>export FOOBAR=foobar; IFS=b echo ${FOOBAR}
I was expecting to see
foo ar
but I see
foobar
Why?
The IFS hasnt yet taken effect. add another ";":
FOOBAR=foobar IFS=b; echo ${FOOBAR}
In man bash section SIMPLE COMMAND EXPANSION
you can read (abbreviated):
When a simple command is executed
The words that the parser has marked as variable assignments (those preceding the command name) are saved for later processing.
The words that are not variable assignments or redirections are expanded.
...
The text after the = in each variable assignment ... [are] assigned to the variable.
so the IFS=b is done after expanding $FOOBAR.
[edit]I removed the technically incorrect answer.
http://tldp.org/LDP/abs/html/internalvariables.html
"This variable determines how Bash recognizes fields, or word boundaries, when it interprets character strings."
Related
To rephrase - I want to use Bash command substitution and string substitution in the same line.
My actual commands are longer, but the ridiculous use of echo here is just a "substitution" for shortness and acts the same - with same errors ;)
I know we can use a Bash command to produce it's output string as a parameter for another command like this:
echo "$(echo "aahahah</ddd>")"
aahahah</ddd>
I also know we can remove last known part of a string like this:
var="aahahah</ddd>"; echo "${var%</ddd>}"
aahahah
I am trying to write a command where one command gives a string output, where I want to remove last part, which is known.
echo "${$(echo "aahahah</ddd>")%</ddd>}"
-bash: ${$(echo "aahahah</ddd>")%</ddd>}: bad substitution
It might be the order of things happening or substitution only works on variables or hardcoded strings. But I suspect just me missing something and it is possible.
How do I make it work?
Why doesn't it work?
When a dollar sign as in $word or equivalently ${word} is used, it asks for word's content. This is called parameter expansion, as per man bash.
You may write var="aahahah</ddd>"; echo "${var%</ddd>}": That expands var and performs a special suffix operation before returning the value.
However, you may not write echo "${$(echo "aahahah</ddd>")%</ddd>}" because there is nothing to expand once $(echo "aahahah</ddd>") is evaluated.
From man bash (my emphasis):
${parameter%word}
Remove matching suffix pattern. The word is expanded to produce a
pattern just as in pathname expansion. If the pattern
matches a trailing portion of the expanded value of parameter, then
the result of the expansion is the expanded value of parameter
with the shortest matching pattern (the ''%'' case) or the longest matching pattern (the ''%%'' case) deleted.
Combine your commands like this
var=$(echo "aahahah</ddd>")
echo ${var/'</ddd>'}
For each of two examples below I'll try to explain what result I expected and what I got instead. I'm hoping for you to help me understand why I was wrong.
1)
VAR1=VAR2
$VAR1=FOO
result: -bash: VAR2=FOO: command not found
In the second line, $VAR1 gets expanded to VAR2, but why does Bash interpret the resulting VAR2=FOO as a command name rather than a variable assignment?
2)
'VAR=FOO'
result: -bash: VAR=FOO: command not found
Why do the quotes make Bash treat the variable assignment as a command name?
Could you please describe, step by step, how Bash processes my two examples?
How best to indirectly assign variables is adequately answered in other Q&A entries in this knowledgebase. Among those:
Indirect variable assignment in bash
Saving function output into a variable named in an argument
If that's what you actually intend to ask, then this question should be closed as a duplicate. I'm going to make a contrary assumption and focus on the literal question -- why your other approaches failed -- below.
What does the POSIX sh language specify as a valid assignment? Why does $var1=foo or 'var=foo' fail?
Background: On the POSIX sh specification
The POSIX shell command language specification is very specific about what constitutes an assignment, as quoted below:
4.21 Variable Assignment
In the shell command language, a word consisting of the following parts:
varname=value
When used in a context where assignment is defined to occur and at no other time, the value (representing a word or field) shall be assigned as the value of the variable denoted by varname.
The varname and value parts shall meet the requirements for a name and a word, respectively, except that they are delimited by the embedded unquoted equals-sign, in addition to other delimiters.
Also, from section 2.9.1, on Simple Commands, with emphasis added:
The words that are recognized as variable assignments or redirections according to Shell Grammar Rules are saved for processing in steps 3 and 4.
The words that are not variable assignments or redirections shall be expanded. If any fields remain following their expansion, the first field shall be considered the command name and remaining fields are the arguments for the command.
Redirections shall be performed as described in Redirection.
Each variable assignment shall be expanded for tilde expansion, parameter expansion, command substitution, arithmetic expansion, and quote removal prior to assigning the value.
Also, from the grammar:
If all the characters preceding '=' form a valid name (see the Base Definitions volume of IEEE Std 1003.1-2001, Section 3.230, Name), the token ASSIGNMENT_WORD shall be returned. (Quoted characters cannot participate in forming a valid name.)
Note from this:
The command must be recognized as an assignment at the very beginning of the parsing sequence, before any expansions (or quote removal!) have taken place.
The name must be a valid name. Literal quotes are not part of a valid variable name.
The equals sign must be unquoted. In your second example, the entire string was quoted.
Assignments are recognized before tilde expansion, parameter expansion, command substitution, etc.
Why $var1=foo fails to act as an assignment
As given in the grammar, all characters before the = in an assignment must be valid characters within a variable name for an assignment to be recognized. $ is not a valid character in a name. Because assignments are recognized in step 1 of simple command processing, before expansion takes place, the literal text $var1, not the value of that variable, is used for this matching.
Why 'var=foo' fails to act as an assignment
First, all characters before the = must be valid in variable names, and ' is not valid in a variable name.
Second, an assignment is only recognized if the = is not quoted.
1)
VAR1=VAR2
$VAR1=FOO
You want to use a variable name contained in a variable for the assignment. Bash syntax does not allow this. However, there is an easy workaround :
VAR1=VAR2
declare "$VAR1"=FOO
It works with local and export too.
2)
By using single quotes (double quotes would yield the same result), you are telling Bash that what is inside is a string and to treat it as a single entity. Since it is the first item on the line, Bash tries to find an alias, or shell builtin, or an executable file in its PATH, that would be named VAR=FOO. Not finding it, it tells you there is no such command.
An assignment is not a normal command. To perform an assignment contained in a quote, you would need to use eval, like so :
eval "$VAR1=FOO" # But please don't do that in real life
Most experienced bash programmers would probably tell you to avoid eval, as it has serious drawbacks, and I am giving it as an example just to recommend against its use : while in the example above it would not involve any security risk or error potential because the value of VAR1 is known and safe, there are many cases where an arbitrary (i.e. user-supplied) value could cause a crash or unexpected behavior. Quoting inside an eval statement is also more difficult and reduces readability.
You declare VAR2 earlier in the program, right?
If you are trying to assign the value of VAR2 to VAR1, then you need to make sure and use $ in front of VAR2, like so:
VAR1=$VAR2
That will set the value of VAR2 equal to VAR1, because when you utilize the $, you are saying that value that is stored in the variable. Otherwise it doesn't recognize it as a variable.
Basically, a variable that doesn't have a $ in front of it will be interpreted as a command. Any word will. That's why we have the $ to clarify "hey this is a variable".
Consider the following two assignments.
$ a="foo bar"
$ b=$a
$ b=foo bar
bash: bar: command not found
Why does the second assignment work fine? How is the second command any different from the third command?
I was hoping the second assignment to fail because
b=$a
would expand to
b=foo bar
Since $a is not within double-quotes, foo bar is not quoted, therefore field-splitting should occur (as per my understanding) which would result in b=foo to be considered an assignment and bar to be a command that cannot be found.
Summary: I was expecting the 2nd command to fail for the same reason that caused the 3rd command to fail. Why does the 2nd command succeed?
I went through the POSIX but I am unable to find anything that specifies that field splitting won't occur after parameter expansion that occurs in an assignment.
I mean anywhere else field splitting would occur for an unquoted parameter after parameter expansion. For example,
$ a="foo bar"
$ printf "[%s] [%s]\n" $a
[foo] [bar]
See Section 2.6.5.
After parameter expansion (Parameter Expansion), command substitution (Command Substitution), and arithmetic expansion (Arithmetic Expansion), the shell shall scan the results of expansions and substitutions that did not occur in double-quotes for field splitting and multiple fields can result.
So which part of the POSIX standard prevents field splitting when parameter expansion occurs in an assignment statement?
In 2.9.1, "Simple Commands":
The words that are recognized as variable assignments or redirections according to Shell Grammar Rules are saved for processing in steps 3 and 4.
Step 2 -- which is explicitly skipped in this case per the above text -- reiterates that it ignores assignments when performing expansion and field splitting:
The words that are not variable assignments or redirections shall be expanded. If any fields remain following their expansion, the first field shall be considered the command name and remaining fields are the arguments for the command.
Thus, it's step 2 that determines the command to run (based on contents other than variable assignments and redirections), which addresses the b=$a case given in your question.
Step 4 performs other expansions -- "tilde expansion, parameter expansion, command substitution, arithmetic expansion, and quote removal" -- for assignments. Notably, field splitting is not a member of this set. Indeed, it's explicit in 2.6 that none of these create multiple words in and of themselves:
Tilde expansions, parameter expansions, command substitutions, arithmetic expansions, and quote removals that occur within a single word expand to a single field. It is only field splitting or pathname expansion that can create multiple fields from a single word. The single exception to this rule is the expansion of the special parameter '#' within double-quotes, as described in Special Parameters.
can someone explain me with this code
data=$(date +"%Y-%m-%dS%H:%M:%S")
name="/home/cft/"$data"_test.tar"
touch $name
works, creating a new .tar file but this code doesn't work
data=$(date +"%Y-%m-%dS%H:%M:%S")
name= "/home/cft/"$data"_test.tar"
touch $name
and gives me this error: no such file or directory?
why the space between = and inverted commas creates this error?
Shell allows you to provide per-command environment overrides by prefixing the command with one or more variable assignments.
name= "/home/cft/"$data"_test.tar"
asks the shell to run the program named /home/cft/2013-10-08S12:00:00_test.tar (for example) with the value of name set to the empty string in its environment.
(In your case, the error occurs because the named tar file either doesn't exist or, if it does, is not an executable file.)
A variable assignment is identified by having no whitespace after the equal sign.
(name = whatever, of course, is simply a command called name with two string arguments, = and whatever.)
You can't have whitespace between the equal sign and the definition.
http://www.tldp.org/LDP/abs/html/varassignment.html
There is no theory behind this. It's just a decision the language designers made, and which the parser enforces.
In BASH (and other Bourne type shells like zsh and Kornshell), the equal sign cannot have spaces around it when setting variables.
Good:
$ foo="bar"
Bad:
$ foo= "bar"
$ foo = "bar"
There's no real reason that would prevent spaces from being used. Other programming languages have no problems with this. It's just the syntax of the shell itself.
The reason might be related to the original Bourne shell parsing where the shell would break up a command line based upon whitespace. That would make foo=bar a single parameter instead of two or three (depending if you have white space on both sides or just one side of the equal sign). The shell could see the = sign, and know this parameter is an assignment.
The shell parameter parsing is very primitive in many ways. Whitespace is very important. The shell has to be small and fast in order to be responsive. That means stripping down unessential things like complex line parsing.
Inverted commas I believe you mean quotation marks. Double quotes are used to override the breaking out of parameters over white space:
Bad:
$ foo=this is a test
bash: is: command not found
Good:
$ foo="this is a test"
Double quotes allow interpolation. Single quotes don't:
$ foo="bar"
$ echo "The value of foo is $foo"
The value of foo is bar
$ echo 'The value of foo is $foo'
The value of foo is $foo.
If you start out with single quotes, you can put double quotes inside. If you have single quotes, you can put double quotes inside.
$ foo="bar"
$ echo "The value of foo is '$foo'"
The value of foo is 'bar'
$ echo 'The value of foo is "$foo"'
The value of foo is "$foo"
This means you didn't have to unquote $data. However, you would have to put curly braces around it because underscores are legal characters in variable names. Thus, you want to make sure that the shell understand that the variable is $data and not $data_backup:
name="/home/cft/${data}_test.tar"
A contrived example... given
FOO="/foo/bar/baz"
this works (in bash)
BAR=$(basename $FOO) # result is BAR="baz"
BAZ=${BAR:0:1} # result is BAZ="b"
this doesn't
BAZ=${$(basename $FOO):0:1} # result is bad substitution
My question is which rule causes this [subshell substitution] to evaluate incorrectly? And what is the correct way, if any, to do this in 1 hop?
First off, note that when you say this:
BAR=$(basename $FOO) # result is BAR="baz"
BAZ=${BAR:0:1} # result is BAZ="b"
the first bit in the construct for BAZ is BAR and not the value that you want to take the first character of. So even if bash allowed variable names to contain arbitrary characters your result in the second expression wouldn't be what you want.
However, as to the rule that's preventing this, allow me to quote from the bash man page:
DEFINITIONS
The following definitions are used throughout the rest of this docu‐
ment.
blank A space or tab.
word A sequence of characters considered as a single unit by the
shell. Also known as a token.
name A word consisting only of alphanumeric characters and under‐
scores, and beginning with an alphabetic character or an under‐
score. Also referred to as an identifier.
Then a bit later:
PARAMETERS
A parameter is an entity that stores values. It can be a name, a num‐
ber, or one of the special characters listed below under Special Param‐
eters. A variable is a parameter denoted by a name. A variable has a
value and zero or more attributes. Attributes are assigned using the
declare builtin command (see declare below in SHELL BUILTIN COMMANDS).
And later when it defines the syntax you're asking about:
${parameter:offset:length}
Substring Expansion. Expands to up to length characters of
parameter starting at the character specified by offset.
So the rules as articulated in the manpage say that the ${foo:x:y} construct must have a parameter as the first part, and that a parameter can only be a name, a number, or one of the few special parameter characters. $(basename $FOO) is not one of the allowed possibilities for a parameter.
As for a way to do this in one assignment, use a pipe to other commands as mentioned in other responses.
Modified forms of parameter substitution such as ${parameter#word} can only modify a parameter, not an arbitrary word.
In this case, you might pipe the output of basename to a dd command, like
BAR=$(basename -- "$FOO" | dd bs=1 count=1 2>/dev/null)
(If you want a higher count, increase count and not bs, otherwise you may get fewer bytes than requested.)
In the general case, there is no way to do things like this in one assignment.
It fails because ${BAR:0:1} is a variable expansion. Bash expects to see a variable name after ${, not a value.
I'm not aware of a way to do it in a single expression.
As others have said, the first parameter of ${} needs to be a variable name. But you can use another subshell to approximate what you're trying to do.
Instead of:
BAZ=${$(basename $FOO):0:1} # result is bad substitution
Use:
BAZ=$(_TMP=$(basename $FOO); echo ${_TMP:0:1}) # this works
A contrived solution for your contrived example:
BAZ=$(expr $(basename $FOO) : '\(.\)')
as in
$ FOO=/abc/def/ghi/jkl
$ BAZ=$(expr $(basename $FOO) : '\(.\)')
$ echo $BAZ
j
${string:0:1},string must be a variable name
for example:
FOO="/foo/bar/baz"
baz="foo"
BAZ=eval echo '${'"$(basename $FOO)"':0:1}'
echo $BAZ
the result is 'f'