I would like to modify input parameters of an SH script (it begins with #!/bin/sh). I found some solutions but they don't work here and need bash. They give me bad substitution error. so I look for a solution that works in SH (or whatever it is called)
The bash_params could be like "_learn _vil=bar _meet=foo". Here "_learn" acts as a flag. I want to set some variabels based on this flag and then remove it so that I can set other variables with eval.
Also, if you know better approaches please let me know
case $bash_params in
*"_learn"*) # learn is enabled
_learn_sp=True
tt="_learn"
bash_params="${bash_params%"$tt"}" # it doesn't work
bash_params="${bash_params/_learn//}" # this gives Bad substitution error
_lsp=False
;;
eval ${bash_params}
Pipe to sed.
bash_params=$(echo "$bash_params" | sed "s/$tt//")
Related
I am trying to do string substitution in bash, want to understand it better.
I crafted a success case like this:
a=abc_de_f
var=$a
echo ${var//_/-}
outout is abc-de-f. This works.
However, the following script fails:
a=abc_de_f
echo ${$a//_/-}
The error message is ${$a//_/-}: bad substitution.
It seems like related to how we can use a variable in substitution. Why this fails? How bash handles variables in this case?
Also, what is the best practice to handle escape characters in bash string substitution?
In the second case, you don't need the second $ as a is the string.
a=abc_de_f
echo ${a//_/-}
If you wanted to add a level of indirection, you can use ! before the variable as in
a=abc_de_f
b=a
echo ${b//_/-}
will output a, while
echo ${!b//_/-}
will output abc-de-f.
See here for a discussion on the art of escaping in BASH
In bash from the CLI I can do:
$ ERR_TYPE=$"OVERLOAD"
$ echo $ERR_TYPE
OVERLOAD
$ read ${ERR_TYPE}_ERROR
1234
$ echo $OVERLOAD_ERROR
1234
This works great to set my variable name dynamically; in a script it doesn't work. I tried:
#!/bin/env bash
ERR_TYPE=("${ERR_TYPE[#]}" "OVERLOAD" "PANIC" "FATAL")
for i in "${ERR_TYPE[#]}"
do
sh -c $(echo ${i}_ERROR=$"1234")
done
echo $OVERLOAD_ERROR # output is blank
# I also tried these:
# ${i}_ERROR=$(echo ${i}_ERROR=$"1234") # command not found
# read ${i}_ERROR=$(echo ${i}_ERROR=$"1234") # it never terminates
How would I set a variable as I do from CLI, but in a script? thanks
When you use dynamic variables names instead of associative arrays, you really need to question your approach.
err_type=("OVERLOAD" "PANIC" "FATAL")
declare -A error
for type in "${err_type[#]}"; do
error[$type]=1234
done
Nevertheless, in bash you'd use declare:
declare "${i}_error=1234"
Your approach fails because you spawn a new shell, passing the command OVERLOAD_ERROR=1234, and then the shell exits. Your current shell is not affected at all.
Get out of the habit of using ALLCAPSVARNAMES. One day you'll write PATH=... and then wonder why your script is broken.
If the variable will hold a number, you can use let.
#!/bin/bash
ERR_TYPE=("OVERLOAD" "PANIC" "FATAL")
j=0
for i in "${ERR_TYPE[#]}"
do
let ${i}_ERROR=1000+j++
done
echo $OVERLOAD_ERROR
echo $PANIC_ERROR
echo $FATAL_ERROR
This outputs:
1000
1001
1002
I'd use eval.
I think this would be considered bad practice though (it had some thing to do with the fact that eval is "evil" because it allows bad input or something):
eval "${i}_ERROR=1234"
I'am new in Linux and I want to write a bash script that can read in a file name of a directory that starts with LED + some numbers.(Ex.: LED5.5.002)
In that directory there is only one file that will starts with LED. The problem is that this file will every time be updated, so the next time it will be for example LED6.5.012 and counting.
I searched and tried a little bit and came to this solution:
export fspec=/home/led/LED*
LedV=`basename $fspec`
echo $LedV
If I give in those commands one by one in my terminal it works fine, LedV= LED5.5.002 but if i run it in a bash scripts it gives the result: LedV = LED*
I search after another solution:
a=/home/led/LED*
LedV=$(basename $a)
echo $LedV
but here again the same, if i give it in one by one it's ok but in a script: LedV = LED*.
It's probably something small but because of my lack of knowledge over Linux I cannot find it. So can someone tell what is wrong?
Thanks! Jan
Shell expansions don't happen on scalar assignments, so in
varname=foo*
the expansion of "$varname" will literally be "foo*". It's more confusing when you consider that echo $varname (or in your case basename $varname; either way without the double quotes) will cause the expansion itself to be treated as a glob, so you may well think the variable contains all those filenames.
Array expansions are another story. You might just want
fspec=( /path/LED* )
echo "${fspec[0]##*/}" # A parameter expansion to strip off the dirname
That will work fine for bash. Since POSIX sh doesn't have arrays like this, I like to give an alternative approach:
for fspec in /path/LED*; do
break
done
echo "${fspec##*/}"
pwd
/usr/local/src
ls -1 /usr/local/src/mysql*
/usr/local/src/mysql-cluster-gpl-7.3.4-linux-glibc2.5-x86_64.tar.gz
/usr/local/src/mysql-dump_test_all_dbs.sql
if you only have 1 file, you will only get 1 result
MyFile=`ls -1 /home/led/LED*`
I am experimenting with wildcards in bash and tried to list all the files that start with "xyz" but does not end with ".TXT" but getting incorrect results.
Here is the command that I tried:
$ ls -l xyz*[!\.TXT]
It is not listing the files with names "xyz" and "xyzTXT" that I have in my directory. However, it lists "xyz1", "xyz123".
It seems like adding [!\.TXT] after "xyz*" made the shell look for something that start with "xyz" and has at least one character after it.
Any ideas why it is happening and how to correct this command? I know it can be achieved using other commands but I am especially interested in knowing why it is failing and if it can done just using wildcards.
These commands will do what you want
shopt -s extglob
ls -l xyz!(*.TXT)
shopt -u extglob
The reason why your command doesn't work is beacause xyz*[!\.TXT] which is equivalent to xyz*[!\.TX] means xyz followed by any sequence of character (*) and finally a character in set {!,\,.,T,X} so matches 'xyzwhateveryouwant!' 'xyzwhateveryouwant\' 'xyzwhateveryouwant.' 'xyzwhateveryouwantT' 'xyzwhateveryouwantX'
EDIT: where whateveryouwant does not contain any of !\.TX
I don't think this is doable with only wildcards.
Your command isn't working because it means:
Match everything that has xyz followed by whatever you want and it must not end with sequent character: \, .,T and X. The second T doesn't count as far as what you have inside [] is read as a family of character and not as a string as you thought.
You don't either need to 'escape' . as long as it has no special meaning inside a wildcard.
At least, this is my knowledge of wildcards.
The colon command is a null command.
The : construct is also useful in the conditional setting of variables. For example,
: ${var:=value}
Without the :, the shell would try to evaluate $var as a command. <=???
I don't quite understand the last sentence in above statement. Can anyone give me some details?
Thank you
Try
var=badcommand
$var
you will get
bash: badcommand: command not found
Try
var=
${var:=badcommand}
and you will get the same.
The shell (e.g. bash) always tries to run the first word on each command line as a command, even after doing variable expansion.
The only exception to this is
var=value
which the shell treats specially.
The trick in the example you provide is that ${var:=value} works anywhere on a command line, e.g.
# set newvar to somevalue if it isn't already set
echo ${newvar:=somevalue}
# show that newvar has been set by the above command
echo $newvar
But we don't really even want to echo the value, so we want something better than
echo ${newvar:=somevalue}.
The : command lets us do the assignment without any other action.
I suppose what the man page writers meant was
: ${var:=value}
Can be used as a short cut instead of say
if [ -z "$var" ]; then
var=value
fi
${var} on its own executes the command stored in $var. Adding substitution parameters does not change this, so you use : to neutralize this.
Try this:
$ help :
:: :
Null command.
No effect; the command does nothing.
Exit Status:
Always succeeds.