shell scripting quotation - shell

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"

Related

Renaming the file Directory which contains Space based on CSV in Shell

I need to rename the files inside the folder that has a space in it eg(Deco/main library/file1.txt )
code:
while IFS="," read orig new pat
do
mv -v $pat$new $pat$orig
done < new.csv
csv file:
newname,file1.txt,Deco/main\\\ library/
error:
mv: invalid option -- '\'
Welcome to Stackoverflow!
First: Use quotes around the use of variables. That means except in very rare occasions, you always should use "$foo" instead of $foo because if you are using the latter, the shell is supposed (and will) interpret spaces in the variables as word delimiters which you rarely want. Especially in your case you do not want it.
Second: Your CSV file seems to contain backslashes to quote the spaces. And some additional step seems to have added another level of quotation so than now you end up with three backslashes and a space for each original space. If this really is the case (please double check if what you wrote in your question is correct, otherwise my answer doesn't fit), you need to unquote this before you can use it.
There are security issues involved in using eval, so do not use it lightly (this disclaimer is necessary whenever proposing to use eval), but if you have trust in the input you are handling to not contain any nastinesses, then you can do this using this code:
while IFS="," read orig new pat
do
eval eval mv -v "$pat$new" "$pat$orig"
done < new.csv
Using this, two levels of quotation are evaluated (that's what eval does) before the mv command is executed.
I strongly suggest to do a dry run by adding echo before the mv first. Then instead of executing your commands they are merely printed first.

Bad substitution on bash substringing operation

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 ${}

Way to create multiline comments in Bash?

I have recently started studying shell script and I'd like to be able to comment out a set of lines in a shell script. I mean like it is in case of C/Java :
/* comment1
comment2
comment3
*/`
How could I do that?
Use : ' to open and ' to close.
For example:
: '
This is a
very neat comment
in bash
'
Multiline comment in bash
: <<'END_COMMENT'
This is a heredoc (<<) redirected to a NOP command (:).
The single quotes around END_COMMENT are important,
because it disables variable resolving and command resolving
within these lines. Without the single-quotes around END_COMMENT,
the following two $() `` commands would get executed:
$(gibberish command)
`rm -fr mydir`
comment1
comment2
comment3
END_COMMENT
Note: I updated this answer based on comments and other answers, so comments prior to May 22nd 2020 may no longer apply. Also I noticed today that some IDE's like VS Code and PyCharm do not recognize a HEREDOC marker that contains spaces, whereas bash has no problem with it, so I'm updating this answer again.
Bash does not provide a builtin syntax for multi-line comment but there are hacks using existing bash syntax that "happen to work now".
Personally I think the simplest (ie least noisy, least weird, easiest to type, most explicit) is to use a quoted HEREDOC, but make it obvious what you are doing, and use the same HEREDOC marker everywhere:
<<'###BLOCK-COMMENT'
line 1
line 2
line 3
line 4
###BLOCK-COMMENT
Single-quoting the HEREDOC marker avoids some shell parsing side-effects, such as weird subsitutions that would cause crash or output, and even parsing of the marker itself. So the single-quotes give you more freedom on the open-close comment marker.
For example the following uses a triple hash which kind of suggests multi-line comment in bash. This would crash the script if the single quotes were absent. Even if you remove ###, the FOO{} would crash the script (or cause bad substitution to be printed if no set -e) if it weren't for the single quotes:
set -e
<<'###BLOCK-COMMENT'
something something ${FOO{}} something
more comment
###BLOCK-COMMENT
ls
You could of course just use
set -e
<<'###'
something something ${FOO{}} something
more comment
###
ls
but the intent of this is definitely less clear to a reader unfamiliar with this trickery.
Note my original answer used '### BLOCK COMMENT', which is fine if you use vanilla vi/vim but today I noticed that PyCharm and VS Code don't recognize the closing marker if it has spaces.
Nowadays any good editor allows you to press ctrl-/ or similar, to un/comment the selection. Everyone definitely understands this:
# something something ${FOO{}} something
# more comment
# yet another line of comment
although admittedly, this is not nearly as convenient as the block comment above if you want to re-fill your paragraphs.
There are surely other techniques, but there doesn't seem to be a "conventional" way to do it. It would be nice if ###> and ###< could be added to bash to indicate start and end of comment block, seems like it could be pretty straightforward.
After reading the other answers here I came up with the below, which IMHO makes it really clear it's a comment. Especially suitable for in-script usage info:
<< ////
Usage:
This script launches a spaceship to the moon. It's doing so by
leveraging the power of the Fifth Element, AKA Leeloo.
Will only work if you're Bruce Willis or a relative of Milla Jovovich.
////
As a programmer, the sequence of slashes immediately registers in my brain as a comment (even though slashes are normally used for line comments).
Of course, "////" is just a string; the number of slashes in the prefix and the suffix must be equal.
I tried the chosen answer, but found when I ran a shell script having it, the whole thing was getting printed to screen (similar to how jupyter notebooks print out everything in '''xx''' quotes) and there was an error message at end. It wasn't doing anything, but: scary. Then I realised while editing it that single-quotes can span multiple lines. So.. lets just assign the block to a variable.
x='
echo "these lines will all become comments."
echo "just make sure you don_t use single-quotes!"
ls -l
date
'
what's your opinion on this one?
function giveitauniquename()
{
so this is a comment
echo "there's no need to further escape apostrophes/etc if you are commenting your code this way"
the drawback is it will be stored in memory as a function as long as your script runs unless you explicitly unset it
only valid-ish bash allowed inside for instance these would not work without the "pound" signs:
1, for #((
2, this #wouldn't work either
function giveitadifferentuniquename()
{
echo nestable
}
}
Here's how I do multiline comments in bash.
This mechanism has two advantages that I appreciate. One is that comments can be nested. The other is that blocks can be enabled by simply commenting out the initiating line.
#!/bin/bash
# : <<'####.block.A'
echo "foo {" 1>&2
fn data1
echo "foo }" 1>&2
: <<'####.block.B'
fn data2 || exit
exit 1
####.block.B
echo "can't happen" 1>&2
####.block.A
In the example above the "B" block is commented out, but the parts of the "A" block that are not the "B" block are not commented out.
Running that example will produce this output:
foo {
./example: line 5: fn: command not found
foo }
can't happen
Simple solution, not much smart:
Temporarily block a part of a script:
if false; then
while you respect syntax a bit, please
do write here (almost) whatever you want.
but when you are
done # write
fi
A bit sophisticated version:
time_of_debug=false # Let's set this variable at the beginning of a script
if $time_of_debug; then # in a middle of the script
echo I keep this code aside until there is the time of debug!
fi
in plain bash
to comment out
a block of code
i do
:||{
block
of code
}

Executing a command in a string in bash

I have a string that contains a command that I want to execute in a bash script. How can I do that?
Sorry for so basic question but I am new in bash.
This is my code:
echo "What is the path to save the result files?"
read out_path
end_cm1=$"fastqc -o "$out_path$" --noextract -fastq "$files1
And I want to execute the instruction that is in the end_cm1 variable.
First, you don't have to put that command in a string at all: you can just do this:
fastqc -o "$out_path" --noextract -fastq $files1
(And I'd recommend putting $out_path in quotes here in case the path has a space in it. I've not put $files1 in quotes because your variable is plural so I assume there's more than one; you should beware spaces in those file names also.)
Second, the answer to the question you asked is eval:
eval $end_cm1
You just have a slight syntax issue in your string:
end_cm1="fastqc -o $out_path --noextract -fastq $files1"
$enc_cm1
Having said that, #ams is right about not needing to assign this
to a string in the first place, and about the risks involved
in not quoting $files1.

ls command in UNIX

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

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