Random Image - Hide Name

[NoFear]

Ok,
for the login of my page, I want to include a random image that the user must type in the code displayed in that image to enter the login. However, I want the name of that image to be hidden from the user, so as to not allow a stubborn person from collecting all the names of the images and their values over the course of a few hours so that they can program their bot anyway. How do I make a random image's name hidden from the source code?

[altexis]

just an idea....
copy the original image file to a temporary directory and give it a random name... use the copy of this image for your page and delete it later 🙂

[suntra]

Originally posted by altexis
just an idea....
copy the original image file to a temporary directory and give it a random name... use the copy of this image for your page and delete it later 🙂

Don't forget to associate the random name with the word inside the image... !

[NoFear]

Originally posted by altexis
just an idea....
copy the original image file to a temporary directory and give it a random name... use the copy of this image for your page and delete it later 🙂

I am doing this. However, this brings up another problem.

If $randomimage is being accessed by one person, and somebody else accesses the site, $randomimage is overwritten, and if the entire image hasn't been delivered, the first person gets a messed up image.

[ereptur]

hmmm... try this, copy the image like this
$time = time();
copy ($original,$time.jpg);
print "<input type="hidden" value="$time.jpg" />";

After you validate the text being entered, use an unlink to destroy the image. it can never to be used again since time is a pretty unique assignment.

[NoFear]

Thanks for the idea. Though I'm not quite sure how it works, I'll look into it. Here's a (very small) snipet of what I've done so far:

copy ($pictures[0], "random".$randomimagenumber.".jpg");
echo "<img src = \"random".$randomimagenumber.".jpg\">";

what this does is copies one of the pictures from an array of pictures, and names it "random#.jpg". While the rest isn't really my concern, there is one part that is.....

It seems that when $picture[0] is the same picture as the "random.jpg" that it's trying to copy to, I get this error message:

Warning: copy(babe.jpg) [function.copy]: failed to create stream: No such file or directory in C:\Program Files\Apache Group\Apache2\htdocs\Test Stuff\randomimage.php on line 82

I don't get this error as long as random#.jpg IS NOT THE SAME PICTURE as $picture[0].

How do I get rid of this error message? Why is this error message coming in the first place?

[altexis]

Seems like the original file is not at the place you request it from. Check for that... also check that the denstination directory has the necessary write privileges

[NoFear]

I know it has write privledges, and I know the files are there, otherwise it would never work to begin with. After thinking about it more, I think that it's not that the image that it's simply displaying the image that was previously there since the copy failed, but my question then is why the copy failed, and how I can prevent this?

[ereptur]

What I do is unlink the file first.

For example:

@unlink("random.$randomnumber.jpg");

then:

copy($picture[0],"random.$randomnumber.jpg");

See if you still get the error message then!

[NoFear]

Originally posted by ereptur
What I do is unlink the file first.

For example:

@unlink("random.$randomnumber.jpg");

then:

copy($picture[0],"random.$randomnumber.jpg");

See if you still get the error message then!

Awesome! What a great idea. So simple, yet I still didn't think of it. I'll see if this takes care of my problems as soon as I get home from work tonight.

Thanks for the idea. I'll report back on the outcome

[NoFear]

nope, didn't work 🙁

Now I'm just getting this error message, plus the other one from time to time...

Warning: unlink("random6.jpg") [function.unlink]: Invalid argument in C:\Program Files\Apache Group\Apache2\htdocs\Test Stuff\randomimage.php on line 86

I tried unlink in a separate php script and it worked perfectly, but for some reason when I use it in this code, I get this warning.

[altexis]

smash the bardard!!!! when it sees the hammer coming at it .. it will definitely work!

hhmmm.... try a few simple stuff then....
try using the functions one by one... giving only a few parameters at the beginning and go on with more and more complicated parameters...
then try to add some more of the original code around it
make lots of little tests before you completely add the whole code as it originally is

this way usually shows the source of problems

one more thing... before you copy or unlink use the file_exists() function!! I think some machines need you to do that before you copy or delete files

[NoFear]

hmm, that's odd. I'll remember that in the future though.

I figured out what my orignal error message was for. It seems that my array had a couple lines of mistakes in it... (feel free to flame me now, I deserve it, makin' you all think so hard about such a small mistake 🙁 ).

However, I still don't know what that unlink error is all about. That I know is fine, as I directly pulled that out into a new script as is, and it worked fine there (it was just the image name and not the variables for the image name, though). Very weird.

[altexis]

small mistakes are always the hardest to find
all experienced programmers have done 100's of them.. if someone "flames" you for such thing 99% he is a newbie 🙂

[NoFear]

Originally posted by altexis
small mistakes are always the hardest to find
all experienced programmers have done 100's of them.. if someone "flames" you for such thing 99% he is a newbie 🙂

😃 😃 😃
If this programmer is really "experienced", he should've done more than 100's of them! LoL

Wait... maybe that's just me... and I don't even consider myself "experienced" (at least not to what I want to be, anyway)

[altexis]

...after all... what does "experienced" means?
probably experienced in mistakes 🙂