This question already has answers here:
How do I properly quote this bash pipeline for watch?
(2 answers)
Closed 3 years ago.
How can I echo this entire piece of code to .bashrc without leaving out a single character?
# automatic logging of terminal input/output
test "$(ps -ocommand= -p $PPID | awk '{print $1}')" == 'script' || (script -f -q /home/user/.logs/terminal/manjaro/$(date +"%Y-%m- %d_%H:%M:%S")_terminal.log)
When I attempt to enter the following into terminal:
echo "the above code" >> ~/.bashrc
I get the following appended to .bashrc which is nothing like "the above code", its short about 45 or so characters.
# automatic logging of terminal input/output
test script == 'script' || (script -f -q /home/user/.logs/terminal/manjaro/2019-05- 08_09:09:19_terminal.log)
As you can see, it's leaving out A LOT of the original code. I understand this has a lot to do with the number of different quotations and placement, but without altering my code much, or at least to the point where it can still function as its intended, how can I go about getting this to echo to the file properly?
Thank you for every nanosecond of your time.
Wrap your echo'd string with single quotes ' instead of double "
Related
This question already has an answer here:
Tilde not expanded when quoting on the right hand side of a Bash variable assignment [duplicate]
(1 answer)
Closed 1 year ago.
I have a function which should loop through every customer in an array and execute a grep to that directory.
test(){
set -x;
export customers=( customer1 customer2 customer3 );
export repo_path="~/repodir";
export output_path='/tmp';
for i in "${customers[#]}"
do
echo "${repo_path}/PEC/repo-${i}/build.yml"
grep 'link: ' $repo_path/PEC/repo-$i/build.yml | cut -d '/' -f 2 | sed 's/.git//g'
done | sort -u > ${output_path}/repos.txt
echo "${output_path}/repos.txt";
}
For some reason I get the following error message:
grep: ~/repodir/PEC/customer1/build.yml: No such file or directory
But when I check that exact same path I can see and read the file...
The first echo command also doesn't seem to be executing.
When I replace grep 'link: ' $repo_path/PEC/repo-$i/build.yml with grep 'link: ' ~/repodir/PEC/repo-$i/build.yml it does work.
I have tried various ways to define the variable like ${repo_path}, adding different types of quotes, ... So I basically don't know what I can do to make it work anymore.
$HOME is an environment variable, that is set to contain the home folder of the current user. The terminal session should have access to this environment variable at all times, unless this has been removed.
~ is a shell expansion symbol, one of the symbols that is processed before the actual command is performed. ~ alone expands to the value of $HOME.
Your code,
export repo_path="~/repodir";
the ~ is in a string and may not be processed, if you want to use a ~ try escaping the character. For readability using the $HOME variable would be simpler.
This question already has answers here:
Capturing multiple line output into a Bash variable
(7 answers)
Closed 3 years ago.
I have a small script where I appended the output of linux mpstat to a log file.
#/bin/bash
CPU_USAGE=$(mpstat)
echo $CPU_USAGE >> temp.log
The problem is that the output of mpstat on the terminal is formatted properly in 3 lines like so
However, the output to the file is all in one line.
How do I format the output like the one on the terminal?
Just quote the variable so it is not seen as several different parameters to be printed one after the other:
echo "$CPU_USAGE" >> temp.log
You could just directly pipe the output to the file:
#!/bin/bash
mpstat >> temp.log
If you must store it in a variable, then quote it like:
#!/bin/bash
CPU_USAGE=$(mpstat)
echo "$CPU_USAGE" >> temp.log
Otherwise, bash will not interpret the newlines as part of the message to echo, but the whole output as a list of short strings to output.
This question already has answers here:
A variable modified inside a while loop is not remembered
(8 answers)
Closed 3 years ago.
I have s simple bash script as follow:
interval=""
cat conf.param|\
while read param
do
item_=$(echo $param|cut -d "=" -f1)
case ${item_} in
interval)
interval=$(echo $param|cut -d "=" -f2)
echo $interval
;;
method)
method=$(echo $param|cut -d "=" -f2)
;;
esac
done
echo "${interval}"
It reads the contents of a file and stores them in different variables. the issue is that the variables are not set properly inside case segment. I put two echos. The first one (inside case) displays the interval value correctly which is '2', but the second one just after the esac statement displays nothing! it shows an empty blank line.
the conf.param is a simple text file. it has more lines I only printed two lines:
interval=2
method="POST"
Your problem is that using a pipe ("cat conf.param | while read param"), you call a second shell which can not export its variables to the calling one. See this example:
interval=""
cat tmp.txt | while read param; do ##### DON'T DO THIS
interval="A$interval"
done
echo "First attempt: $interval"
interval=""
while read param; do
interval="A$interval"
done < tmp.txt ##### BUT DO THIS INSTEAD
echo "Redirection attempt: $interval"
The file tmp.txt contains 4 lines; the output of the script is:
First attempt:
Redirection attempt: AAAA
As you see, the first attempt retains the old value of interval. The second/redirection attempt instead works because no new process is created.
This question already has answers here:
Variables overwriting text problem with "echo" in Bash
(2 answers)
Closed 4 years ago.
I wrote a script to change .CSV to json.
#!/bin/bash
exec 0< example.csv
while IFS=, read name element input decrease
do
echo "${element}decrease: ${decrease}test"
done
the example.csv I paste here
name1,A,11,12
name2,B,13,14
But the output is really weird...
testrease: 12
testrease: 14
As u can see, The test rewrite Adecrease and Bdecrease, makes them to testrease.
I can't believe it!! So I tried with out exec 0< example.csv, type them in stdin, this time I got what I want
name1,A,11,12
Adecrease: 12test
So I guess maybe there are some characters in example.csv I can't see which makes this problem. I use cat -v example.csv
name1,A,11,12^M
name2,B,13,14^M
Nothing strange and I stuck here.
I am very new to shell script, so if anyone can give me some suggestions I will be really thrilled!!
Thank u, #chepner! tag wiki saves me another hour on this stupid question.
And here is the solution from wiki:
Check whether your script or data has DOS style end-of-line characters.
Use cat -v yourfile or echo "$yourvariable" | cat -v.
DOS carriage returns will show up as ^M after each line.
If you find them, delete them using dos2unix (a.k.a. fromdos) or tr -d '\r'.
This question already has answers here:
Capturing multiple line output into a Bash variable
(7 answers)
Closed 5 years ago.
I'm trying to capture a block of text into a variable, with newlines maintained, then echo it.
However, the newlines don't seemed to be maintained when I am either capturing the text or displaying it.
Any ideas regarding how I can accomplish this?
Example:
#!/bin/bash
read -d '' my_var <<"BLOCK"
this
is
a
test
BLOCK
echo $my_var
Output:
this is a test
Desired output:
this
is
a
test
You need to add " quotes around your variable.
echo "$my_var"