I Have been going crazy over this simple problem...I just can't figure it out!!

Somone wanna take a shot at it?

i have a parse error, unexpected T_STRING in line 129 :


$fontString http://<input type='text' width='15' name='image' value=\"$image\"> $fontString<a href='http://$image' target='blank'>Click To Preview</a>



can anyone tell me whats wrong in there?

Sorry I'm a bit confused with that snippet of code :P
Can you post the surrounding code as well?
I don't understand why there is $fontString http:// at the beginning

im trying to make this in my site man look this pic:
http://i61.photobucket.com/albums/h44/Vanhell06/MIo.jpg

here the whole code:

<tr><td width=50% bgcolor=$tableColor>

$fontString

Personal Image:

</td><td width=50% bgcolor=$tableColor>

$fontString http://<input type='text' width='15' name='image' value=\"$image\"> $fontString<a href='http://$image' target='blank'>Click To Preview</a>

</td></tr>
<tr><td width=50% bgcolor=$tableColor>

$fontString

Sizes:

</td><td width=50% bgcolor=$tableColor>
$fontString
Width:<input type='text' width='5' name='width' value=\"$width\" size='3'> Height:<input type='text' width='5' name='height' value=\"$height\" size='3'></td></tr>
<tr><td bgcolor=$tableColor colspan=2>
<tr><td width=50% bgcolor=$tableColor>

$fontString

Personal Image Preview:

</td><td width=50% bgcolor=$tableColor>");
if($image == NULL) echo("
$fontString Image Preview cannot Be shown beause you have no link currently typed in.");
else echo("
<img src='http://$image' width='$width' height='$height'></img>");
echo("
</td></tr><br>

thanks for helping me dude.

unexpected T_STRING generally means that it's found text that's not normal code outside of " " quotes or ()'s. The code you've shown us I can't see anything wrong with it. But not having the entire code in front of me, it's hard to tell.
What I will ask though, I presume the line in error is the one under "Personal Image" in your example. Are you echoing this, or outputting by closing and re-opening <?php ?> tags? Might be worth just checking all your quoting, and that you've escaped everything you need to around that area.
It's also worth noting, and that to minimize the risk of getting these sorts of errors when you're outputting a lot of HTML in a PHP script to output all your HTML outside <?php ?> tags, and use <?=$var_name?> to put in variable names. This method allows you to use as many quotes, brackets, and the likes, without it upsetting PHP. Using echo(".."); might be good for outputting just things like single variables, words, etc, but for large amount of content, try close the PHP tag, put your content in, and open the tag again to close your if statement or whatever you have.

Hope this helps.