see what's wrong here:
ereg('(^[0-1231]$).jpg$',$file_name)
[] meens a group of characters, so in your case
0,1,2 and 3 are valid characters. you haven't defined any
modifer like ?,*,+ or{}, so one of this characters has to
be found exactly one time. you're using ^ outside the [] so it meens the
beginning of the string.
in your case none of these characters inside the [] can be found at the
beginning of the string.
then you use $ after the []. $ meens the end of the string. none of the
characters in the [] matches at the end of the string.
so this would be right:
ereg('_[0-9]{4}\.jpg$', $file_name);
so this meens:
the beginning of the string doesn't matter, because we have not specified ^
at the beginning.
there has to be an underscore, followed by 4 characters between 0 and 9,
followed by an dot,
followed by j, followd by p, followed by g. g has to be at the end of the
string, because of the $.
or you can use:
ereg('^\.*_[0-9]{4}\.jpg$', $file_name);
this will meen :
any characters at the beginning between 0 and unlimited times, then followed
by an underscore,
followed by 4 characters between 0 and 9, followed by a dot, followed by
jpg. same as above
though. But the * is a real performance eater so it could be slightly faster
if you're using the first example.
"Jas" <[EMAIL PROTECTED]> schrieb im Newsbeitrag
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> I hate to say it but that didn't work, I have been trying different
> variations of the same ereg('_(^90-9{4}$).jpg$',$file_names) and nothing
> seems to work for me, I have also been looking at the ereg and preg_ereg
> functions but they don't seem to make sense to me, here is the code as a
> whole if this helps:
> // query directory and place results in select box
> $dir_name = "/path/to/images/directory/on/server/"; // path to directory
on
> server
> $dir = opendir($dir_name); // open the directory in question
> $file_lost .= "<p><FORM METHOD=\"post\" ACTION=\"done.php3\"
NAME=\"ad01\">
> <SELECT NAME=\"image_path\">";
> while ($file_names = readdir($dir)) {
> if ($file_names != "." && $file_names !=".." &&
ereg('_(^[0-9]{4}.jpg$)',
> $file_names)) // filter my contents
> {
> $file_lost .= "<OPTION VALUE=\"$file_names\"
> NAME=\"$file_names\">$file_names</OPTION>";
> }
> }
> $file_lost .= "</SELECT><br><br><INPUT TYPE=\"submit\" NAME=\"submit\"
> VALUE=\"select\"></FORM></p>";
> closedir($dir);
> What I am trying to accomplish is to list the contents of a directory in
> select box but I want to filter out any files that dont meet this criteria
> *_4444.jpg and nothing is working for me, any help or good tutorials on
> strings would be great.
> Jas
> "Erik Price" <[EMAIL PROTECTED]> wrote in message
> [EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> >
> > On Thursday, April 11, 2002, at 05:59 AM, jas wrote:
> >
> > > Is this a correct string to show only files that look like so:
> > > *_2222.jpg
> > > if ($file_names != "." && $file_names !=".." &&
> > > ereg('(^[0-1231]$).jpg$',$file_name))
> > > Any help would be great.
> >
> > preg_match(/^_[0-9]{4,4}\.jpg$/, $file_name) should match any string
> > that starts with an underscore, is followed by exactly four digits, and
> > then a ".jpg". It will not match anything but this exact string.
> >
> >
> > Erik
> >
> >
> >
> >
> > ----
> >
> > Erik Price
> > Web Developer Temp
> > Media Lab, H.H. Brown
> > [EMAIL PROTECTED]
> >
>
>
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