bash environment-variables string – How to Get Variable Name from Command Output

bashenvironment-variablesstring

I have some string, and want to split on colon ":" assigning on variable with name from left part and value from right part. For example:

echo "Title: Some title" | sed 's/:.*//'

gives me wanted variable name "Title", and:

echo "Title: Some title" | sed 's/.*: //'

gives me the value "Some title".

My problem is how to assign variable named "Title" by the output of above command.

I can assign the value like $(echo "Title: Some title" | sed 's/.*: //) but can't find how to assign the name

Best Answer

Depending on how this is going to be used and your bash version, you would probably be better off using an associative array.

Failing that, I suggest you fake one.

For the sake of completeness, you can actually do this using eval. Note that eval is dangerous and you shouldn't use this unless you have to.

eval "${string/: /=\"}\""

Related Solutions

Shell – Expansion of a shell variable and effect of glob and split on it

Variable expansion (the standard term is parameter expansion, and it's also sometimes called variable substitution) basically means replacing the variable by its value. More precisely, it means replacing the $VARIABLE construct (or ${VARIABLE} or ${VARIABLE#TEXT} or other constructs) by some other text which is built from the value of the variable. This other text is the expansion of the variable.

The expansion process goes as follows. (I only discuss the common case, some shell settings and extensions modify the behavior.)

Take the value of the variable, which is a string. If the variable is not defined, use the empty string.
If the construct includes a transformation, apply it. For example, if the construct is ${VARIABLE#TEXT}, and the value of the variable begins with TEXT, remove TEXT from the beginning of the value.
If the context calls for a single word (for example within double quotes, or in the right-hand side of an assignment, or inside a here document), stop here. Otherwise continue with the next steps.
Split the value into separate words at each sequence of whitespace. (The variable IFS can be changed to split at characters other than whitespace.) The result is thus no longer a string, but a list of strings. This list can be empty if the value contained only whitespace.
Treat each element of the list as a file name wildcard pattern, i.e. a glob. If the pattern matches some files, it is replaces by the list of matching file names, otherwise it is left alone.

For example, suppose that the variable foo contains a* b* c* and the current directory contains the files bar, baz and paz. Then ${foo#??} is expanded as follows:

The value of the variable is the 8-character string a* b* c*.
#?? means strip off the first two characters, resulting in the 6-character string b* c* (with an initial space).
If the expansion is in a list context (i.e. not in double quotes or other similar context), continue.
Split the string into whitespace-delimited words, resulting in a list of two-strings: b* and c*.
The string b*, interpreted as a pattern, matches two files: bar and baz. The string c* matches no file so it is left alone. The result is a list of three strings: bar, baz, c*.

For example echo ${foo#??} prints bar baz c* (the command echo joins its arguments with a space in between).

For more details, see:

Parameter expansion in the POSIX standard, followed by field splitting and pathname expansion
Shell parameter expansion in the bash manual, followed by word splitting and filename expansion
$VAR vs ${VAR} and to quote or not to quote
When is double-quoting necessary?

Bash – In bash, how to call a variable and wrap quotes around its value as part of a command without causing the variable name to become a string literal

Just for completeness, you don't need all those (") nor the final $(echo ...). Here's the simplified version of your assignments that produce the same effect:

STARTIME=$(date +"%T")
ENDTIME="$STARTIME today + 10 seconds"
CALL="date -d '$ENDTIME' +'%H:%M:%S'"

Note how you don't need to quote when doing var=$(...) but you do usually with var="many words":

a=$(echo 'a    b'); echo "$a" # result: a    b

Inside (") a (') has no special significance, and vice-versa, eg:

a="that's nice"; echo "$a" # result: that's nice
a='that "is nice'; echo "$a" # result: that "is nice

Best Answer

Related Solutions

Shell – Expansion of a shell variable and effect of glob and split on it

Bash – In bash, how to call a variable and wrap quotes around its value as part of a command without causing the variable name to become a string literal

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