myDir = 'apple'
If I have the above variable, what functionality can I use to use it within a command? (I'm guessing it's some type of substitution - so I'd like to know what it's called if so)
How could I use the above variable to do an ls /home/applefruit, as obviously ls /home/$myDirfruit does not work.
First of all, your variable declaration is not right. There must be no spaces around = in declaration:
myDir='apple'
Now, ls /home/$myDirfruit did not work because myDirfruit is being treated as the variable name instead of just myDir. You need to use {} to enclose the variable name when the name is being followed by valid variable name constituent character:
ls /home/${myDir}fruit
would be expanded to:
ls /home/applefruit
Also if you have spaces in variable name e.g. myDir='foo bar', use quotes around variable:
ls /home/"${myDir}"fruit
Related
I want to define variable in shell script as:
value1 = 40 (this can be number or character)
and want to use as in a text like:
$value1_position.xyz (I basically want 40_position.xyz)
How do I do this?
this should do:
${value1}_position.xyz
beware that the variable should be declared with this syntax
value1=40
notice the absence of spaces around the =
To define a variable, simply make sure there are no spaces between the variable name and value
value1=40
To use that variable in bash substitution, creating what you want, use the $ replacement symbol like so:
${value1}_position.xyz
To append that to your text file
echo "${value1}_position.xyz" >> file.txt
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"
I have a variable that contains this kind of string :
var='$FOO/bar/baz*'
and I want to replace the variable $FOO by its content. However, when i do
var=$(eval "echo $var")
The variable is replaced, but the star is also replaced so that var now contains every possible match in my filesystem (as if i pressed tab in a shell).
for example, if $FOO contains /home, var will contain "/home/bar/baz1.sh /home/bar/baz2.sh /home/bar/baz.conf"
How do i replace the variable without expanding wildcards ?
Turn off globbing in bash, then reenable it.
set -f
var="$FOO/bar/baz*"
set +f
Just drop the quotes:
var=$FOO/bar/baz/*
Globs are not expanded on the RHS of a variable assignment.
I noticed that in shell script when we declare a variable, the preceding dollar sign is not needed, although when we want to access this variable later we should add a dollar sign in front of this variable name.
just like:
#!/bin/sh
VAR_1=Hello
VAR_2=Unix
echo "$VAR_1 $VAR_2"
This is different from other languages, like Perl we will always have the preceding dollar sign with the variable name, I just want to know any good reason for shell script to do it in this way, or it's just a convention...?
Shell is a different language than Perl is a different language than C++ is a different language than Python. You can add "with different rules" to each of the languages.
In shell an identifier like VAR_1 names a variable, the dollar sign is used to invoke expansion. $var is replaced with var's content; ${var:-foo} is replaced with var's content if it is set and with the word foo if the variable isn't set. Expansion works on non-variables as well, e.g. you can chain expansion like ${${var##*/}%.*} should leave only a file base name if var contains a file name with full path and extension.
In Perl the sigil in front of the variable tells Perl how to interpret the identifier: $var is a scalar, #var an array, %var a hash etc.
In Ruby the sigil in front of the varible tells Ruby its scope: var is a local variable, $var is a global one, #var is an instance variable of an object and ##var is a class variable.
In C++ we don't have sigils in front of variable names.
Etc.
In the shell, the $ sign is not part of the variable name. It just tells the shell to replace the following word with the contents of the variable with the same name, i.e. $foo means "insert the contents of the variable foo here".
This is not used when assigning to the variable because there you explicitly don't want to insert the old contents; you want to use the variable itself (in some ways this is similar to dereferencing pointers).
It's basically a syntactical convention.
DOS/.bat file syntax works the same way.
1) to create a variable, no metacharacter.
2) to "dereference" the contents of the variable, use the metacharacter.
DOS:
set VAR=123
echo %VAR%
I have a script which contains the following line:
propFile="${0%/*}/anteater.properties"
What does "${0%/*}" mean?
This command gives a path to the script - but there is a spaces at path and script can't find this file - how to deal with it?
The % operator in variable expansion removes the matching suffix pattern given to it. So ${0%/*} takes the variable $0, and removes all matching /* at the end. This is equivalent to the command dirname, which, when given a path, returns the parent directory of that path.
In order to deal with spaces in bash variable, whenever expanding the variable (i.e. whenever you write $var), you should quote it. In short, always use "$var" instead of just $var.
Consider reading shell parameter expansion and variable quoting in the bash manual to learn more about these two subjects.
strips the suffix matching /*, i.e. everything after last slash including the slash itself.
quote it wherever you use it (cat "$propFile").