else... I can't figure it out.

Anon

I just posted this on another board. Bad form perhaps. I'm wearing a permanent groove between me an my hosting company doing a lot trial and failure stuff.

I'm loosing it.

With some help, I've been able to get the first part of this to work. But I need to build an else factor.

How do I build in an else statement?

$file="/path and file name";
$fd = fopen ($file, "r");
$contents = file($file);
$number = count($contents);
for($i=0;$i<$number;$i++) { if(strstr($contents[$i],$REMOTE_USER)) print ("Instructional Greeting One"); }

Background on usage:
User accesses .htpasswd protected page. User name is checked against an external file. If found... Instructional Greeting One. If not found... Instructional Greeting Two.

Thanks,
averick

ace21

http://www.phpbuilder.com/manual/control-structures.php

Ace21

[deleted]

Learn to use indenting!

Anon

Thanks vincent,
I'll try to indent it. Maybe that'll make it work.

Either way... thanks for the input. I appreciate you taking the time help a clueless newbie like myself.

averick

[deleted]

Indenting won't "make it work" but it will make it a lot clearer so you can see where the else would go.

Putting everything on one line just confuses you to the point where you don't understand what's going on anymore.

Anon

Given. And point taken. But the problem is that I don't understand how I'm supposed to introduce an else into this thing. I've got the manual and I'll keep reading.

As near as I can understand it, something like this should work.

$file="/path and file name";
$fd = fopen ($file, "r");$contents = file($file);
$number = count($contents);
for($i=0;$i<$number;$i++) {
if(strstr($contents[$i],$REMOTE_USER))

{print ("Instructional Greeting ONE");}
else
{print ("Instructional Greeting TWO");}
}

But what I get is:
If REMOTE_USER name is found in external file, eight instances of Greeting ONE are printed, with one instance of Greeting Two in the middle.

If REMOTE_USER name isn't found in the external file, nine instances of Greeting TWO are printed.

I've tried every alteration I can think of. Everything else just gives a parse error.

Anon

Perhaps (if there's so few lines in the file) you could show us
the contents of it plus one of the "matching" REMOTE_USER
values.

Anon

Thanks Kirk,
I got back here as fast as I could.

Thanks to a few other kind folks, this problem has been solved and delivered to me in whole.

User accesses .htpasswd protected page. PHP checks external, unrelated file for an instance of the REMOTE_USER name. If found, it displays instructional greeting 1. If not found, it prints instructional greeting 2.

Solution below for others to benefit from.

$file="/path and file name";
$f = 0;
$contents = file($file);
foreach($contents as $key => $val) {
if(strstr($val, $REMOTE_USER)) {
$f = 1;
}
}
if($f) {
print "Instructional Greeting ONE";
}
else {
print "Instructional Greeting TWO";
}

Thanks go to several different folks. Delivery of final, complete solution was by freddydoesphp -- phpworld message board.

Anon

Oh, duh! Of course: you were looping throught the contents of the file, and printing the comparison at each time throught.

I might suggest you enhance your solution by breaking out of the loop as soon as you find a match. No need to keep on looking if you know the answer is Yes.