Re: Inner Join or 2nd Query...?

[Mark Sargent <powderkeg@xxxxxxxxxxxxxxxx>]

Richard Lynch wrote:

> On Thu, May 12, 2005 12:17 am, Mark Sargent said:
>  
>
> > Hi All,
> >
> > ok, this revised code produces the error below,
> >
> > <?php
> >    
>
> You took out the mysql_connect completely???
>
> Don't do that.
>
>  
>
> > mysql_select_db("status",$db);
> > $maker_result = mysql_query("SELECT Makers.maker_id Makers.maker_detail
> > FROM Makers",$db);
> >    
>
> if (!$maker_result) die(mysql_error());
>
>  
>
> > $maker_num = mysql_num_rows($maker_result);    //Line 40
> > ?>
> > <select name="slct_maker">
> > <?php
> > for ($i=0; $i<$maker_num; $i++){
> > $maker_myrow=mysql_fetch_array($maker_result);
> > $maker=mysql_result($maker_result,$i,"maker_detail");
> > $maker_id=mysql_result($maker_result,$i,"maker_id");
> > echo "<option value=\"$maker_id\">$maker</option><br>";
> > }
> > ?>
> > </select>
> >
> >
> > *Warning*: mysql_num_rows(): supplied argument is not a valid MySQL
> > result resource in */var/www/html/products.php* on line *40
> >
> > Cheers.
> >
> > Mark Sargent.
> > *
> >
> > --
> > PHP General Mailing List (http://www.php.net/)
> > To unsubscribe, visit: http://www.php.net/unsub.php
> >
> >
> >    
>
>
>  
Hi All,

all is well.

1st Query

<?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);
?>
<select name="slct_product_type">
<?php
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>

2nd Query

<?php
$maker_result = mysql_query("SELECT Makers.maker_id, Makers.maker_detail 
FROM Makers",$db);
if (!$maker_result) die(mysql_error());
?>
<select name="slct_maker">
<?php
$maker_num = mysql_num_rows($maker_result);
for ($i=0; $i<$maker_num; $i++){
$myrow=mysql_fetch_array($maker_result);
$maker=mysql_result($maker_result,$i,"maker_detail");
$maker_id=mysql_result($maker_result,$i,"maker_id");
echo "<option value=\"$maker_id\">$maker</option><br>";
}
?>
</select>

Thanx for everyone's patience. Cheers.

Mark Sargent.

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