Couple of Shell Script Questions

mathias1979 · 06-16-2005, 09:40 AM

Two quick questions about shell, or c-shell scripting. Is there a command that will read the names of all files in a particular directory into an array so that I may loop through those files with my script without having to manually enter their names? Also, how would I go about reading just a portion of a character string into a new variable. For instance, if I had the character string "file02" stored in a variable...how could I read just the last two characters ('02') into a new variable?

-Matt

JediGuy_bob · 06-16-2005, 10:41 AM

You can always use a bash script using ls and cut to get the names and have perl or python take that and store it as an array.

perfect_circle · 06-16-2005, 11:10 AM

Quote:

Originally posted by mathias1979
Two quick questions about shell, or c-shell scripting. Is there a command that will read the names of all files in a particular directory into an array so that I may loop through those files with my script without having to manually enter their names? Also, how would I go about reading just a portion of a character string into a new variable. For instance, if I had the character string "file02" stored in a variable...how could I read just the last two characters ('02') into a new variable?

-Matt

I only know bash (sh in most cases is a link to bash)

Question one:

Code:

for name in *; do ......done

be carefull and protect $name variable with " " because the filenames may contain spaces

question two:

first approach: (cutting off a part from the begging of a string that fits in a pattern)

Code:

skalkoto@darkstar:~$ var="file02"
skalkoto@darkstar:~$ var1=${var#file}
skalkoto@darkstar:~$ echo $var1
02
skalkoto@darkstar:~$

second approach: (using the string position)

Code:

skalkoto@darkstar:~$ var="file02"
skalkoto@darkstar:~$ var2=${var:4}
skalkoto@darkstar:~$ echo $var2
02
skalkoto@darkstar:~$

mathias1979 · 06-16-2005, 03:22 PM

Quote:

Originally posted by perfect_circle

Code:

for name in *; do ......done

be carefull and protect $name variable with " " because the filenames may contain spaces

second approach: (using the string position)

Code:

skalkoto@darkstar:~$ var="file02" skalkoto@darkstar:~$ var2=${var:4} skalkoto@darkstar:~$ echo $var2 02 skalkoto@darkstar:~$

Both worked like charm. Thanks!

-Matt