for bool in $jobdummyjob1 $jobdummyjob2 $jobdummyjob3
do
echo "Boolean Value is $bool"
if "$bool" ; then
echo "$alljobs"
curl -X POST https://jenkins.xxxxxxxxx.com/job/$bool/build --user xxxxxxx.yyyyyyy#xxxxxx.com:ewfwedf3f234523555235235235235235
fi
done
All I need is to take somehow the names of jobdummyjob1, 2, 3 and put them in the URL as a string. Those vars are booleans so when I do this I get true or false in the URL. I do not need the variable value, but its name.
First I run the 'for' and I go through each object. Each object contains boolean value. Then, I do the true/false check and if true, I need to get the string name of the same variable and put it in the URL . This is a Jenkins job.
You can use variable indirection:
for name in jobdummyjob1 jobdummyjob2 jobdummyjob3
do
bool=${!name}
echo "Boolean Value is $bool"
if "$bool" ; then
echo "$alljobs"
curl -X POST https://jenkins.xxxxxxxx.com/job/"$name"/build --user xxxxxxxx.xxxxxxxx#xxxxxxxx.com:xxxxxxxx
fi
done
But it's cleaner to use an associative array:
declare -A bools
bools=([jobdummyjob1]=true [jobdummyjob2]=false [jobdummyjob3]=true)
for name in "${!bools[#]}" ; do
bool=${bools[$name]}
if ...
done
Related
"$name" yields 2 different values: ?
#!/bin/bash
name=whoami
$name # $name yields {user} being me is "jdl"
date >> /home/$name/crondate.txt # $name yields "whoami"
Would like to have the path using "jdl"?
The value of name is whoami.
In your first example, it is not $name itself that yields jdl; that's the output of the command whoami when it is executed.
If you were to run
echo "$name"
then you would see the value of the variable itself.
If you want to put jdl in name you should use :
name=$(whoami)
and if you want to print the value echo $name
actually what your code does :
name=whoami
$name # replace name to whoami and execute it
date >> /home/$name/crondate.txt # replace name to whoami and the path is /home/whoami/crondate.txt
I have a script that is (supposed to be) assigning a dynamic variable name (s1, s2, s3, ...) to a directory path:
savedir() {
declare -i n=1
sn=s$n
while test "${!sn}" != ""; do
n=$n+1
sn=s$n
done
declare $sn=$PWD
echo "SAVED ($sn): ${!sn}"
}
The idea is that the user is in a directory they'd like to recall later on and can save it to a shell variable by typing 'savedir'. It -does- in fact write out the echo statement successfully: if I'm in the directory /home/mrjones and type 'savedir', the script returns:
SAVED (s1): /home/mrjones
...and I can further type:
echo $sn
and the script returns:
s1
...but typing either...
> echo $s1
...or
echo ${!sn}
...both return nothing (empty strings). What I want, in case it's not obvious, is this:
echo $s1
/home/mrjones
Any help is greatly appreciated! [apologies for the formatting...]
To set a variable using a name stored in another variable I use printf -v, in this example:
printf -v "$sn" '%s' "$PWD"
declare here is creating a variable local to the function, which doesn't seem to be what you want. Quoting from help declare:
When used in a function, declare makes NAMEs local, as with the local
command. The -g option suppresses this behavior.
so you can either try the -g or with the printf
Use an array instead.
savedir() {
s+=("$PWD")
echo "SAVED (s[$((${#s[#]}-1))]): ${s[${#s[#]}-1]}"
}
I want to have a function in bash, which create a Dictionary as a local variable. Fill the Dictionary with one element and then return this dictionary as output.
Is the following code correct?
function Dictionary_Builder ()
{
local The_Dictionary
unset The_Dictionary
declare -A The_Dictionary
The_Dictionary+=(["A_Key"]="A_Word")
return $The_Dictionary
}
How can I access to the output of the function above? Can I use the following command in bash?
The_Output_Dictionary=Dictionary_Builder()
To capture output of a command or function, use command substitution:
The_Output_Dictionary=$(Dictionary_Builder)
and output the value to return, i.e. replace return with echo. You can't easily return a structure, though, but you might try returning a string that declares it (see below).
There's no need to use local and unset in the function. declare creates a local variable inside a function unless told otherwise by -g. The newly created variable is always empty.
To add a single element to an empty variable, you can assign it directly, no + is needed:
The_Dictionary=([A_Key]=A_Word)
In other words
#!/bin/bash
Dictionary_Builder () {
declare -A The_Dictionary=([A_Key]=A_Word)
echo "([${!The_Dictionary[#]}]=${The_Dictionary[#]})"
}
declare -A The_Output_Dictionary="$(Dictionary_Builder)"
echo key: ${!The_Output_Dictionary[#]}
echo value: ${The_Output_Dictionary[#]}
For multiple keys and values, you need to loop over the dictionary:
Dictionary_Builder () {
declare -A The_Dictionary=([A_Key]=A_Word
[Second]=Third)
echo '('
for key in "${!The_Dictionary[#]}" ; do
echo "[$key]=${The_Dictionary[$key]}"
done
echo ')'
}
declare -A The_Output_Dictionary="$(Dictionary_Builder)"
for key in "${!The_Output_Dictionary[#]}" ; do
echo key: $key, value: ${The_Output_Dictionary[$key]}
done
The answer by #choroba is what I was looking for. However, my dictionary values also had white spaces in them and the above answer didn't work outright. What worked was a minor variation of the above answer.
#!/bin/bash
function Dictionary_Builder() {
declare -A dict=(['title']="Title of the song"
['artist']="Artist of the song"
['album']="Album of the song"
)
echo '('
for key in "${!dict[#]}" ; do
echo "['$key']='${dict[$key]}'"
done
echo ')'
}
declare -A Output_Dictionary="$(Dictionary_Builder)"
for key in "${!Output_Dictionary[#]}" ; do
echo "${key}: '"${Output_Dictionary[$key]}"'"
done
Note the extra single quotes on the 2nd echo line which made it possible to output values with whitespaces in them.
I have a ksh script that reads a profile script with a number of sessions defined. Each session defines its own parameters as such:
SESSION_ONE_USER=...
SESSION_ONE_PWD=...
SESSION_TWO_USER=...
...
The script gets the SESSION parameter from the command line, but I simply set it for the example.
I want to let the SESSION parameter value define part of another parameter name, that I need the value from, like:
SESSION="SESSION_ONE"
USER=${${SESSION}_USER}
PASS=${${SESSION}_PWD}
That gives me a compile error.
I also tried
GET_USER_PARAM(){
echo ${SESSION}_USER
}
echo $`GET_USER_PARAM`
But that returns $SESSION_ONE_USER
I want it to return the value of the parameter SESSION_ONE_USER instead.
Does anyone have any solutions?
This is what eval is for:
SESSION=SESSION_ONE
eval echo \$${SESSION}_USER
should display the value of $SESSION_ONE_USER.
Don't monkey with variable names, use associative arrays instead
typeset -A users
typeset -A pwd
session=SESSION_ONE
users[$session]=joe
pwd[$session]=secret
for key in "${!users[#]}"; do
echo "user for session $key is ${users[$key]}"
echo "pwd for session $key is ${pwd[$key]}"
done
Try this:
SESSION="SESSION_ONE"
SESSION_ONE_USER="foo"
SESSION_ONE_PWD="bar"
SESSION_USER=${SESSION}_USER
SESSION_PWD=${SESSION}_PWD
USER=${!SESSION_USER}
PASS=${!SESSION_PWD}
echo $USER
echo $PASS
The "!" does a level of indirection. See Shell Parameter Expansion.
If this is ksh, then this is a job for nameref
alias nameref='typeset -n'
Example Solution
function session_parameters { set -u
typeset session=${1:?session name}
nameref user=SESSION_${session}_USER
nameref pass=SESSION_${session}_PASS
print session=$session user=$user pass=$pass
}
SESSION_ONE_USER="User1"
SESSION_ONE_PASS="Pass1"
SESSION_TWO_USER="User2"
SESSION_TWO_PASS="Pass2"
for s in ONE TWO THREE; do
session_parameters $s
done
Sample output
session=ONE user=User1 pass=Pass1
session=TWO user=User2 pass=Pass2
test_session_parameters[12]: session_parameters: line 5:
SESSION_THREE_USER: parameter not set
Note the usage of set -u to force the error message on line 3.
nameref usage: (from the builtin help text)
NAME
typeset - declare or display variables with attributes
SYNOPSIS
typeset [ options ] [name[=value]...]
-n
Name reference.
The value is the name of a variable that name references. name cannot contain a ... Cannot be use with any other options.
I want to save the variable name and its contents easily from my script.
Currently :-
LOGFILE=/root/log.txt
TEST=/file/path
echo "TEST : ${TEST}" >> ${LOGFILE}
Desired :-
LOGFILE=/root/log.txt
function save()
{
echo "$1 : $1" >> ${LOGFILE}
}
TEST=/file/path
save TEST
Obviously the above save function just saves TEST : TEST
Want I want it to save is TEST : /file/path
Can this be done? How? Many thanks in advance!
You want to use Variable Indirection. Also, don't use the function keyword, it is not POSIX and also not necessary as long as you have () at the end of your function name.
LOGFILE=/root/log.txt
save()
{
echo "$1 : ${!1}" >> ${LOGFILE}
}
TEST=/file/path
save TEST
Proof of Concept
$ TEST=foo; save(){ echo "$1 : ${!1}"; }; save TEST
TEST : foo
Yes, using indirect expansion:
echo "$1 : ${!1}"
Quoting from Bash reference manual:
The basic form of parameter expansion is ${parameter} [...] If the first character of parameter is an exclamation point (!), a level of variable indirection is introduced. Bash uses the value of the variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value is used in the rest of the substitution, rather than the value of parameter itself. This is known as indirect expansion
Consider using the printenv function. It does exactly what it says on the tin, prints your environment. It can also take parameters
$ printenv
SSH_AGENT_PID=2068
TERM=xterm
SHELL=/bin/bash
LANG=en_US.UTF-8
HISTCONTROL=ignoreboth
...etc
You could do printenv and then grep for any vars you know you have defined and be done in two lines, such as:
$printenv | grep "VARNAME1\|VARNAME2"
VARNAME1=foo
VARNAME2=bar