Passing information to variables within sseveral quotations

[jensendw]

I have tried everything i can think of, read everything on php syntax i can find and still cant make this work, perhaps someone here will know a better way the problem is displayed below what im trying to do is take the html from $image and insert it into a database (that is not the problem) the $manuf variable which is supposed to be a number whihc is a previously assigned one dosnt show up it simply shows up as $manuf in the html source anyways heres the code.

$image =('<CENTER><IMG SRC="http://www.deceptiononline.com/~ameister/$manuf.jpg" WIDTH="100" HEIGHT="74.6666666666667" BORDER="0"></CENTER></A></TD>');

i have already tried surrounding the $manuf with {} dosnt work also tried "" still dostn work, im sure theres a simple way i just cant find it... Thanks if anyone can help with this...

[jesus_hairdo]

try

$image ="<CENTER><IMG SRC=\"http://www.deceptiononline.com/~ameister/".$manuf.".jpg\" WIDTH=\"100\" HEIGHT=\"74.6666666666667\" BORDER=\"0\"></CENTER></A></TD>"; 

See if thats any better.

The problem was that you were using a single quote around the whole string. Single quotes DONT parse a variable so

$variable = "image";
$image = '$variable';
print $image;

will ouput

$image

where as

$variable = "image";
$image = "$variable";
print $image;

will output

image