I have some bash code that i am running in a zsh shell on macos.
Code seems to work fine when i run in the shell, but when i run the file from my path, it only brings back the initial and not the initial+surname, any help most appreciated.
user="Steve Thomas"
dbuser=$user
initial="${user%"${user#?}"}"
userie=( $( echo $dbuser | cut -d' ' -f1- ) )
userlastname=${userie[2]}
fulluser="${initial}${userlastname}"
echo $fulluser
when run in shell i get what is expected
SThomas
and when i run as file.sh from path i get..
S
Not sure what i am doing wrong here, please advise.
In bash, arrays are indexed from 0, not 1 like in zsh. So, to make the script work in bash, change line 5 to
userlastname=${userie[1]}
You can even make it universal for both the shells:
startindex=2
if [[ $BASH_VERSION ]] ; then
startindex=1
fi
...
userlastname=${userie[startindex]}
Related
Two questions about the same thing I think...
Question one:
Is it possible to have a bash script run with default parameters/options? ...in the sense if someone were to run the script:
./somescript.sh
it would actually run with ./somescript.sh | tee /tmp/build.txt?
Question two:
Would it also possible to prepend the script with defaults? For example, if you were to run the script ./somescript.sh
it would actually run
script -q -c "./somescript.sh" /tmp/build.txt | aha > /tmp/build.html?
Any help or guidance is very much appreciated.
You need a wrapper script that handles all such scenario for you.
For example, your wrapper script can take parameters that helps you decide.
./wrapper_script.sh --input /tmp/build.txt --ouput /tmp/build.html
By default, --input and --output can be set to values you want when they are empty.
You can use the builtin $# to know how many arguments you have and take action based on that. If you want to do your second part, for example, you could do something like
if [[ $# -eq 0 ]]; then
script -q -c "$0 /tmp/build.txt | aha /tmp/build.html
exit
fi
# do everything if you have at least one argument
Though this will have problems if your script/path have spaces, so you're probably better putting the real path to your script in the script command instead of $0
You can also use exec instead of running the command and exiting, but make sure you have your quotes in the right place:
if [[ $# -eq 0 ]]; then
exec script -q -c "$0 /tmp/build.txt | aha /tmp/build.html"
fi
# do everything when you have at least 1 argument
in the default shell
the for loop given below
for ((i=$llimit; i<=$ulimit; i++));
do
echo $i
done;
it throws error "'((' is not expected"
but when switching to the bash shell
the for loop works fine
is there a way to change shell inside a shellscript
or any other solution as this for loop is inside a shell script
EDIT:
this is hte shell script
#!/bin/bash
nav_var=`sqlplus -s tcs384160/tcs#1234 <<\EOF
set pagesize 0 feedback off verify off heading off echo off
select max(sequence#) from v$archived_log where applied='YES' and thread#=2 and dest_id=2;
exit;
EOF`
echo $nav_var;
ulimit=`expr $nav_var - 30`;
llimit=`expr $ulimit - 200`;
for ((i=$llimit; i<=$ulimit; i++));
do ls -l arch_aceprod_2_${i}_743034701.arc;
done;
The C-style for loop you've used is a bashism.
Change the line
for ((i=$llimit; i<=$ulimit; i++));
to
for i in $(seq $llimit $ulimit);
and it would work well with both sh and bash.
EDIT: If you don't have seq, you could change the loop as:
i=$llimit
while [ $i -le $ulimit ]; do
echo "Do something here"
let i=i+1
done
By "default shell" I assume you mean /bin/sh? Is there a line starting "#!" at the top of the script?
Bash is pretty much backwards compatible with sh. If you put "#!/bin/bash" (without the quotes) as the first line this should get the whole thing to run under bash.
try another for loop syntax
for counter in {$llimit..$ulimit}
do
your logic
done
this works for all type of shells.
Or #!bin/bash will also work in your case
I read the answer for this issue from this link
in Stackoverflow.com. But I am so new in writing shell script that I did something wrong. The following are my scripts:
testscript:
#!/bin/csh -f
pid=$(ps -opid= -C csh testscript1)
while [ -d /proc/$pid ] ; do
sleep 1
done && csh testscript2
exit
testscript1:
#!/bin/csh -f
/usr/bin/firefox
exit
testscript2:
#!/bin/csh -f
echo Done
exit
The purpose is for testscript to call testscript1 first; once testscript1 already finish (which means the firefox called in script1 is closed) testscript will call testscript2. However I got this result after running testscript:
$ csh testscript
Illegal variable name.
Please help me with this issue. Thanks ahead.
I believe this line is not CSH:
pid=$(ps -opid= -C csh testscript1)
In general in csh you define variables like this:
set pid=...
I am not sure what the $() syntax is, perhaps back ticks woudl work as a replacement:
set pid=`ps -opid= -C csh testscript1`
Perhaps you didn't notice that the scripts you found were written for bash, not csh, but
you're trying to process them with the csh interpreter.
It looks like you've misunderstood what the original code was trying to do -- it was
intended to monitor an already-existing process, by looking up its process id using the process name.
You seem to be trying to start the first process from inside the ps command. But
in that case, there's no need for you to do anything so complicated -- all you need
is:
#!/bin/csh
csh testscript1
csh testscript2
Unless you go out of your way to run one of the scripts in the background,
the second script will not run until the first script is finished.
Although this has nothing to do with your problem, csh is more oriented toward
interactive use; for script writing, it's considered a poor choice, so you might be
better off learning bash instead.
Try,
below script will check testscript1's pid, if it is not found then it will execute testscirpt2
sp=$(ps -ef | grep testscript1 | grep -v grep | awk '{print $2}')
/bin/ls -l /proc/ | grep $sp > /dev/null 2>&1 && sleep 0 || /bin/csh testscript2
I'm trying some staff that is working perfectly when I write it in the regular shell, but when I include it in a bash script file, it doesn't.
First example:
m=`date +%m`
m_1=$((m-1))
echo $m_1
This gives me the value of the last month (actual minus one), but doesn't work if its executed from a script.
Second example:
m=6
m=$m"t"
echo m
This returns "6t" in the shell (concatenates $m with "t"), but just gives me "t" when executing from a script.
I assume all these may be answered easily by an experienced Linux user, but I'm just learning as I go.
Thanks in advance.
Re-check your syntax.
Your first code snippet works either from command line, from bash and from sh since your syntax is valid sh. In my opinion you probably have typos in your script file:
~$ m=`date +%m`; m_1=$((m-1)); echo $m_1
4
~$ cat > foo.sh
m=`date +%m`; m_1=$((m-1)); echo $m_1
^C
~$ bash foo.sh
4
~$ sh foo.sh
4
The same can apply to the other snippet with corrections:
~$ m=6; m=$m"t"; echo $m
6t
~$ cat > foo.sh
m=6; m=$m"t"; echo $m
^C
~$ bash foo.sh
6t
~$ sh foo.sh
6t
Make sure the first line of your script is
#!/bin/bash
rather than
#!/bin/sh
Bash will only enable its extended features if explicitly run as bash. If run as sh, it will operate in POSIX compatibility mode.
First of all, it works fine for me in a script, and on the terminal.
Second of all, your last line, echo m will just output "m". I think you meant "$m"..
I'm converting some Windows batch files to Unix scripts using sh. I have problems because some behavior is dependent on the %~dp0 macro available in batch files.
Is there any sh equivalent to this? Any way to obtain the directory where the executing script lives?
The problem (for you) with $0 is that it is set to whatever command line was use to invoke the script, not the location of the script itself. This can make it difficult to get the full path of the directory containing the script which is what you get from %~dp0 in a Windows batch file.
For example, consider the following script, dollar.sh:
#!/bin/bash
echo $0
If you'd run it you'll get the following output:
# ./dollar.sh
./dollar.sh
# /tmp/dollar.sh
/tmp/dollar.sh
So to get the fully qualified directory name of a script I do the following:
cd `dirname $0`
SCRIPTDIR=`pwd`
cd -
This works as follows:
cd to the directory of the script, using either the relative or absolute path from the command line.
Gets the absolute path of this directory and stores it in SCRIPTDIR.
Goes back to the previous working directory using "cd -".
Yes, you can! It's in the arguments. :)
look at
${0}
combining that with
{$var%Pattern}
Remove from $var the shortest part of $Pattern that matches the back end of $var.
what you want is just
${0%/*}
I recommend the Advanced Bash Scripting Guide
(that is also where the above information is from).
Especiall the part on Converting DOS Batch Files to Shell Scripts
might be useful for you. :)
If I have misunderstood you, you may have to combine that with the output of "pwd". Since it only contains the path the script was called with!
Try the following script:
#!/bin/bash
called_path=${0%/*}
stripped=${called_path#[^/]*}
real_path=`pwd`$stripped
echo "called path: $called_path"
echo "stripped: $stripped"
echo "pwd: `pwd`"
echo "real path: $real_path
This needs some work though.
I recommend using Dave Webb's approach unless that is impossible.
In bash under linux you can get the full path to the command with:
readlink /proc/$$/fd/255
and to get the directory:
dir=$(dirname $(readlink /proc/$$/fd/255))
It's ugly, but I have yet to find another way.
I was trying to find the path for a script that was being sourced from another script. And that was my problem, when sourcing the text just gets copied into the calling script, so $0 always returns information about the calling script.
I found a workaround, that only works in bash, $BASH_SOURCE always has the info about the script in which it is referred to. Even if the script is sourced it is correctly resolved to the original (sourced) script.
The correct answer is this one:
How do I determine the location of my script? I want to read some config files from the same place.
It is important to realize that in the general case, this problem has no solution. Any approach you might have heard of, and any approach that will be detailed below, has flaws and will only work in specific cases. First and foremost, try to avoid the problem entirely by not depending on the location of your script!
Before we dive into solutions, let's clear up some misunderstandings. It is important to understand that:
Your script does not actually have a location! Wherever the bytes end up coming from, there is no "one canonical path" for it. Never.
$0 is NOT the answer to your problem. If you think it is, you can either stop reading and write more bugs, or you can accept this and read on.
...
Try this:
${0%/*}
This should work for bash shell:
dir=$(dirname $(readlink -m $BASH_SOURCE))
Test script:
#!/bin/bash
echo $(dirname $(readlink -m $BASH_SOURCE))
Run test:
$ ./somedir/test.sh
/tmp/somedir
$ source ./somedir/test.sh
/tmp/somedir
$ bash ./somedir/test.sh
/tmp/somedir
$ . ./somedir/test.sh
/tmp/somedir
This is a script can get the shell file real path when executed or sourced.
Tested in bash, zsh, ksh, dash.
BTW: you shall clean the verbose code by yourself.
#!/usr/bin/env bash
echo "---------------- GET SELF PATH ----------------"
echo "NOW \$(pwd) >>> $(pwd)"
ORIGINAL_PWD_GETSELFPATHVAR=$(pwd)
echo "NOW \$0 >>> $0"
echo "NOW \$_ >>> $_"
echo "NOW \${0##*/} >>> ${0##*/}"
if test -n "$BASH"; then
echo "RUNNING IN BASH..."
SH_FILE_RUN_PATH_GETSELFPATHVAR=${BASH_SOURCE[0]}
elif test -n "$ZSH_NAME"; then
echo "RUNNING IN ZSH..."
SH_FILE_RUN_PATH_GETSELFPATHVAR=${(%):-%x}
elif test -n "$KSH_VERSION"; then
echo "RUNNING IN KSH..."
SH_FILE_RUN_PATH_GETSELFPATHVAR=${.sh.file}
else
echo "RUNNING IN DASH OR OTHERS ELSE..."
SH_FILE_RUN_PATH_GETSELFPATHVAR=$(lsof -p $$ -Fn0 | tr -d '\0' | grep "${0##*/}" | tail -1 | sed 's/^[^\/]*//g')
fi
echo "EXECUTING FILE PATH: $SH_FILE_RUN_PATH_GETSELFPATHVAR"
cd "$(dirname "$SH_FILE_RUN_PATH_GETSELFPATHVAR")" || return 1
SH_FILE_RUN_BASENAME_GETSELFPATHVAR=$(basename "$SH_FILE_RUN_PATH_GETSELFPATHVAR")
# Iterate down a (possible) chain of symlinks as lsof of macOS doesn't have -f option.
while [ -L "$SH_FILE_RUN_BASENAME_GETSELFPATHVAR" ]; do
SH_FILE_REAL_PATH_GETSELFPATHVAR=$(readlink "$SH_FILE_RUN_BASENAME_GETSELFPATHVAR")
cd "$(dirname "$SH_FILE_REAL_PATH_GETSELFPATHVAR")" || return 1
SH_FILE_RUN_BASENAME_GETSELFPATHVAR=$(basename "$SH_FILE_REAL_PATH_GETSELFPATHVAR")
done
# Compute the canonicalized name by finding the physical path
# for the directory we're in and appending the target file.
SH_SELF_PATH_DIR_RESULT=$(pwd -P)
SH_FILE_REAL_PATH_GETSELFPATHVAR=$SH_SELF_PATH_DIR_RESULT/$SH_FILE_RUN_BASENAME_GETSELFPATHVAR
echo "EXECUTING REAL PATH: $SH_FILE_REAL_PATH_GETSELFPATHVAR"
echo "EXECUTING FILE DIR: $SH_SELF_PATH_DIR_RESULT"
cd "$ORIGINAL_PWD_GETSELFPATHVAR" || return 1
unset ORIGINAL_PWD_GETSELFPATHVAR
unset SH_FILE_RUN_PATH_GETSELFPATHVAR
unset SH_FILE_RUN_BASENAME_GETSELFPATHVAR
unset SH_FILE_REAL_PATH_GETSELFPATHVAR
echo "---------------- GET SELF PATH ----------------"
# USE $SH_SELF_PATH_DIR_RESULT BEBLOW
I have tried $0 before, namely:
dirname $0
and it just returns "." even when the script is being sourced by another script:
. ../somedir/somescript.sh