How to find and replace using sed on records starting with Specific numberReplace non-printable characters in...
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How to find and replace using sed on records starting with Specific number
Replace non-printable characters in perl and sedSed Replace at a specific group of positionsHow to replace variables strings with special characters in sedsearch and Replace substring in mac address in a text fileReplace a pattern in File1 and replace it with corresponding matched pattern + column in File2using sed to replace a string in files which are in different foldersAdd a new line (n) in front of every forward arrow (>)Replace string in specific line at specific position of fixed-length fileStorage disk info - Replace multiple Input values to output fileReplace substring of characters with awk and sed
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
I Have a fixed width file with the below format
012019-06-03070005000799111160300000030XXXXXXX0700000000030
012019-06-03070005000799165030700000030XXXXXXX0700000000030
012019-06-03070005000799175500700000030XXXXXXX0700000000030
022019-06-030007276384I06000000000000207000991755007000000300
022019-06-030007276384I06000000000000107000991755007000000300
012019-06-03070005000799175840700000030XXXXXXX0700000000030
022019-06-030007276384I06000000000000407000991758407000000300
022019-06-030007276384I06000000000000307000991758407000000300
012019-06-03070005000799194080700000030XXXXXXX0700000000030
012019-06-03070005000790035750700000030XXXXXXX0700000000030
012019-06-03070005000790036660700000030XXXXXXX0700000000030
i need work on the lines starting with 02 and have to replace the characters from 20th position to 50th position with "MMMM" preserving the spaces till the 50th position.
My output should look like this
i tried sed -Ee '/^02"s/((.20).{30}/1$(printf "%-30s" MMMM)/"', which replaces on the records starting with 01 and 02 both. but I want to work on records starting with 02.
linux shell-script sed
edited 30 mins ago
msp9011
5,3695 gold badges42 silver badges69 bronze badges
asked 48 mins ago
ArunArun
11 bronze badge
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Arun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
I Have a fixed width file with the below format
012019-06-03070005000799111160300000030XXXXXXX0700000000030
012019-06-03070005000799165030700000030XXXXXXX0700000000030
012019-06-03070005000799175500700000030XXXXXXX0700000000030
022019-06-030007276384I06000000000000207000991755007000000300
022019-06-030007276384I06000000000000107000991755007000000300
012019-06-03070005000799175840700000030XXXXXXX0700000000030
022019-06-030007276384I06000000000000407000991758407000000300
022019-06-030007276384I06000000000000307000991758407000000300
012019-06-03070005000799194080700000030XXXXXXX0700000000030
012019-06-03070005000790035750700000030XXXXXXX0700000000030
012019-06-03070005000790036660700000030XXXXXXX0700000000030
i need work on the lines starting with 02 and have to replace the characters from 20th position to 50th position with "MMMM" preserving the spaces till the 50th position.
My output should look like this
i tried sed -Ee '/^02"s/((.20).{30}/1$(printf "%-30s" MMMM)/"', which replaces on the records starting with 01 and 02 both. but I want to work on records starting with 02.
linux shell-script sed
edited 30 mins ago
msp9011
5,3695 gold badges42 silver badges69 bronze badges
asked 48 mins ago
ArunArun
11 bronze badge
New contributor
Arun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
the desired output should be cross-checked... It's not only doing for line starts with "02".
– msp9011
23 mins ago
add a comment |
I Have a fixed width file with the below format
012019-06-03070005000799111160300000030XXXXXXX0700000000030
012019-06-03070005000799165030700000030XXXXXXX0700000000030
012019-06-03070005000799175500700000030XXXXXXX0700000000030
022019-06-030007276384I06000000000000207000991755007000000300
022019-06-030007276384I06000000000000107000991755007000000300
012019-06-03070005000799175840700000030XXXXXXX0700000000030
022019-06-030007276384I06000000000000407000991758407000000300
022019-06-030007276384I06000000000000307000991758407000000300
012019-06-03070005000799194080700000030XXXXXXX0700000000030
012019-06-03070005000790035750700000030XXXXXXX0700000000030
012019-06-03070005000790036660700000030XXXXXXX0700000000030
i need work on the lines starting with 02 and have to replace the characters from 20th position to 50th position with "MMMM" preserving the spaces till the 50th position.
My output should look like this
i tried sed -Ee '/^02"s/((.20).{30}/1$(printf "%-30s" MMMM)/"', which replaces on the records starting with 01 and 02 both. but I want to work on records starting with 02.
linux shell-script sed
edited 30 mins ago
msp9011
5,3695 gold badges42 silver badges69 bronze badges
asked 48 mins ago
ArunArun
11 bronze badge
New contributor
Arun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I Have a fixed width file with the below format
012019-06-03070005000799111160300000030XXXXXXX0700000000030
012019-06-03070005000799165030700000030XXXXXXX0700000000030
012019-06-03070005000799175500700000030XXXXXXX0700000000030
022019-06-030007276384I06000000000000207000991755007000000300
022019-06-030007276384I06000000000000107000991755007000000300
012019-06-03070005000799175840700000030XXXXXXX0700000000030
022019-06-030007276384I06000000000000407000991758407000000300
022019-06-030007276384I06000000000000307000991758407000000300
012019-06-03070005000799194080700000030XXXXXXX0700000000030
012019-06-03070005000790035750700000030XXXXXXX0700000000030
012019-06-03070005000790036660700000030XXXXXXX0700000000030
i need work on the lines starting with 02 and have to replace the characters from 20th position to 50th position with "MMMM" preserving the spaces till the 50th position.
My output should look like this
i tried sed -Ee '/^02"s/((.20).{30}/1$(printf "%-30s" MMMM)/"', which replaces on the records starting with 01 and 02 both. but I want to work on records starting with 02.
linux shell-script sed
linux shell-script sed
edited 30 mins ago
msp9011
5,3695 gold badges42 silver badges69 bronze badges
asked 48 mins ago
ArunArun
11 bronze badge
New contributor
Arun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 30 mins ago
msp9011
5,3695 gold badges42 silver badges69 bronze badges
asked 48 mins ago
ArunArun
11 bronze badge
New contributor
Arun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 30 mins ago
msp9011
5,3695 gold badges42 silver badges69 bronze badges
edited 30 mins ago
msp9011
5,3695 gold badges42 silver badges69 bronze badges
edited 30 mins ago
msp9011
5,3695 gold badges42 silver badges69 bronze badges
5,3695 gold badges42 silver badges69 bronze badges
asked 48 mins ago
ArunArun
11 bronze badge
New contributor
Arun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 48 mins ago
ArunArun
11 bronze badge
asked 48 mins ago
ArunArun
11 bronze badge
11 bronze badge
New contributor
Arun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Arun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
the desired output should be cross-checked... It's not only doing for line starts with "02".
– msp9011
23 mins ago
add a comment |
the desired output should be cross-checked... It's not only doing for line starts with "02".
– msp9011
23 mins ago
the desired output should be cross-checked... It's not only doing for line starts with "02".
– msp9011
23 mins ago
the desired output should be cross-checked... It's not only doing for line starts with "02".
– msp9011
23 mins ago
add a comment |
1 Answer
1
active
oldest
votes
Try this,
Option 1: As per your desired output
sed "s/(.{21})(.{30})(.*)/1$(printf "%-30s" MMMM)3/" file
012019-06-03070005000MMMM 00000030
012019-06-03070005000MMMM 00000030
012019-06-03070005000MMMM 00000030
022019-06-03000727638MMMM 7000000300
022019-06-03000727638MMMM 7000000300
012019-06-03070005000MMMM 00000030
022019-06-03000727638MMMM 7000000300
022019-06-03000727638MMMM 7000000300
012019-06-03070005000MMMM 00000030
012019-06-03070005000MMMM 00000030
012019-06-03070005000MMMM 00000030
(.{21})first 21 characters will be stored in1
(.{30})next 30 charecters will be stored in2
(.{30})remaining stored in3
- Then it will substitute 123 with 1$(printf "%-30s" MMMM)3
Option 2: As per your context
sed "/^02/ s/(.{19})(.{31})(.*)/1$(printf "%-30s" MMMM)3/" file
012019-06-03070005000799111160300000030XXXXXXX0700000000030
012019-06-03070005000799165030700000030XXXXXXX0700000000030
012019-06-03070005000799175500700000030XXXXXXX0700000000030
022019-06-030007276MMMM 07000000300
022019-06-030007276MMMM 07000000300
012019-06-03070005000799175840700000030XXXXXXX0700000000030
022019-06-030007276MMMM 07000000300
022019-06-030007276MMMM 07000000300
012019-06-03070005000799194080700000030XXXXXXX0700000000030
012019-06-03070005000790035750700000030XXXXXXX0700000000030
012019-06-03070005000790036660700000030XXXXXXX0700000000030
/^02/will do the replacement only if the line starts with "02"
answered 10 mins ago
msp9011msp9011
5,3695 gold badges42 silver badges69 bronze badges
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Try this,
Option 1: As per your desired output
sed "s/(.{21})(.{30})(.*)/1$(printf "%-30s" MMMM)3/" file
012019-06-03070005000MMMM 00000030
012019-06-03070005000MMMM 00000030
012019-06-03070005000MMMM 00000030
022019-06-03000727638MMMM 7000000300
022019-06-03000727638MMMM 7000000300
012019-06-03070005000MMMM 00000030
022019-06-03000727638MMMM 7000000300
022019-06-03000727638MMMM 7000000300
012019-06-03070005000MMMM 00000030
012019-06-03070005000MMMM 00000030
012019-06-03070005000MMMM 00000030
(.{21})first 21 characters will be stored in1
(.{30})next 30 charecters will be stored in2
(.{30})remaining stored in3
- Then it will substitute 123 with 1$(printf "%-30s" MMMM)3
Option 2: As per your context
sed "/^02/ s/(.{19})(.{31})(.*)/1$(printf "%-30s" MMMM)3/" file
012019-06-03070005000799111160300000030XXXXXXX0700000000030
012019-06-03070005000799165030700000030XXXXXXX0700000000030
012019-06-03070005000799175500700000030XXXXXXX0700000000030
022019-06-030007276MMMM 07000000300
022019-06-030007276MMMM 07000000300
012019-06-03070005000799175840700000030XXXXXXX0700000000030
022019-06-030007276MMMM 07000000300
022019-06-030007276MMMM 07000000300
012019-06-03070005000799194080700000030XXXXXXX0700000000030
012019-06-03070005000790035750700000030XXXXXXX0700000000030
012019-06-03070005000790036660700000030XXXXXXX0700000000030
/^02/will do the replacement only if the line starts with "02"
answered 10 mins ago
msp9011msp9011
5,3695 gold badges42 silver badges69 bronze badges
add a comment |
Try this,
Option 1: As per your desired output
sed "s/(.{21})(.{30})(.*)/1$(printf "%-30s" MMMM)3/" file
012019-06-03070005000MMMM 00000030
012019-06-03070005000MMMM 00000030
012019-06-03070005000MMMM 00000030
022019-06-03000727638MMMM 7000000300
022019-06-03000727638MMMM 7000000300
012019-06-03070005000MMMM 00000030
022019-06-03000727638MMMM 7000000300
022019-06-03000727638MMMM 7000000300
012019-06-03070005000MMMM 00000030
012019-06-03070005000MMMM 00000030
012019-06-03070005000MMMM 00000030
(.{21})first 21 characters will be stored in1
(.{30})next 30 charecters will be stored in2
(.{30})remaining stored in3
- Then it will substitute 123 with 1$(printf "%-30s" MMMM)3
Option 2: As per your context
sed "/^02/ s/(.{19})(.{31})(.*)/1$(printf "%-30s" MMMM)3/" file
012019-06-03070005000799111160300000030XXXXXXX0700000000030
012019-06-03070005000799165030700000030XXXXXXX0700000000030
012019-06-03070005000799175500700000030XXXXXXX0700000000030
022019-06-030007276MMMM 07000000300
022019-06-030007276MMMM 07000000300
012019-06-03070005000799175840700000030XXXXXXX0700000000030
022019-06-030007276MMMM 07000000300
022019-06-030007276MMMM 07000000300
012019-06-03070005000799194080700000030XXXXXXX0700000000030
012019-06-03070005000790035750700000030XXXXXXX0700000000030
012019-06-03070005000790036660700000030XXXXXXX0700000000030
/^02/will do the replacement only if the line starts with "02"
answered 10 mins ago
msp9011msp9011
5,3695 gold badges42 silver badges69 bronze badges
add a comment |
Try this,
Option 1: As per your desired output
sed "s/(.{21})(.{30})(.*)/1$(printf "%-30s" MMMM)3/" file
012019-06-03070005000MMMM 00000030
012019-06-03070005000MMMM 00000030
012019-06-03070005000MMMM 00000030
022019-06-03000727638MMMM 7000000300
022019-06-03000727638MMMM 7000000300
012019-06-03070005000MMMM 00000030
022019-06-03000727638MMMM 7000000300
022019-06-03000727638MMMM 7000000300
012019-06-03070005000MMMM 00000030
012019-06-03070005000MMMM 00000030
012019-06-03070005000MMMM 00000030
(.{21})first 21 characters will be stored in1
(.{30})next 30 charecters will be stored in2
(.{30})remaining stored in3
- Then it will substitute 123 with 1$(printf "%-30s" MMMM)3
Option 2: As per your context
sed "/^02/ s/(.{19})(.{31})(.*)/1$(printf "%-30s" MMMM)3/" file
012019-06-03070005000799111160300000030XXXXXXX0700000000030
012019-06-03070005000799165030700000030XXXXXXX0700000000030
012019-06-03070005000799175500700000030XXXXXXX0700000000030
022019-06-030007276MMMM 07000000300
022019-06-030007276MMMM 07000000300
012019-06-03070005000799175840700000030XXXXXXX0700000000030
022019-06-030007276MMMM 07000000300
022019-06-030007276MMMM 07000000300
012019-06-03070005000799194080700000030XXXXXXX0700000000030
012019-06-03070005000790035750700000030XXXXXXX0700000000030
012019-06-03070005000790036660700000030XXXXXXX0700000000030
/^02/will do the replacement only if the line starts with "02"
answered 10 mins ago
msp9011msp9011
5,3695 gold badges42 silver badges69 bronze badges
Try this,
Option 1: As per your desired output
sed "s/(.{21})(.{30})(.*)/1$(printf "%-30s" MMMM)3/" file
012019-06-03070005000MMMM 00000030
012019-06-03070005000MMMM 00000030
012019-06-03070005000MMMM 00000030
022019-06-03000727638MMMM 7000000300
022019-06-03000727638MMMM 7000000300
012019-06-03070005000MMMM 00000030
022019-06-03000727638MMMM 7000000300
022019-06-03000727638MMMM 7000000300
012019-06-03070005000MMMM 00000030
012019-06-03070005000MMMM 00000030
012019-06-03070005000MMMM 00000030
(.{21})first 21 characters will be stored in1
(.{30})next 30 charecters will be stored in2
(.{30})remaining stored in3
- Then it will substitute 123 with 1$(printf "%-30s" MMMM)3
Option 2: As per your context
sed "/^02/ s/(.{19})(.{31})(.*)/1$(printf "%-30s" MMMM)3/" file
012019-06-03070005000799111160300000030XXXXXXX0700000000030
012019-06-03070005000799165030700000030XXXXXXX0700000000030
012019-06-03070005000799175500700000030XXXXXXX0700000000030
022019-06-030007276MMMM 07000000300
022019-06-030007276MMMM 07000000300
012019-06-03070005000799175840700000030XXXXXXX0700000000030
022019-06-030007276MMMM 07000000300
022019-06-030007276MMMM 07000000300
012019-06-03070005000799194080700000030XXXXXXX0700000000030
012019-06-03070005000790035750700000030XXXXXXX0700000000030
012019-06-03070005000790036660700000030XXXXXXX0700000000030
/^02/will do the replacement only if the line starts with "02"
answered 10 mins ago
msp9011msp9011
5,3695 gold badges42 silver badges69 bronze badges
edited 2 mins ago
answered 10 mins ago
msp9011msp9011
5,3695 gold badges42 silver badges69 bronze badges
answered 10 mins ago
msp9011msp9011
5,3695 gold badges42 silver badges69 bronze badges
answered 10 mins ago
msp9011msp9011
5,3695 gold badges42 silver badges69 bronze badges
5,3695 gold badges42 silver badges69 bronze badges
add a comment |
add a comment |
Arun is a new contributor. Be nice, and check out our Code of Conduct.
Arun is a new contributor. Be nice, and check out our Code of Conduct.
Arun is a new contributor. Be nice, and check out our Code of Conduct.
Arun is a new contributor. Be nice, and check out our Code of Conduct.
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the desired output should be cross-checked... It's not only doing for line starts with "02".
– msp9011
23 mins ago