query within a results table field

OwenTheSamoan

I'm creating a site where a search form will produce a table of other members of my site. This table has different fields which tells a little about the member
Example: Age, Location, Gender, etc....

One of the fields will contain a camera icon if the member has a photo of themself in their profile. The below code is what I am using for this. The code works perfectly. The problem is when I try to place the code within the table field I continually get parse errors.

<?
$sql_username_check=mysql_query("select username from images where username ='$username'");

$username_check=mysql_num_rows($sql_username_check);

if($username_check>0){
echo "<img border=0 src=http://mysite.com/images/picture.jpg width=29 height=25>";
}else{
echo "";}
?>

Below is an example of how I am setting up my table results. All the other results work fine. Can anyone show me how to properly insert the above code to work in a field within the table? Thanks.

<td>".$result['state']."</td><td>".$result['gender'].
"</td><td >".$result['age']."</td</tr>\n";

stolzyboy

instead of writing everything with echo just hop in and out of php, you will create less parse errors that way for example:

instead of:

<?
echo "<tr><td width=100>" . $var . "</td></tr>"
?>

use

<?
some other php stuff
?>
<tr><td width="100"><? echo $var; ?></td></tr>
<?
end of other php stuff
?>

and by the way owen i have been really busy to start on that image thing for you, haven't forgotten though, will start as soon as possible

OwenTheSamoan

Hey, how's it going? I was just about to send you a personal email saying that I found a solution to that other problem last night but I see you're on line here now. I do appreciate you wanting to help.
I see what you're saying about the echo with this other problem. I wasn't getting parse problems with the other fields, just the field with the script that would produce the icon. So are you saying I should set a $var to = the script and then just echo the script in the field?

stolzyboy

ok i won't do that image thing then, fine, lol, j/k, but here is your code you supplied and rewritten the way i would write it

<? 
$sql_username_check=mysql_query("select username from images where username ='$username'"); 

$username_check=mysql_num_rows($sql_username_check
); 

if($username_check>0){ 
?>
<img border="0" src="http://mysite.com/images/picture.jpg" width="29" height="25">
<?
}
else
{ 
}
?>

then your other code

<?
php code if desired
?>
<tr>
<td><? echo $result['state']; ?></td>
<td><? echo $result['gender']; ?></td>
<td><? echo $result['age']; ?></td>
</tr>
<?
more php code if desired
?>

my way it is easier to do your html formatting and not worry about parsing problems with your php and it isn't any slower/faster than staying in php

OwenTheSamoan

wouldn't work. I have other scripts that are a part of the table where I alternate the rows of results with color and a pagination script. So I couldn't seperate the php tags like you suggested. Any other ideas? Also for username shouldn't I be using

$result['username'] instead of $username as a part of the code?

stolzyboy

if username is pulled from a database and you haven't set it equal to $username you would have to use $result['username'], if it is coming from a form you can use $username, and as far as the other problems i would have to see the whole part(with your other script you say you can't hop out of and the part where you are receiving errors) you are talking about to see if i can figure out something else

OwenTheSamoan

okay, I figured out how to do it. Here it is in case you are curious.

echo "<td><a href=http://mysite.com/page.php?username=".$result['username'].">".$result['username']."</a></td><td align=center>".$result['sex'].
"</td><td align=center>".$result['state']."</td><td align=center>".$result['religion].
"</td><td align=center>".$result['age']."</td>";

$username = $result['username'];

$sql_username_check=mysql_query("select username from photo_table where username ='$username '");

$username _check=mysql_num_rows($sql_username_check);

if($username _check>0){
echo "<td align=center><img border=0 src=http://mysite.com/images/cameraicon.jpg width=29 height=25></td></tr>\n";
}else{
echo "<td align=center></td>"No Photo"</tr>\n"; }

}
echo"</table>\n";