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_${DATE}.pdf`
If you echo the value of file you will get
Report_execution_032916.pdf Summary_execution_032916.pdf
as the answer
There were some other scripts which were invoked before the control reaches the line of codes which I mentioned in the question. In one such script there is a code
typeset -u Model
This sets the value of the variable model always to uppercase which was the reason this error was thrown
ls: cannot access REPORT_execution_032916_.pdf: No such file or directory
I am sorry that
i couldn't provide a minimal,complete and verifiable code
Related
A config file that the last line contains data that I want to assign everything to the RIGHT of the = sign into a variable that I can display and call later in the script.
Example: /path/to/magic.conf:
foo
bar
ThisOption=foo.bar.address:location.555
What would be the best method in a bash shell script to read the last line of the file and assign everything to the right of the equal sign? In this case, foo.bar.address:location.555.
The last line always has what I want to target and there will only ever be a single = sign in the file that happens to be the last line.
Google and searching here yielded many close but non-relative results with using sed/awk but I couldn't come up with exactly what I'm looking for.
Use sed:
variable=$(sed -n 's/^ThisOption=//p' /path/to/magic.conf)
echo "The option is: $variable")
This works by finding and removing the ThisOption= marker at the start of the line, and printing the result.
IMPORTANT: This method absolutely requires that the file be trusted 100%. As mentioned in the comments, anytime you "eval" code without any sanitization there are grave risks (a la "rm -rf /" magnitude - don't run that...)
Pure, simple bash. (well...using the tail utility :-) )
The advantage of this method, is that it only requires you to know that it will be the last line of the file, it does not require you to know any information about that line (such as what the variable to the left of the = sign will be - information that you'd need in order to use the sed option)
assignment_line=$(tail -n 1 /path/to/magic.conf)
eval ${assignment_line}
var_name=${assignment_line%%=*}
var_to_give_that_value=${!var_name}
Of course, if the var that you want to have the value is the one that is listed on the left side of the "=" in the file then you can skip the last assignment and just use "${!var_name}" wherever you need it.
This may be poorly titled as I'm not fully sure what the process is called.
Basically I want to get only the last part of a symlink path, and I'm trying to use the same method I use with PWD.
For example:
if I do
PWD
it prints
/opt/ct/mydir
if I do
echo ${PWD##*/}
it prints only the last part
mydir
So using that design I can do
readlink mysymlink
which gives
/opt/ct/somedir
and I can do
TMP=$(readlink mysymlink)
echo ${TMP##*/}
and it will print
somedir
So now how can I combine that last part into one line like
TMP=$(readlink mysymlink && echo ${TMP##*/})
???
The example I show gives me 2 concatenated results.. one with the full path and one with just the part I want. I only want that last directory.
I also tried
TMP=${ $(readlink mysymlink)##*/}
to no avail
Variable substitution suffixes can only be used with variables, not command substitutions. You either have to set the variable and modify it in separate statements, as in your first attempt, or use additional command substitutions:
TMP=$(basename $(readlink))
I'm dealing with a pipeline of predominantly shell and Perl files, all of which pass parameters (paths) to the next. I decided it would be better to use a single file to store all the paths and just call that for every file. The issue is I am using awk to grab the files at the beginning of each file, and it's turning out to be a lot of repetition.
My question is: I do not know if there is a way to store key-value pairs in a file so shell can natively do something with the key and return the value? It needs to access an external file, because the pipeline uses many scripts and a map in a specific file would result in parameters being passed everywhere. Is there some little quirk I do not know of that performs a map function on an external file?
You can make a file of env var assignments and source that file as need, ie.
$ cat myEnvFile
path1=/x/y/z
path2=/w/xy
path3=/r/s/t
otherOpt1="-x"
Inside your script you can source with either . myEnvFile or the more versbose version of the same feature sourc myEnvFile (assuming bash shell) , i.e.
$cat myScript
#!/bin/bash
. /path/to/myEnvFile
# main logic below
....
# references to defined var
if [[ -d $path2 ]] ; then
cd $path2
else
echo "no pa4h2=$path2 found, can't continue" 1>&1
exit 1
fi
Based on how you've described your problem this should work well, and provide a-one-stop-shop for all of your variable settings.
IHTH
In bash, there's mapfile, but that reads the lines of a file into a numerically-indexed array. To read a whitespace-separated file into an associative array, I would
declare -A map
while read key value; do
map[$key]=$value
done < filename
However this sounds like an XY problem. Can you give us an example (in code) of what you're actually doing? When I see long piplines of grep|awk|sed, there's usually a way to simplify. For example, is passing data by parameters better than passing via stdout|stdin?
In other words, I'm questioning your statement "I decided it would be better..."
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"
(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.