Replace comma with space in shell script - linux

Replace comma with space using a shell script
Given the following input:
Test,10.10.10.10,"80,22,3306",connect
I need to get below output using a bash script
Test 10.10.10.10 "80,22,3306" connect

If you have gawk, you can use FPAT (field pattern), setting it to a regular expression.
awk -v FPAT='([^,]+)|(\"[^"]+\")' '{ for(i=1;i<=NF;i++) { printf "%s ",$i } }' <<< "Test,10.10.10.10,\"80,22,3306\",connect"
We set FPAT to separate the text based on anything that is not a comma and also data enclosed in quotation marks as as well as anything that is not a quotation mark. We then print all the fields with a spaces in between.

Considering if your Input_file is same as shown sample then following sed may help you in same too.
sed 's/\(.[^,]*\),\([^,]*\),\(".*"\),\(.*\)/\1 \2 \3 \4/g' Input_file

Assuming you can read your input from the file, this works
#!/usr/bin/bash
while read -r line;do
declare -a begin=$(echo $line | awk -F'"' '{print $1}' | tr "," " " )
declare -a end=$(echo $line |awk -F'"' '{print $3}' | tr "," " " )
declare -a middle=$(echo $line | awk -F'"' '{print $2}' )
echo "${begin[#]} \"${middle[#]}\" ${end[#]}"
done < connect_file
Edit: I see,that you want to keep the commas between port numbers. I have edited the script.

echo Test,10.10.10.10,\"80,22,3306\",connect|awk '{sub(/,/," ")gsub(/,"80,22,3306",/," \4280,22,3306\42 ")}1'
Test 10.10.10.10 "80,22,3306" connect

Related

Using awk to separate an output containing a tab and a "/" separators into a delimited format

I'll appreciate help in converting this output to a pipe delimited
I have the following output
abcde1234 /path/A/file1
test23455 /path/B/file2345
But I would like in
abcde1234|file1
test23455|file2345
In awk, If you set FS as [[:blank:]]+/|/ you can print the first and last fields:
awk -v FS='[[:blank:]]+/|/' -v OFS='|' '{print $1, $NF}' file
abcde1234|file1
test23455|file2345
Here is a one-liner awk solution:
awk -v FS='[ \t].*/' -v OFS='|' '{$1=$1}1' file
and, a sed one-liner:
sed 's%[[:blank:]].*/%|%' file
and a pure bash one
while read -r; do echo "${REPLY%%[[:blank:]]*}|${REPLY##*/}"; done < file
try to use cut 🤷🏻‍♀️.
abcde1234 /path/A/file1
test23455 /path/B/file2345
while IFS= read -r line; do
value1=$(echo $line | cut -d ' ' -f1)
value2=$(echo $line | cut -d '/' -f4)
printf "$value1 $value2\n"
done < <(cat list)

How to extract words between two characters in linux?

I have the following stored in a file named tmp.txt
user/config/jars/content-config-factory-3.2.0.0.jar
I need to store this word to a variable -
$variable=content-config-factory
I have written the following
while read line
do
var=$(echo $line | awk 'BEGIN{FS="\/"; OFS=" "} {print $NF}' )
var=$(echo $var | awk 'BEGIN{FS="-"; OFS=" "} {print $(1)}' )
echo $var
done < tmp.txt
This returns the result "content" instead of "content-config-factory".
Can anyone please tell me how to extract a word between two characters from a string efficiently.
An awk solution would be like
awk -F/ '{sub("-[^-]+$", "", $NF); print $NF}
Test
$ echo "user/config/jars/content-config-factory-3.2.0.0.jar" | awk -F/ '{sub("-[^-]+$", "", $NF); print $NF}'
content-config-factory
You can try this way also and get your expected result
variable=$(sed 's:.*/\(.*\)-.*:\1:' FileName)
echo $variable
OutPut :
content-config-factory
You could use grep,
grep -oP '(?<=/)[^/]*(?=-\d+\.)' file
Example:
$ var=$(echo 'user/config/jars/content-config-factory-3.2.0.0.jar' | grep -oP '(?<=/)[^/]*(?=-\d+\.)')
$ echo "$var"
content-config-factory

Linux/Unix bash basic script awk/sed

I'm working on bash script.
var=$(ls -t1 | head -n1);
cat $var | sed 's/"//g' > latest.csv
cat latest.csv | sed -e 's/^\|$/"/g' -e 's/,/","/g' > from_epos.csv
echo "LATEST: $var";
Here's the whole script, it's meant to delete all quotation mark from current file and add new one, between each field.
INPUT:
"sku","item","price","qty"
5135,"ITEM1",1.79,5
5338,"ITEM2",1.39,5
5318,"ITEM3",1.09,5
5235,"ITEM4",1.09,5
9706,"ITEM5",1.99,5
OUTPUT:
"sku","item","price","qty"
"5135","ITEM1","1.79","5
"
"5338","ITEM2","1.39","5
"
"5318","ITEM3","1.09","5
"
"5235","ITEM4","1.09","5
"
"9706","ITEM5","1.09","5
"
My ideal output is:
"sku","item","price","qty"
"5135","ITEM1","1.79","5"
"5338","ITEM2","1.39","5"
"5318","ITEM3","1.09","5"
"5235","ITEM4","1.09","5"
"9706","ITEM5","1.99","5"
It seems like it's entering random character between line in current output like
" and quotation mark is between CR and LF.
What's the problem and how to get it to my ideal vision?
Thanks,
Adam
awk 'BEGIN{FS=OFS=","}{gsub(/\"/,"");gsub(/[^,]+/,"\"&\"")}1' input
Solution using sed:
sed -e 's/"//g; s/,/","/g; s/^/"/; s/$/"/'
Long-piped-commented version:
sed -e 's/"//g' | # removes all quotations
sed -e 's/,/","/g' | # changes all colons to ","
sed -e 's/^/"/; s/$/"/' # puts quotations in the start and end of each line
awk can do all this in one command:
awk -F"," 'NR>1{for(i=1; i<=NF; i++) {if (!($i ~ /^"/)) printf("\"%s\"",$i);
else printf("%s",$i); if (i<NF) printf(","); else print "";}}' latest.csv
EDIT:
Try this awk: (modified from JS's suggested command)
awk 'BEGIN{FS=OFS=","}{gsub(/\"/,"");gsub(/[^,\r]+/,"\"&\"")}1'
OR
awk -F"[,\r]" 'NR==1{print} NR>1{for(i=1; i<NF; i++) {if (!($i ~ /^"/))
printf("\"%s\"",$i); else printf("%s",$i); if (i<NF-1) printf(",");
else print "";}}'

bash, extract string from text file with space delimiter

I have a text files with a line like this in them:
MC exp. sig-250-0 events & $0.98 \pm 0.15$ & $3.57 \pm 0.23$ \\
sig-250-0 is something that can change from file to file (but I always know what it is for each file). There are lines before and above this, but the string "MC exp. sig-250-0 events" is unique in the file.
For a particular file, is there a good way to extract the second number 3.57 in the above example using bash?
use awk for this:
awk '/MC exp. sig-250-0/ {print $10}' your.txt
Note that this will print: $3.57 - with the leading $, if you don't like this, pipe the output to tr:
awk '/MC exp. sig-250-0/ {print $10}' your.txt | tr -d '$'
In comments you wrote that you need to call it in a script like this:
while read p ; do
echo $p,awk '/MC exp. sig-$p/ {print $10}' filename | tr -d '$'
done < grid.txt
Note that you need a sub shell $() for the awk pipe. Like this:
echo "$p",$(awk '/MC exp. sig-$p/ {print $10}' filename | tr -d '$')
If you want to pass a shell variable to the awk pattern use the following syntax:
awk -v p="MC exp. sig-$p" '/p/ {print $10}' a.txt | tr -d '$'
More lines would've been nice but I guess you would like to have a simple use awk.
awk '{print $N}' $file
If you don't tell awk what kind of field-separator it has to use it will use just a space ' '. Now you just have to count how many fields you have got to get your field you want to get. In your case it would be 10.
awk '{print $10}' file.txt
$3.57
Don't want the $?
Pipe your awk result to cut:
awk '{print $10}' foo | cut -d $ -f2
-d will use the $ als field-separator and -f will select the second field.
If you know you always have the same number of fields, then
#!/bin/bash
file=$1
key=$2
while read -ra f; do
if [[ "${f[0]} ${f[1]} ${f[2]} ${f[3]}" == "MC exp. $key events" ]]; then
echo ${f[9]}
fi
done < "$file"

Remove a specific character using awk or sed

I have a command output from which I want to remove the double quotes ".
Command:
strings -a libAddressDoctor5.so |\
grep EngineVersion |\
awk '{if(NR==2)print}' |\
awk '{print$2}'
Output:
EngineVersion="5.2.5.624"
I'd like to know how to remove unwanted characters with awk or sed.
Use sed's substitution: sed 's/"//g'
s/X/Y/ replaces X with Y.
g means all occurrences should be replaced, not just the first one.
Using just awk you could do (I also shortened some of your piping):
strings -a libAddressDoctor5.so | awk '/EngineVersion/ { if(NR==2) { gsub("\"",""); print $2 } }'
I can't verify it for you because I don't know your exact input, but the following works:
echo "Blah EngineVersion=\"123\"" | awk '/EngineVersion/ { gsub("\"",""); print $2 }'
See also this question on removing single quotes.
tr can be more concise for removing characters than sed or awk, especially when you want to remove multiple different characters from a string.
Removing double quotes:
echo '"Hi"' | tr -d \"
# Prints Hi without quotes
Removing different kinds of brackets:
echo '[{Hi}]' | tr -d {}[]
# Prints Hi without brackets
-d stands for "delete".

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