(Sorry for the confusion. Previous $ sign occurred when I tried to simplify the actual problem. Thanks for correcting the question)
I wanted to split a directory name on underscores (ex: dir_to_split="my_test_dir") like this:
my_dir=($dir_to_split)
var=$(echo $my_dir | awk -F"_" '{print $1,$2,$3}')
set -- $var
splited_1=$1
splited_2=$2
splited_3=$3
now using these splited_x is causing me errors. ex.
myprograme $splited_1 $splited_2 $splited_3
Can anyone please help me with this ? Thank you....
(Rewritten after updated question.)
What kind of errors do you get? I find it useful to add set -x to the top of my shell scripts when debugging, this lets the shell print all commands it executes so you can pinpoint the line where problems begin.
Are you sure that $dir_to_split is actually set? Does it contain spaces or tabs? Does it contain two underscores? I don't see any other problems right now.
There are in-shell methods of splitting a variable such as:
dir="my_test_dir"
OIFS="$IFS"
IFS="-"
set --
IFS="$OIFS"
See also this SO question.
Related
I am trying to get this simple function working:
p4edit(){
p4 edit ${$1:25}
}
I read the other popular bad substitution question on SO and it did not seem to help me or be related to my problem. What am I doing wrong here? I want to cut off the first 25 characters of the argument provided to my function.
I have noticed a simple echo ${"test":3} fails the same way, but this succeeds:
test="test"
echo ${test:3}
I am just running this in a bash instance.
You have too much money! (too many dollar signs). Use:
p4edit(){
p4 edit ${1:25}
}
To extract the 25th-and-onwards characters from $1.
Why two times a $ ?
p4edit(){
echo ${1:25}
}
works fine for me. String functions in bash are a bit tricky, since they are not really consistent. But ${} already defines, that you are looking for a variable. So only submit the name to it. There are some stringfunctions with ${#var} but as far as I know, there is never a $ inside a ${}
Apologies for a seemingly inane question. But I have spent the whole day trying to figure it out and it drives me up the walls. I'm trying to write a seemingly simple bash script that would take a list of files in the directory from ls, replace part of the file names using sed, get unique names from the list and pass them onto some command. Like so:
inputs=`ls *.ext`
echo $inputs
test1_R1.ext test1_R2.ext test2_R1.ext test2_R2.ext
Now I would like to put it through sed to replace 1.ext and 2.ext with * to get test1_R* etc. Then I'd like to remove resulting duplicates by running sort -u to arrive to the following $outputs variable:
echo $outputs
test1_R* test2_R*
And pass this onto a command, like so
cat $outputs
I can do something like this in a command line:
ls *.ext | sed s/..ext/\*/g | sort -u
But if I try to assign the above to a variable in the script it just returns the output from the ls. I have tried several ways to do it: including the whole pipe in the script. Running each command separately and assigning it to a variable, then passing that variable to the next command and writing the outputs to files then passing the file to the next command. But so far none of this managed to achieve what I aimed to. I think my problem lies in (except general cluelessness aroung bash scripting) inability to run seq on a variable within script. There seems to be a lot of advice around in how to pass variables to pattern or replacement string in sed, but they all seem to take files as input. But I understand that it might not be the proper way of doing it anyway. Therefore I would really appreciate if someone could suggest an elegant way to achieve, what I'm trying to.
Many thanks!
Update 2/06/2014
Hi Barmar, thanks for your answer. Can't say it solved the problem, but it helped pin-pointing it. Seems like the problem is in me using the asterisk. I have to say, I'm very puzzled. The actual file names I've got are:
test1_R1.fastq.gz test1_R2.fastq.gz test2_R1.fastq.gz test2_R2.fastq.gz
If I'm using the code you suggested, which seems to me the right way do to it:
ins=$(ls *.fastq.gz | sed 's/..fastq.gz/\*/g' | sort -u)
Sed doesn't seem to do anything and I'm getting the output of ls:
test1_R1.fastq.gz test1_R2.fastq.gz test2_R1.fastq.gz test2_R2.fastq.gz
Now if I replace that backslash with anything else, the sed works, but it also returns whatever character I'm putting in front (or after) the asteriks:
ins=$(ls *.fastq.gz | sed 's/..fastq.gz/"*/g' | sort -u)
test1_R"* test2_R"*
That's odd enough, but surely I can just put an "R" in front of the asteriks and then replace R in the search pattern string, right? Wrong! If I do that whichever way: 's/R..fastq.gz/R*/g' 's/...fastq.gz/R*/g' 's/[A-Z]..fastq.gz/R*/g' I'm back to the original names! And even if I end up with something like test1_RR* test2_RR* and try to run it through sed again and replace "_R" for "_" or "RR" for "R", I'm having no luck and I'm back to the original names. And yet I can replace the rest of the file name no problem, just not to get me test1_R* I need.
I have a feeling I should be escaping that * in some very clever way, but nothing I've tried seems to work. Thanks again for your help!
This is how you capture the result of the whole pipeline in a variable:
var=$(ls *.ext | sed s/..ext/\*/g | sort -u)
I have written a small script with which I take the name of a File.
#objectname
echo "objectname"
read ON
Can't get simpler.
I do some processing with the file I get.
gpg -c --no-use-agent "$ON"
For example if I have a file a.exe --> It will encrypt it and give me a file with a different md5 and an extension. Now, the file looks this way a.exe.gpg
Now, if I give it a bind the name of the file directly.
like this for example:
Taken from : this link
# This works
fileName='a.exe.gpg'
md5sum=$(md5sum ${fileName})
echo $md5sum
it returns it properly.
What if I want to do it dynamically.
This is what I tried:
#does not work
gpg -c --no-use-agent "$ON"
fileName= `$ON.gpg`
md5sum= $(md5sum ${fileName})
echo $md5sum
I get this bug here: upload.sh: 1: upload.sh: Fire.exe.gpg: not found and the program does not exit.
May I ask where exactly is the mistake I am doing?
The error is here:
fileName= `$ON.gpg`
There should be no space after =. (Also look at the next line.)
You used back-quotes, which execute $ON.gpg rather than simply evaluating it. Back-quotes are the same as $(...) but less elegant. Use double-quotes for this.
Read Greg's wiki entry on quotes for an ultra-detailed explanation with opinionated commentary. :-)
Be careful when making assignments in shell script. Don't use spaces in any sides of the operator=. Try the following:
fileName="$ON.gpg"
md5sum=$(md5sum ${fileName})
Note that the variable and the assignment operator= are together with no space.
Also, when you use backticks as `expression`, it will be executed by shell like using $(expression), as pointed by user ghoti.
You goofed on fixing the filename.
fileName="$ON.gpg"
I am trying to make heads or tails of a shell script. Could someone please explain this line?
$FILEDIR is a directory containing files. F is a marker in an array of files that is returned from this command:
files=$( find $FILEDIR -type f | grep -v .rpmsave\$ | grep -v .swp\$ )
The confusing line is within a for loop.
for f in $files; do
target=${f:${#FILEDIR}}
<<do some more stuff>>
done
I've never seen the colon, and the hash before in a shell script for loop. I haven't been able to find any documentation on them... could someone try and enlighten me? I'd appreciate it.
There are no arrays involved here. POSIX sh doesn't have arrays (assuming you're not using another shell based upon the tags).
The colon indicates a Bash/Ksh substring expansion. These are also not POSIX. The # prefix expands to the number of characters in the parameter. I imagine they intended to chop off the directory part and assign it to target.
To explain the rest of that: first find is run and hilariously piped into two greps which do what could have been done with find alone (except breaking on possible filenames containing newlines), and the output saved into files. This is also something that can't really be done correctly if restricted only to POSIX tools, but there are better ways.
Next, files is expanded unquoted and mutalated by the shell in more ridiculous ways for the for loop to iterate over the meaningless results. If the rest of the script is this bad, probably throw it out and start over. There's no way that will do what's expected.
The colon can be as a substring. So:
A=abcdefg
echo ${A:4}
will print the output:
efg
I'm not sure why they would use a file directory as the 2nd parameter though...
If you are having problems understanding the for loop section, try http://www.dreamsyssoft.com/unix-shell-scripting/loop-tutorial.php
I have to ls command to get the details of certain types of files. The file name has a specific format. The first two words followed by the date on which the file was generated
e.g.:
Report_execution_032916.pdf
Report_execution_033016.pdf
Word summary can also come in place of report.
e.g.:
Summary_execution_032916.pdf
Hence in my shell script I put these line of codes
DATE=`date +%m%d%y`
Model=Report
file=`ls ${Model}_execution_*${DATE}_*.pdf`
But the value of Model always gets resolved to 'REPORT' and hence I get:
ls: cannot access REPORT_execution_*032916_*.pdf: No such file or directory
I am stuck at how the resolution of Model is happening here.
I can't reproduce the exact code here. Hence I have changed some variable names. Initially I had used the variable name type instead of Model. But Model is the on which I use in my actual code
You've changed your script to use Model=Report and ${Model} and you've said you have typeset -u Model in your script. The -u option to the typeset command (instead of declare — they're synonyms) means "convert the strings assigned to all upper-case".
-u When the variable is assigned a value, all lower-case characters are converted to upper-case. The lower-case attribute is disabled.
That explains the upper-case REPORT in the variable expansion. You can demonstrate by writing:
Model=Report
echo "Model=[${Model}]"
It would echo Model=[REPORT] because of the typeset -u Model.
Don't use the -u option if you don't want it.
You should probably fix your glob expression too:
file=$(ls ${Model}_execution_*${DATE}*.pdf)
Using $(…) instead of backticks is generally a good idea.
And, as a general point, learn how to Debug a Bash Script and always provide an MCVE (How to create a Minimal, Complete, and Verifiable Example?) so that we can see what your problem is more easily.
Some things to look at:
type is usually a reserved word, though it won't break your script, I suggest you to change that variable name to something else.
You are missing an $ before {DATE}, and you have an extra _ after it. If the date is the last part of the name, then there's no point in having an * at the end either. The file definition should be:
file=`ls ${type}_execution_*${DATE}.pdf`
Try debugging your code by parts: instead of doing an ls, do an echo of each variable, see what comes out, and trace the problem back to its origin.
As #DevSolar pointed out you may have problems parsing the output of ls.
As a workaround
ls | grep `date +%m%d%y` | grep "_execution_" | grep -E 'Report|Summary'
filters the ls output afterwards.
touch 'Summary_execution_032916.pdf'
DATE=`date +%m%d%y`
Model=Summary
file=`ls ${Model}_execution_*${DATE}*.pdf`
worked just fine on
GNU bash, version 4.3.11(1)-release (x86_64-pc-linux-gnu)
Part of question:
But the value of Model always gets resolved to 'REPORT' and hence I get:
This is due to the fact that in your script you have exported Model=Report
Part of question:
ls: cannot access REPORT_execution_*032916_*.pdf: No such file or directory
No such file our directory issue is due to the additional "_" and additional "*"s that you have put in your 3rd line.
Remove it and the error will be gone. Though, Model will still resolve to Report
Original 3rd line :
file=`ls ${Model}_execution_*${DATE}_*.pdf`
Change it to
file=`ls ${Model}_execution_${DATE}.pdf`
Above change will resolve the could not found issue.
Part of question
I am stuck at how the resolution of Model is happening here.
I am not sure what you are trying to achieve, but if you are trying to populate the file parameter with file name with anything_exection_someDate.pdf, then you can write your script as
DATE=`date +%m%d%y`
file=`ls *_execution_${DATE}.pdf`
If you echo the value of file you will get
Report_execution_032916.pdf Summary_execution_032916.pdf
as the answer
There were some other scripts which were invoked before the control reaches the line of codes which I mentioned in the question. In one such script there is a code
typeset -u Model
This sets the value of the variable model always to uppercase which was the reason this error was thrown
ls: cannot access REPORT_execution_032916_.pdf: No such file or directory
I am sorry that
i couldn't provide a minimal,complete and verifiable code