I am new to shell script. I am sourcing a file, which is created in Windows and has carriage returns, using the source command. After I source when I append some characters to it, it always comes to the start of the line.
test.dat (which has carriage return at end):
testVar=value123
testScript.sh (sources above file):
source test.dat
echo $testVar got it
The output I get is
got it23
How can I remove the '\r' from the variable?
yet another solution uses tr:
echo $testVar | tr -d '\r'
cat myscript | tr -d '\r'
the option -d stands for delete.
You can use sed as follows:
MY_NEW_VAR=$(echo $testVar | sed -e 's/\r//g')
echo ${MY_NEW_VAR} got it
By the way, try to do a dos2unix on your data file.
Because the file you source ends lines with carriage returns, the contents of $testVar are likely to look like this:
$ printf '%q\n' "$testVar"
$'value123\r'
(The first line's $ is the shell prompt; the second line's $ is from the %q formatting string, indicating $'' quoting.)
To get rid of the carriage return, you can use shell parameter expansion and ANSI-C quoting (requires Bash):
testVar=${testVar//$'\r'}
Which should result in
$ printf '%q\n' "$testVar"
value123
use this command on your script file after copying it to Linux/Unix
perl -pi -e 's/\r//' scriptfilename
Pipe to sed -e 's/[\r\n]//g' to remove both Carriage Returns (\r) and Line Feeds (\n) from each text line.
for a pure shell solution without calling external program:
NL=$'\n' # define a variable to reference 'newline'
testVar=${testVar%$NL} # removes trailing 'NL' from string
Related
I am trying to create a string eg (file1.txt,file2.txt,file3.txt).
All these 3 files names are within a file.
ls file*.txt > lstfiles.txt
while read filename; do
filename+=$line","
done <lstfiles.txt
This returns me with output:
file1.txt,file2.txt,file3.txt,
How can I find the last iteration of the loop so I dont add another comma at the end.
Required output:
file1.txt,file2.txt,file3.txt
For your use case I would rather get rid of the while loop and combine sedand tr commands like so:
sed -e '$ ! s/$/,/g' lstfiles.txt | tr -d '\n'
Where sed command replace each line endings execept the last one with a comma and tr command remove the linebreaks.
Probably avoid using ls in scripts though.
printf '%s,' file*.txt |
sed 's/,$/\n/'
assuming your sed recognizes \n to be a newline, and copes with input which doesn't have a final newline.
I am new to shell script. I am sourcing a file, which is created in Windows and has carriage returns, using the source command. After I source when I append some characters to it, it always comes to the start of the line.
test.dat (which has carriage return at end):
testVar=value123
testScript.sh (sources above file):
source test.dat
echo $testVar got it
The output I get is
got it23
How can I remove the '\r' from the variable?
yet another solution uses tr:
echo $testVar | tr -d '\r'
cat myscript | tr -d '\r'
the option -d stands for delete.
You can use sed as follows:
MY_NEW_VAR=$(echo $testVar | sed -e 's/\r//g')
echo ${MY_NEW_VAR} got it
By the way, try to do a dos2unix on your data file.
Because the file you source ends lines with carriage returns, the contents of $testVar are likely to look like this:
$ printf '%q\n' "$testVar"
$'value123\r'
(The first line's $ is the shell prompt; the second line's $ is from the %q formatting string, indicating $'' quoting.)
To get rid of the carriage return, you can use shell parameter expansion and ANSI-C quoting (requires Bash):
testVar=${testVar//$'\r'}
Which should result in
$ printf '%q\n' "$testVar"
value123
use this command on your script file after copying it to Linux/Unix
perl -pi -e 's/\r//' scriptfilename
Pipe to sed -e 's/[\r\n]//g' to remove both Carriage Returns (\r) and Line Feeds (\n) from each text line.
for a pure shell solution without calling external program:
NL=$'\n' # define a variable to reference 'newline'
testVar=${testVar%$NL} # removes trailing 'NL' from string
I am new to shell script. I am sourcing a file, which is created in Windows and has carriage returns, using the source command. After I source when I append some characters to it, it always comes to the start of the line.
test.dat (which has carriage return at end):
testVar=value123
testScript.sh (sources above file):
source test.dat
echo $testVar got it
The output I get is
got it23
How can I remove the '\r' from the variable?
yet another solution uses tr:
echo $testVar | tr -d '\r'
cat myscript | tr -d '\r'
the option -d stands for delete.
You can use sed as follows:
MY_NEW_VAR=$(echo $testVar | sed -e 's/\r//g')
echo ${MY_NEW_VAR} got it
By the way, try to do a dos2unix on your data file.
Because the file you source ends lines with carriage returns, the contents of $testVar are likely to look like this:
$ printf '%q\n' "$testVar"
$'value123\r'
(The first line's $ is the shell prompt; the second line's $ is from the %q formatting string, indicating $'' quoting.)
To get rid of the carriage return, you can use shell parameter expansion and ANSI-C quoting (requires Bash):
testVar=${testVar//$'\r'}
Which should result in
$ printf '%q\n' "$testVar"
value123
use this command on your script file after copying it to Linux/Unix
perl -pi -e 's/\r//' scriptfilename
Pipe to sed -e 's/[\r\n]//g' to remove both Carriage Returns (\r) and Line Feeds (\n) from each text line.
for a pure shell solution without calling external program:
NL=$'\n' # define a variable to reference 'newline'
testVar=${testVar%$NL} # removes trailing 'NL' from string
This question already has answers here:
Sed Insert Multiple Lines
(3 answers)
Closed 1 year ago.
I have a multiline variable that I captured from STDOUT.
I want to insert an echo command using this multiline variable to line 15 in another script (target).
#!/bin/bash
TEST=`cat foo`
echo "$TEST"
sed -i "15i echo \"$TEST\" > someotherfile" target
Contents of foo :
apples
oranges
bananas
carrots
I thought the sed command read in line feeds, which I confirmed my foo has:
user#test$ cat foo | tr -cd '\n' | wc -c
4
When I run my test.sh script, I see what's in $TEST, but am getting an error for the sed command:
user#test$ ./test.sh
apples
oranges
bananas
carrots
sed: -e expression #1, char 18: unknown command: `o'
What am I doing wrong?
Thanks in advance.
GNU sed is assumed, as implied by the syntax used in the question.
#!/bin/bash
# Read contents of file 'foo' into shell variable $test.
test=$(<foo)
# \-escape the newlines in $test for use in Sed.
testEscapedForSed=${test//$'\n'/\\$'\n'}
sed -i "15i echo \"$testEscapedForSed\" > someotherfile" target
Your problem was that passing multi-line strings to sed functions such as i (insert) requires the newlines embedded in those strings to be \-escaped, so that sed knows where the string ends and additional commands, if any, start.
A (nonstandard) parameter expansion is used to replace all newlines in $test with themselves prefixed by \, using ANSI C-quoted string $'\n' to generate actual newline chars.
Also note:
I've renamed TEST to test, because all-uppercase shell-variable names should be avoided.
I've used modern command-substitution syntax $(..) in lieu of legacy syntax `...`.
$(<foo) is a slightly more efficient - although nonstandard - way of reading the content of a file at once.
Try:
Solution1:
awk 'NR==15{print;system("cat foo");next} 1' Input_file
No need to get the complete file into a variable, we could simply print it whichever line of Input_file you want to print it.
Solution2:
line=15; sed -e "${line}r foo" target
Or (in script mode)
cat script.ksh
line=15;
sed -e "${line}r foo" target
Where you could change the number of line where you want to insert the lines from another file.
The i command in sed inserts the lines of text that end with a newline, up until a line that doesn't end with a backslash. The a and c commands are similar. Classic sed doesn't like the first line to appear on the same line as the i command; GNU sed isn't as fussy.
If you were writing the command manually, you'd need to write:
15i\
echo "apples\
oranges\
bananas\
carrots" > someotherfile
At issue now is "how do you want to create this given the file foo contains the list of names?". Sometimes, using sed to generate the sed script is useful. However, it can also be intricate if you need to get backslashes at the ends of lines which are subject to an i (or a or c) command, and it is simpler to circumvent the problem.
{
echo "15i\\"
sed -e '1s/^/echo "/' -e 's/$/\\/' -e '$s/\\$/" > someotherfile/' foo
} | sed -f /dev/stdin target
GNU sed can read its script from standard input using -f -; BSD (macOS) sed doesn't like that, but you can use -f /dev/stdin instead (which also works with GNU sed), at least on systems where there is a /dev/stdin.
Interesting issue.
As already mentioned the whole story for sed to be able to insert multiline text in another file is that this new multiline text must have actually literral \n for line breaks.
So we can use sed to convert real new line chars to literal \n:
$ a=$(tr '\n' '\\' <file3 |sed 's#[\]$##' |sed "s#[\]#\0n#g")
#Alternative: a=$(sed "s#[\]#\0n#g" <(sed 's#[\]$##' <(tr '\n' '\\' <file3)))
$ echo "$a"
apples\noranges\nbananas\ncarrots
How this translation works:
* First we replace all new lines with a single backslash using tr
* Then we remove the backslash from the end of the string
* Then we replace all other backslashes with backaslash and n char.
Since now variable $a contains literal \n between lines, sed will translate them back to actuall new lines:
$ cat file4
Line1
line2
line3
$ sed "2i $a" file4
Line1
apples
oranges
bananas
carrots
line2
line3
Result:
Mutliline replacement can be done with two commands:
$ a=$(tr '\n' '\\' <file3 |sed 's#[\]$##' |sed "s#[\]#\0n#g")
$ sed "2i $a" file4
sed 2i means insert a text before line2. 2a can be used in order to insert something after line2.
Remark:
According to this post which seems to be a duplicate, translation of new lines to literal \n seems that can be done with just :
a=$(echo ${a} | tr '\n' "\\n")
But this method never worked in my system.
Remark2:
The sed operation sed "2i $a" = insert variable $a before line 2 , can be also expressed as sed "1 s/.*/\0\n$a/" = replace all chars of first line with the same chars \0 plus a new line \n plus the contents of variable $a => insert $a after line1 = insert $a before line2.
I am new to shell script. I am sourcing a file, which is created in Windows and has carriage returns, using the source command. After I source when I append some characters to it, it always comes to the start of the line.
test.dat (which has carriage return at end):
testVar=value123
testScript.sh (sources above file):
source test.dat
echo $testVar got it
The output I get is
got it23
How can I remove the '\r' from the variable?
yet another solution uses tr:
echo $testVar | tr -d '\r'
cat myscript | tr -d '\r'
the option -d stands for delete.
You can use sed as follows:
MY_NEW_VAR=$(echo $testVar | sed -e 's/\r//g')
echo ${MY_NEW_VAR} got it
By the way, try to do a dos2unix on your data file.
Because the file you source ends lines with carriage returns, the contents of $testVar are likely to look like this:
$ printf '%q\n' "$testVar"
$'value123\r'
(The first line's $ is the shell prompt; the second line's $ is from the %q formatting string, indicating $'' quoting.)
To get rid of the carriage return, you can use shell parameter expansion and ANSI-C quoting (requires Bash):
testVar=${testVar//$'\r'}
Which should result in
$ printf '%q\n' "$testVar"
value123
use this command on your script file after copying it to Linux/Unix
perl -pi -e 's/\r//' scriptfilename
Pipe to sed -e 's/[\r\n]//g' to remove both Carriage Returns (\r) and Line Feeds (\n) from each text line.
for a pure shell solution without calling external program:
NL=$'\n' # define a variable to reference 'newline'
testVar=${testVar%$NL} # removes trailing 'NL' from string