_connect("localhost", "root", "grunger");
>> mysql_select_db("status",$db);
>> $result = mysql_query("SELECT ProductTypes.product_type_detail,
>> ProductTypes.product_type_id FROM ProductTypes",$db);
>> $num = mysql_num_rows($result);
>> for ($i=0; $i<$num; $i++){
>> $myrow=mysql_fetch_array($result);
>> $product_type=mysql_result($result,$i,"product_type_detail");
>> $product_type_id=mysql_result($result,$i,"product_type_id");
>> echo "<option value=\"$product_type_id\">$product_type</option><br>";
>> }
>> ?>
>> </select>
>> </center>
>> </td></tr>
>> </table>
>> <p>
>> <p>
>> <p>
>> <hr><center>email: status1@zae.att.ne.jp<p>
>> Telephone: 03-5209-1777<p>
>> Fax: 03-5209-2539
>> </center>
>> <hr>
>> </body>
>> </html>
>>
>
>
>
Hi All,
I'll try this differently(I hope). With the below code, variables are
named/set twice ($db, $num, $result, $myrow) etc. In ASP, from what I
remember, that would be a no no. Currently, this code only populates the
1st select box. Coupla Qs. Do I have to connect to the db again, in the
second block of code..? Should the variables in the 2nd block of code,
be named differently to the 1st block.? In ASP the query variables would
be named uniquely, identifying to each query like so, prod_query_num,
prod_query_result and so forth. I must admit, my recollection of ASP is
very vague. Anyway, I hope this explains what I'm looking to understand.
Cheers.
Mark Sargent.
<html>
<body>
<hr>
<h1 align=center>JUMBO STATUS</h1><p>
<center>Used Hardware Specialist</center>
<hr>
<h2 align=center>Admin</h2>
<table align="center" border="2">
<tr>
<td><h3 align=center>PRODUCT TYPE</h3></td>
</tr>
<tr><td><center>
<select name="poduct_type">
<?php
$db = mysql_connect("localhost", "root", "grunger");
mysql_select_db("status",$db);
$result = mysql_query("SELECT ProductTypes.product_type_detail,
ProductTypes.product_type_id FROM ProductTypes",$db);
$num = mysql_num_rows($result);
for ($i=0; $i<$num; $i++){
$myrow=mysql_fetch_array($result);
$product_type=mysql_result($result,$i,"product_type_detail");
$product_type_id=mysql_result($result,$i,"product_type_id");
$maker=mysql_result($result,$i,"maker_detail");
$maker_id=mysql_result($result,$i,"maker_id");
echo "<option value=\"$product_type_id\">$product_type</option><br>";
}
?>
</select>
</center>
</td></tr>
</table>
<table align="center" border="2">
<tr>
<td><h3 align=center>Maker</h3></td>
</tr>
<tr><td><center>
<select name="slect_maker">
<?php
$db = mysql_connect("localhost", "root", "grunger");
mysql_select_db("status",$db);
$result = mysql_query("SELECT Makers.maker_id Makers.maker_detail FROM
Makers",$db);
$num = mysql_num_rows($result);
for ($i=0; $i<$num; $i++){
$myrow=mysql_fetch_array($result);
$maker=mysql_result($result,$i,"maker_detail");
$maker_id=mysql_result($result,$i,"maker_id");
echo "<option value=\"$maker_id\">$maker</option><br>";
}
?>
</select>
</center>
</td></tr>
</table>
<p>
<p>
<p>
<hr><center>email: ???????????<p>
Telephone:????????<p>
Fax: ?????????
</center>
<hr>
</body>
</html>
[Back to original message]
|