Re: show_source question

You could just append the files into a string, then show the string instead. Like if(strlen($php)>0){ $php = $DOCUMENT_ROOT . $php; if (ereg("(\.php)$",$php) && !ereg("\/\.\.", $php)){ $show=file("head.php"); $show.=file("$php"); $show.=file("foot.php"); echo "<font class=\"source\">\n\n"; show_source("$show"); echo "\n\n</font>\n"; } This would give you the full source, and would need to be in addition to actually 'requiring' the files. If your show_source() function is set to take a file, you'd need to change it to take a string instead. Or save $show as a file and send the name or handle (I'm not sure which it expects) to the function. Mike "Eriol" <eriol@eriol.com> wrote in message news:018801c24257$9bf66780$f32d5b40@eriol... > I've got the following bit of code to work in showing the actual PHP file source > code, but it doesn't show include() or require() files and I'd like it to.. > > -------------- source.php -------------- > > if(strlen($php)>0){ > $php = $DOCUMENT_ROOT . $php; > if (ereg("(\.php)$",$php) && !ereg("\/\.\.", $php)){ > require("head.php"); > echo "<font class=\"source\">\n\n"; > show_source("$php"); > echo "\n\n</font>\n"; > require("foot.php"); > } > else{ > require("head.php"); > echo "<font class=\"s\">\n"; > echo " There's been an error..\n"; > echo "</font>\n"; > require("foot.php"); > } > exit; > } > > ---------------------------- > > I've checked the manual and google, but I'm not too sure the keywords I'd used > were correct as I didn't find any results.. I looked at highlight_file() but it > would seem to cause the same output as show_source() neglecting to include the > php code within head.php and foot.php.. > > Is there a function I'm missing that would somehow allow me to add the head.php > and foot.php code within the source code output page? Will I need to add > something to the head.php and foot.php files to have them included? > > I'm a bit of a php newbie, so sorry if my terminology isn't correct.. TIA.. > > Take care.. peace.. > eriol >