You need to understand how "variable expansion" works: lets assume we have a variable assigned (i suggest you try the examples yourself at the shell prompt and play around a bit with them to get familiar):
The point of shell variables is that you cannot use them directly, like in other programming languages:
This (or similar constructs) would work in other languages, but in the shell you use the variable "indirectly", once you have assigned it, through a set of quasi-functions. This is "variable expansion". The most basic expansion is:
which will expand to the content of the variable "var". This content is replaced at the command line and then the command line is executed. For instance:
Keep in mind this mechanism when we discuss more complicated expansions. The next in the list are these:
The first one ("#") takes the content of the variable, then takes the regexp, expands that and if it matches the beginning of the content, the matching part is cut off. Sounds complicated? OK, here is an example with our variable from above:
The complete content would be "abc/def/ghi". The regexp ("?") means "any one single character", which is deleted from the beginning, therefore leaving the first character out. Notice, that this DOES NOT CHANGE the variable at all:
The opposite of "#" is "%", which works the same, but takes away from the content at the end instead of the beginning. Also notice the "*", which means "any number of any characters". In case you wonder: yes, these are the same characters you can use as filemasks when issuing a "ls -l <mask>". Whatever you can use there you can use here:
You might wonder what the difference between "#" and "##" and "%" and "%%" respectively is. Try out the following and notice the difference:
The one is always the shortest possible match the other the longest possible match. For matches which only occur once there is no difference.
There are a lot of other interesting and powerful expansions: you can replace one substring with another:
and a lot of other things. Check out "man ksh" for reference.
I'd like to show you another trick: nested expansions. You can use an expansion inside another expansion, even if it uses the same variable (because - you know already - the variables content itself is not changed!). This, finally, will do what you look for:
Remember what this gives:
Correct: everything save for the last character. Now, let us use this as the regexp we want to take away from the beginning of the content. Obviously this will match most of the content and only leave the last character, yes?
And because this will always be true you can use this regexp every time, regardless of what the content of "var" is, for the last characters - or more characters, if you modify it a bit:
I hope this helps.
bakunin
These 3 Users Gave Thanks to bakunin For This Post:
Hi All,
We are in the process of Migrating from AIX 4 to Solaris 10 and getting a Few Errors.
I have been programming in shell but could never establish muself as an expert, hence please need you help.
I am Getting Bad Substitution error in my script, I have isolated the issue and its... (6 Replies)
hi,
i created a shell script having the following content:
#! /usr/bin/ksh
FROM="myemail@domain.com"
MAILTO="someemail@domain"
SUBJECT="TEST"
BODY="/export/home/adshocker/body.txt"
ATTACH="/export/home/adshocker/attach.prog"
echo $ATTACH
ATTACH_NAME="${ATTACH##*/}"
echo $ATTACH_NAME... (5 Replies)
Hello,
In bash I can use the following:
TMP=12345
MID=${TMP:1:1}
the expected result is: 2
but when using KSH I'm getting a ''bad substitution" error.
What is the correct syntaxin ksh?
Thanks (2 Replies)
#!/bin/bash
a1=( win 12,01,02,03,04 )
a2=( pre 04,05,06 )
a3=( msn 06,07,08,09 )
Given the above arrays, I want the script to return/echo the following in a loop;
win
12,01,02,03,04
pre
04,05,06,07
msn
06,07,08,09
But I can't get it to do as such.
I've tried; (2 Replies)
Hi Gurus,
I am working with a korn shell script. I should replace in a very great file the character ";" with a space.
Example:
2750;~
2734;~
2778;~
2751;~
2751;~
2752;~
what the fastest method is? Sed? Awk?
Speed is dead main point, Seen the dimensions of the files
Thanks (6 Replies)
Hi All,
I'm building a new shell script but i'm facing a problem with one line which is giving "bad substitution" error. Please assist
script lines:
#!/bin/sh
printf "%s: " "Occurrence DATE (YYYYMMDD)"; read DATE
shortdate=${DATE#??}
o/p:
./test1: bad substitution
This command is... (2 Replies)
Why I get bad replace when using eval?
$ map0=( "0" "0000" "0")
$ i=0
$ eval echo \${map$i}
0000
$ a=`eval echo \${map$i}` !!!error happens!!!
bash: ${map$i}: bad substitution
How to resolve it ?
Thanks! (5 Replies)
I have script data.sh which has following error.
Script Name : data.sh
#!/bin/sh
infile=$1
len=${#infile}
echo $len
texfile=${infile:0:$len-4}
echo $texfile
run command
./data.sh acb.xml
I get following error message: (5 Replies)
Hi I'm using ksh.
And i'm trying to get the substring like below.
but giving the following error
#!/bin/ksh
foo=teststring
bar=${foo:0:5}
echo $bar
And the error is
./sbstr_test.sh: bar=${foo:0:5}: bad substitution
what is wrong in this script. Please correct me
... (3 Replies)
Cant undestand :) why i have an error on line 2.it is working on my other boxes
#!/bin/bash
ret=$(echo Q | timeout 5 openssl s_client connect "${1`hostname`}:${2-443}" -ssl3 2> /dev/null)
if echo "${ret}" | grep -q 'Protocol.*SSLv3'; then
if echo "${ret}" | grep -q 'Cipher.*0000'; then
... (7 Replies)