I have two variable or files with strings like
a="p|q"
b="x y z"
b variable has three characters separate by space.
my question is how to generate output as using a shell command in Linux with separator as "|"
p|q|x
p|q|y
p|q|z
Not quite sure if I understand your question correctly. The following for-loop will give you the output you want:
for i in $b; do echo "$a|$i"; done
You can do it without a loop, but it's not recommended in general:
printf "$a|%s\n" $b
$a is expanded before the command is run, so your format string ends up as 'p|q|%s\n'
$b is also expanded and since it is unquoted, each word becomes a separate argument
So the command you run is:
printf 'p|q|%s\n' x y z
which produces the output you wanted.
Note that this is not ideal because:
In general, putting variables into a printf format string can lead to unexpected output (e.g. if the variable contains characters like %s).
Deliberately using unquoted variables reads like a typo to a lot of people, so you probably need to add a comment to explain that you're being clever.
Related
I'm trying to create function that will print message bound to variable in certain color. The message variable is passed as argument of this function. The problem is that I'm getting only text up to first space (only first word of message). My script looks like this:
#!/usr/bash
lbGREEN='\e[1;92m'
NC='\e[0m'
normalMessage="Everything fine"
echo_message() {
echo -e ${lbGREEN}$1${NC}
}
echo_message $normalMessage
My output is:
Everything
As Inian pointed out in comments, your problem is unquoted variable expansion
echo_message $normalMessage
becomes
echo_message Everything fine
once the variable expands, meaning that each word of your input string is getting read in as a separate argument. When this happens $1=Everything and $2=fine.
This is fixed by double-quoting your variable, which allows expansion, but will mean the result of the expansion will still be read as one argument.
echo_message "$normalMessage"
becomes
echo_message "Everything fine"
Like this $1=Everything fine
In the future, I recommend using https://www.shellcheck.net/, or the CLI version of shellcheck, it will highlight all kinds of common bash gotcha's, included unquoted expansion.
For me, I had to change the header for "#!/bin/bash", but apparently that is not the problem for you.
In your echo you are printing only the first word with the $1, if you change it to $2 you will print the second word (parameter) and so on.
You can pass the name inside quotes or print all the parameters with $#
Solution 1 (with $#):
lbGREEN='\e[1;92m'
NC='\e[0m'
normalMessage="Everything fine"
echo_message() {
echo -e ${lbGREEN}$#${NC}
}
echo_message $normalMessage
Solution 2 (with quotes):
lbGREEN='\e[1;92m'
NC='\e[0m'
normalMessage="Everything fine"
echo_message() {
echo -e ${lbGREEN}$1${NC}
}
echo_message "$normalMessage"
You should get a look to https://stackoverflow.com/a/6212408/1428602
IMHO, $1 only return the 1st word, so you have to use a loop or try with $*
You've got the quoting wrong.
If you want to simulate the behaviour of echo, your function should accept multiple parameters, and print them all. Currently it's only evaluating the first parameter, so I suggest using $* instead. You also need to enclose the argument in double quotes to protect any special characters:
echo_message() {
echo -e "${lbGREEN}$*${NC}"
}
The special variable $* expands to all the arguments, separated by spaces (or more accurately, the first character of $IFS, which is usually a space character). Note that you almost always want "$#" instead of "$*", and this is one of the rare occasions where the latter is also correct, though with slightly different semantics if IFS is set to a non-standard value.
Now the function supports multiple arguments, and prints them all in green, separated by spaces. However, I would recommend that you also quote the argument when calling the function:
echo_message "$normalMessage"
While spaces in $normalMessage will now be treated correctly, other special characters like ! will still require the quotes.
I've been trying to figure this out, unfortunately I can't. I am trying to create a function that finds the ';' character, puts four spaces before it and then and puts the code back together in a neat sentence. I've been cracking at this for a bit, and can't figure out a couple of things. I can't get the output to display what I want it to. I've tried finding the index of the ';' character and it seems I'm going about it the wrong way. The other mistake that I seem to be making is that I'm trying to split in a array in a for loop, and then split the individual words in the array by letter but I can't figure out how to do that either. If someone can give me a pointer this would be greatly appreciated. This is in bash version 4.3.48
#!commentPlacer()
{
arg=($1) #argument
len=${#arg[#]} #length of the argument
comment=; #character to look for in second loop
commaIndex=(${arg[#]#;}) #the attempted index look up
commentSpace=" ;" #the variable being concatenated into the array
for(( count1=0; count1 <= ${#arg[#]}; count1++ )) #search the argument looking for comment space
do if [[ ${arg[count1]} != commentSpace ]] #if no commentSpace variable then
then for (( count2=0; count2 < ${#arg[count1]} ; count2++ )) #loop through again
do if [[ ${arg[count2]} != comment ]] #if no comment
then A=(${arg[#]:0:commaIndex})
A+=(commentSpace)
A+=(${arg[#]commaIndex:-1}) #concatenate array
echo "$A"
fi
done
fi
done
}
If I understand what you want correctly, it's basically to put 4 spaces in front of each ";" in the argument, and print the result. This is actually simple to do in bash with a string substitution:
commentPlacer() {
echo "${1//;/ ;}"
}
The expansion here has the format ${variable//pattern/replacement}, and it gives the contents of the variable, with each occurrence of pattern replaced by replacement. Note that with only a single / before the pattern, it would replace only the first occurrence.
Now, I'm not sure I understand how your script is supposed to work, but I see several things that clearly aren't doing what you expect them to do. Here's a quick summary of the problems I see:
arg=($1) #argument
This doesn't create an array of characters from the first argument. var=(...) treats the thing in ( ) as a list of words, not characters. Since $1 isn't in double-quotes, it'll be split into words based on whitespace (generally spaces, tabs, and linefeeds), and then any of those words that contain wildcards will be expanded to a list of matching filenames. I'm pretty sure this isn't at all what you want (in fact, it's almost never what you want, so variable references should almost always be double-quoted to prevent it). Creating a character array in bash isn't easy, and in general isn't something you want to do. You can access individual characters in a string variable with ${var:index:1}, where index is the character you want (counting from 0).
commaIndex=(${arg[#]#;}) #the attempted index look up
This doesn't do a lookup. The substitution ${var#pattern} gives the value of var with pattern removed from the front (if it matches). If there are multiple possible matches, it uses the shortest one. The variant ${var##pattern} uses the longest possible match. With ${array[#]#pattern}, it'll try to remove the pattern from each element -- and since it's not in double-quotes, the result of that gets word-split and wildcard-expanded as usual. I'm pretty sure this isn't at all what you want.
if [[ ${arg[count1]} != commentSpace ]] #if no commentSpace variable then
Here (and in a number of other places), you're using a variable without $ in front; this doesn't use the variable at all, it just treats "commentSpace" as a static string. Also, in several places it's important to have double-quotes around it, e.g. to keep the spaces in $commentSpace from vanishing due to word splitting. There are some places where it's safe to leave the double-quotes off, but in general it's too hard to keep track of them, so just use double-quotes everywhere.
General suggestions: don't try to write c (or java or whatever) programs in bash; it works too differently, and you have to think differently. Use shellcheck.net to spot common problems (like non-double-quoted variable references). Finally, you can see what bash is doing by putting set -x before a section that doesn't do what you expect; that'll make bash print each line as it executes it, showing the equivalent of what it's executing.
Make a little function using pattern substitution on stdin:
semicolon4s() { while read x; do echo "${x//;/ ;}"; done; }
semicolon4s <<< 'foo;bar;baz'
Output:
foo ;bar ;baz
If you append ^ to a variable, Bash capitalises the first letter of its contents. (Similarly, , sends it to lowercase and doubling-up either of these applies the transformation to the whole string, rather than just the first letter.)
foo="hello world"
echo ${foo^} # Hello world
You can also do ${variable:position:length} to extract a substring:
echo ${foo:0:1} # h
So far, I haven't found a way to combine these without, obviously, creating a temporary variable. Is there a form where I can get just the capitalised first letter out of an arbitrary string?
It does not change the basic limitation you are seeing in terms of not being able to "chain" expansions, but you can assign the result of an expansion to the same variable and do away with the temporary variable.
For instance:
A=text
A="${A^}"
A="${A//x/s}"
echo "$A"
echoes "Test".
No. Parameter expansion operators do not compose, so if you want more than one side effect, you need a temporary variable (which can include overwriting the original value as shown by #fred) or an external tool to process the result of the expansion (as shown by #anubhava).
Your other alternative is to use a different shell that does support more complicated operations, like zsh:
% foo="hello world"
% % print ${(U)${foo:0:1}}
H
You can use tr with substring:
tr [[:lower:]] [[:upper:]] <<< "${foo:0:1}"
H
Suppose I have a bash script, myscript.sh, that takes 2 numbers, N and P, as arguments - i.e., I'd run it like this: sh myscript.sh 3 4
Inside the script, I'd like to run a program that takes a list of comma-separated files as arguments. N denotes the number of elements in that list. I need to construct this list of file paths, differentiated by a sample number running from 1 to N. It's easier to see with an example:
#!/bin/sh
myProgram -p $1 -f /home/folder/sample01.R,/home/folder/sample02.R,/home/folder/sample03.R
I'm struggling to create that comma-separated list of files given only N (3) as input. Is there a good way to create a variable holding the string /home/folder/sample01.R,/home/folder/sample02.R,/home/folder/sample03.R as a function of 3?
I've had success creating a variable holding 01,02,03 using seq, but have not been able to figure out how to attach the file paths to those numbers. (I'm getting stuck due to the fact that brace expansion happens before variable evaluation -- so, setting X='seq -f %02.0f 1 $N' and doing a{$X}b does not give a01b a02b a03b as I'd hoped, and even if it did, I'm not sure how I'd generate a01b,a02b,a03b from that.). Another snag is that I'm not using bash 4, so {00..09} gives a sequence of numbers that are NOT zero-padded (which causes problems). I developed a workaround with Python, but I'd love to be able to do it in bash to avoid having to call an external script.
Any and all suggestions appreciated!
You can construct your string using a number N like this:
N=3
s=""
for ((n=1; n<=N; n++)); do
[[ -n "$s" ]] && s="$s,"
printf -v s "%s/home/folder/sample%02d.R" "$s" "$n"
done
echo "$s"
/home/folder/sample01.R,/home/folder/sample02.R,/home/folder/sample03.R
If I were you, I'd have a look at the printf built-in:
f=$(printf "/home/folder/sample%02d.R" $num)
If you put this in a loop you will have your properly formatted numbers.
To create the comma-separated list, you can just append each filename to a string with a comma:
files="$files, $f"
After the loop, you just need to remove the first character from the string (it's a surplus comma), and you're done.
I'm sure there's a really easy way of doing this. I'm trying to take a file which contains some environment variables and expand it so that those which are known are expanded to their values whereas those which are not are left alone.
For example, if my file contained the following:
${I_EXIST}
${I_ALSO_EXIST}
${I_DONT_EXIST}
this would be expanded to:
existValue
alsoExistValue
${I_DONT_EXIST}
I ideally want to do this as simply as possible so I don't want a complex substitution using sed, awk or perl. I'm thinking of something similar to a "Here" file, but apart from the fact that I can't get the syntax right, it also blanks out anything which does not expand. E.g:
cat <<EOF
> ${I_EXIST}
> ${I_ALSO_EXIST}
> ${I_DONT_EXIST}
EOF
existValue
alsoExistValue
(i.e. the last value expands to nothing)
Update
Should really have made clear that I was thinking about potentially more than one substitution per line. One way I did find to do this, if we're not fussed about the variables appearing in the file as ${MYVAR} but maybe MYVAR will do:
m4 $( env | sed 's/\([A-Za-z0-9]*\)=\([\/A-Za-z_0-9:|%*. -#]*\)/-D\1=\2' ) myfile
This uses the M4 preprocessor to substitute all the pairs in your environment. A couple of caviats here:
Sorry about the reg exp stuff. It looks pretty nasty and I'm sure there are nicer ways of expressing this. I found problems if my env vars had spaces in them or any unusual characters that weren't in the set.
Of course this is a blunt substitution tool (which I was trying to avoid) so variable might get substituted when you didn't want it to happen.
#!/bin/bash
while read a;
do
n=$(eval echo $a)
if [[ "$n" == "" ]]
then
echo $a
else
echo $n
fi
done < input
Using this as input
${HOME}
${nonexistent}
Gives
/home/myuser
${nonexistent}
Easy to read? Maybe not. It is short and works though :-)
while read r; do
echo $(eval echo ${r%\}}:-'$r'\})
done < input
Magic used:
http://www.gnu.org/software/bash/manual/bashref.html#Shell-Parameter-Expansion
Edit: Further explanation, I hope it makes some sense.
We use two techniques; from the above docs:
${parameter:−word} If parameter is
unset or null, the expansion of word
is substituted. Otherwise, the value
of parameter is substituted.
And
${parameter%word}
The word is expanded to produce a
pattern just as in filename expansion.
If the pattern matches a trailing
portion of the expanded value of
parameter, then the result of the
expansion is the value of parameter
with the shortest matching pattern
(the ‘%’ case) or the longest matching
pattern (the ‘%%’ case) deleted. ...
We use the fact that the input is just what we can use in the shell, we have ${FOOBAR} but need ${FOOBAR:-'${FOOBAR}'} (Single quotes to avoid expansion).
# echo ${doesntexist:-Hello}
Hello
# doesexist=World
# echo ${doesexist:-Will not be printed}
World
So what we need to inject is :-'${FOOBAR}'
To achieve this we trim the } at the end, add the string, then put another } back afterwards.
# echo $r
${FOOBAR}
# echo ${r%\}}
${FOOBAR
The final \} isn't really necessary, since it's got no beginning in this case, but it's better to be explicit and escape it. (Much like you would escape echo \* even if echo * without any matching files gives you a literal *).
Edit2: This of course doesn't take into account that you wanted to support multiple variables in a single row; or any rows with other stuff in them.
while read name; do echo "$name = " $(eval echo $name); done < file_with_vars.txt
will echo all variables what know.
e.g.
in my file called vv
${PATH}
${HAVENOT}
${LOCALE}
will print
${PATH} = /usr/local/narwhal/bin:/opt/local/bin:/opt/local/sbin:/usr/bin:/bin:/usr/sbin:/sbin:/usr/local/bin:~/bin
${HAVENOT} =
${LOCALE} = UTF-8
modify the output format as you wish :)