quick stringinstr question

[diveboy]

CODE

$filename = "Dexter-s02-e01er-s01-e02yr.avi"

$dot = StringInStr($filename,".",-1)

$fileext = Stringright($filename,($dot-1))

I'm trying to get the ".avi" from the $filename. the StringInStr function is returning 27 which is correct from the left, but the -1 should have it going from the right, which should give 5 correct ?

thanks

Michael.

[SmOke_N]

CODE

$filename = "Dexter-s02-e01er-s01-e02yr.avi"

$dot = StringInStr($filename,".",-1)

$fileext = Stringright($filename,($dot-1))

I'm trying to get the ".avi" from the $filename. the StringInStr function is returning 27 which is correct from the left, but the -1 should have it going from the right, which should give 5 correct ?

thanks

Michael.

You have -1 in the case sensitivity parameter.

$dot = StringInStr($filename, ".", 0, -1)

[Jos]

CODE

$filename = "Dexter-s02-e01er-s01-e02yr.avi"

$dot = StringInStr($filename,".",-1)

$fileext = Stringright($filename,($dot-1))

I'm trying to get the ".avi" from the $filename. the StringInStr function is returning 27 which is correct from the left, but the -1 should have it going from the right, which should give 5 correct ?

thanks

Michael.

Nope... -1 (when used in the correct parameter position) means "find the first occurrence from the right" but still will returns the character position counting from the left.

Edited January 20, 2007 by JdeB

[diveboy]

crap.

why didn't I see that.

Thanks a million.