Query result into link

[jgsketch]

Hello,
I have been working on this code for a while and can't seem to get it to work. I am trying to add a result into a url. But I think my syntex is wrong. I can't think of how to explain it. Here is the code

<?php

$subcat = mysql_query("SELECT name, id FROM categories
WHERE parent_id = '$catid' ");

while ($subcatrow = mysql_fetch_array($subcat)) 
	{ 
	echo "<a href="http://localhost/template.php?
catid="$subcatrow["id"]"">" . $subcatrow["name"] . 
"</a><br>";
	} 
?>

If any could review my code, that would be great. My problem is located in the echo statement. Thanks for the help.

Jeremy

[Weedpacket]

Now you look at it, the reason it's not working is there in Technicolor®

<?php

$subcat = mysql_query("SELECT name, id FROM categories
WHERE parent_id = '$catid' ");

while ($subcatrow = mysql_fetch_array($subcat)) 
    { 
    echo "<a href=\"http://localhost/template.php?catid=".$subcatrow["id"]."\">" . $subcatrow["name"] . "</a><br>";
    } 
?>

With a judicious use of different kinds of quotes we can be a wee bit tidier:

<?php
$subcat = mysql_query("SELECT name, id FROM categories
WHERE parent_id = '$catid' ");

while ($subcatrow = mysql_fetch_array($subcat)) 
    { 
    echo '<a href="http://localhost/template.php?catid='.$subcatrow['id'].'">'.$subcatrow['name'].'</a><br>';
    } 
?>

Or:

<?php
$subcat = mysql_query("SELECT name, id FROM categories
WHERE parent_id = '$catid' ");

while ($subcatrow = mysql_fetch_array($subcat)) 
    { 
    ?><a href="http://localhost/template.php?catid=<?php echo $subcatrow['id']?>"><?php echo $subcatrow['name']?></a><br><?php
    } 
?>

Or:

<?php
$subcat = mysql_query("SELECT name, id FROM categories
WHERE parent_id = '$catid' ");

while ($subcatrow = mysql_fetch_array($subcat)) 
    { 
    echo "<a href=\"http://localhost/template.php?catid=$subcatrow[id]\">$subcatrow[name]</a><br>";
    } 
?>

Whichever is easiest for you.

[jgsketch]

Thanks for the help. I end up looking so long at the syntex, that I miss the obvious.

Jeremy