Print all the instances of a matching pattern in a file
I've been trying to print all the instances of a matching pattern from file.
Input file:
{"id":"prod123","a":1.3,"c":"xyz","q":2},
{"id":"prod456","a":1.3,"c":"xyz","q":1}]}
{"id":"prod789","a":1.3,"currency":"xyz","q":2},
{"id":"prod101112","a":1.3,"c":"xyz","q":1}]}
I'd want to print everything between "id":" and ",.
Expected output:
prod123
prod456
prod789
prod101112
I'm using the command
grep -Eo 'id":"[^"]+"",*' | grep -Eo '^[^"]+'
Am I missing anything here?
bash awk sed grep
edited Nov 14 '18 at 17:15
oguzismail
3,72031125
asked Nov 14 '18 at 17:09
VenkatVenkat
63
add a comment |
I've been trying to print all the instances of a matching pattern from file.
Input file:
{"id":"prod123","a":1.3,"c":"xyz","q":2},
{"id":"prod456","a":1.3,"c":"xyz","q":1}]}
{"id":"prod789","a":1.3,"currency":"xyz","q":2},
{"id":"prod101112","a":1.3,"c":"xyz","q":1}]}
I'd want to print everything between "id":" and ",.
Expected output:
prod123
prod456
prod789
prod101112
I'm using the command
grep -Eo 'id":"[^"]+"",*' | grep -Eo '^[^"]+'
Am I missing anything here?
bash awk sed grep
edited Nov 14 '18 at 17:15
oguzismail
3,72031125
asked Nov 14 '18 at 17:09
VenkatVenkat
63
2
Do you havejqinstalled?
– anubhava
Nov 14 '18 at 17:18
2
Your output seems to be agrepor whatever on a JSON file. You might be interested in usingjqon your full file. But for this, we should need to see how this thing looks like.
– kvantour
Nov 14 '18 at 17:27
1
Use some JSON parser, whether it bejqor a module in your language of choice. Don't use line-oriented tools likegreporawkto try to parse JSON.
– chepner
Nov 14 '18 at 17:27
add a comment |
I've been trying to print all the instances of a matching pattern from file.
Input file:
{"id":"prod123","a":1.3,"c":"xyz","q":2},
{"id":"prod456","a":1.3,"c":"xyz","q":1}]}
{"id":"prod789","a":1.3,"currency":"xyz","q":2},
{"id":"prod101112","a":1.3,"c":"xyz","q":1}]}
I'd want to print everything between "id":" and ",.
Expected output:
prod123
prod456
prod789
prod101112
I'm using the command
grep -Eo 'id":"[^"]+"",*' | grep -Eo '^[^"]+'
Am I missing anything here?
bash awk sed grep
edited Nov 14 '18 at 17:15
oguzismail
3,72031125
asked Nov 14 '18 at 17:09
VenkatVenkat
63
I've been trying to print all the instances of a matching pattern from file.
Input file:
{"id":"prod123","a":1.3,"c":"xyz","q":2},
{"id":"prod456","a":1.3,"c":"xyz","q":1}]}
{"id":"prod789","a":1.3,"currency":"xyz","q":2},
{"id":"prod101112","a":1.3,"c":"xyz","q":1}]}
I'd want to print everything between "id":" and ",.
Expected output:
prod123
prod456
prod789
prod101112
I'm using the command
grep -Eo 'id":"[^"]+"",*' | grep -Eo '^[^"]+'
Am I missing anything here?
bash awk sed grep
bash awk sed grep
edited Nov 14 '18 at 17:15
oguzismail
3,72031125
asked Nov 14 '18 at 17:09
VenkatVenkat
63
edited Nov 14 '18 at 17:15
oguzismail
3,72031125
asked Nov 14 '18 at 17:09
VenkatVenkat
63
edited Nov 14 '18 at 17:15
oguzismail
3,72031125
edited Nov 14 '18 at 17:15
oguzismail
3,72031125
edited Nov 14 '18 at 17:15
oguzismail
3,72031125
3,72031125
asked Nov 14 '18 at 17:09
VenkatVenkat
63
asked Nov 14 '18 at 17:09
VenkatVenkat
63
asked Nov 14 '18 at 17:09
VenkatVenkat
63
63
2
Do you havejqinstalled?
– anubhava
Nov 14 '18 at 17:18
2
Your output seems to be agrepor whatever on a JSON file. You might be interested in usingjqon your full file. But for this, we should need to see how this thing looks like.
– kvantour
Nov 14 '18 at 17:27
1
Use some JSON parser, whether it bejqor a module in your language of choice. Don't use line-oriented tools likegreporawkto try to parse JSON.
– chepner
Nov 14 '18 at 17:27
add a comment |
2
Do you havejqinstalled?
– anubhava
Nov 14 '18 at 17:18
2
Your output seems to be agrepor whatever on a JSON file. You might be interested in usingjqon your full file. But for this, we should need to see how this thing looks like.
– kvantour
Nov 14 '18 at 17:27
1
Use some JSON parser, whether it bejqor a module in your language of choice. Don't use line-oriented tools likegreporawkto try to parse JSON.
– chepner
Nov 14 '18 at 17:27
2
2
Do you have
jq installed?– anubhava
Nov 14 '18 at 17:18
Do you have
jq installed?– anubhava
Nov 14 '18 at 17:18
2
2
Your output seems to be a
grep or whatever on a JSON file. You might be interested in using jq on your full file. But for this, we should need to see how this thing looks like.– kvantour
Nov 14 '18 at 17:27
Your output seems to be a
grep or whatever on a JSON file. You might be interested in using jq on your full file. But for this, we should need to see how this thing looks like.– kvantour
Nov 14 '18 at 17:27
1
1
Use some JSON parser, whether it be
jq or a module in your language of choice. Don't use line-oriented tools like grep or awk to try to parse JSON.– chepner
Nov 14 '18 at 17:27
Use some JSON parser, whether it be
jq or a module in your language of choice. Don't use line-oriented tools like grep or awk to try to parse JSON.– chepner
Nov 14 '18 at 17:27
add a comment |
1 Answer
1
active
oldest
votes
What went wrong is the place of the comma in the first grep:
grep -Eo 'id.:.[^"]+","' inputfile
You need to do something extra for getting the desired substring.
grep -Eo 'id.:.[^"]+","' inputfile | cut -d: -f2 | grep -Eo '[^",]+'
I used cut, that would be easy for your example input.
cut -d'"' -f4 < inputfile
You have alternatives, like using jq, or
sed -r 's/{"id":"([^"]*).*/1/' inputfile
or using awk (solution now like cut but can be changed easy)
awk -F'"' '{print $4}' inputfile
answered Nov 14 '18 at 20:33
Walter AWalter A
10.8k21031
Thank you very much, this has helped and I got the desired result.
– Venkat
Nov 15 '18 at 5:41
In that case you can accept it: stackoverflow.com/help/someone-answers
– Walter A
Nov 15 '18 at 8:57
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
What went wrong is the place of the comma in the first grep:
grep -Eo 'id.:.[^"]+","' inputfile
You need to do something extra for getting the desired substring.
grep -Eo 'id.:.[^"]+","' inputfile | cut -d: -f2 | grep -Eo '[^",]+'
I used cut, that would be easy for your example input.
cut -d'"' -f4 < inputfile
You have alternatives, like using jq, or
sed -r 's/{"id":"([^"]*).*/1/' inputfile
or using awk (solution now like cut but can be changed easy)
awk -F'"' '{print $4}' inputfile
answered Nov 14 '18 at 20:33
Walter AWalter A
10.8k21031
Thank you very much, this has helped and I got the desired result.
– Venkat
Nov 15 '18 at 5:41
In that case you can accept it: stackoverflow.com/help/someone-answers
– Walter A
Nov 15 '18 at 8:57
add a comment |
What went wrong is the place of the comma in the first grep:
grep -Eo 'id.:.[^"]+","' inputfile
You need to do something extra for getting the desired substring.
grep -Eo 'id.:.[^"]+","' inputfile | cut -d: -f2 | grep -Eo '[^",]+'
I used cut, that would be easy for your example input.
cut -d'"' -f4 < inputfile
You have alternatives, like using jq, or
sed -r 's/{"id":"([^"]*).*/1/' inputfile
or using awk (solution now like cut but can be changed easy)
awk -F'"' '{print $4}' inputfile
answered Nov 14 '18 at 20:33
Walter AWalter A
10.8k21031
Thank you very much, this has helped and I got the desired result.
– Venkat
Nov 15 '18 at 5:41
In that case you can accept it: stackoverflow.com/help/someone-answers
– Walter A
Nov 15 '18 at 8:57
add a comment |
What went wrong is the place of the comma in the first grep:
grep -Eo 'id.:.[^"]+","' inputfile
You need to do something extra for getting the desired substring.
grep -Eo 'id.:.[^"]+","' inputfile | cut -d: -f2 | grep -Eo '[^",]+'
I used cut, that would be easy for your example input.
cut -d'"' -f4 < inputfile
You have alternatives, like using jq, or
sed -r 's/{"id":"([^"]*).*/1/' inputfile
or using awk (solution now like cut but can be changed easy)
awk -F'"' '{print $4}' inputfile
answered Nov 14 '18 at 20:33
Walter AWalter A
10.8k21031
What went wrong is the place of the comma in the first grep:
grep -Eo 'id.:.[^"]+","' inputfile
You need to do something extra for getting the desired substring.
grep -Eo 'id.:.[^"]+","' inputfile | cut -d: -f2 | grep -Eo '[^",]+'
I used cut, that would be easy for your example input.
cut -d'"' -f4 < inputfile
You have alternatives, like using jq, or
sed -r 's/{"id":"([^"]*).*/1/' inputfile
or using awk (solution now like cut but can be changed easy)
awk -F'"' '{print $4}' inputfile
answered Nov 14 '18 at 20:33
Walter AWalter A
10.8k21031
answered Nov 14 '18 at 20:33
Walter AWalter A
10.8k21031
answered Nov 14 '18 at 20:33
Walter AWalter A
10.8k21031
answered Nov 14 '18 at 20:33
Walter AWalter A
10.8k21031
10.8k21031
Thank you very much, this has helped and I got the desired result.
– Venkat
Nov 15 '18 at 5:41
In that case you can accept it: stackoverflow.com/help/someone-answers
– Walter A
Nov 15 '18 at 8:57
add a comment |
Thank you very much, this has helped and I got the desired result.
– Venkat
Nov 15 '18 at 5:41
In that case you can accept it: stackoverflow.com/help/someone-answers
– Walter A
Nov 15 '18 at 8:57
Thank you very much, this has helped and I got the desired result.
– Venkat
Nov 15 '18 at 5:41
Thank you very much, this has helped and I got the desired result.
– Venkat
Nov 15 '18 at 5:41
In that case you can accept it: stackoverflow.com/help/someone-answers
– Walter A
Nov 15 '18 at 8:57
In that case you can accept it: stackoverflow.com/help/someone-answers
– Walter A
Nov 15 '18 at 8:57
add a comment |
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2
Do you have
jqinstalled?– anubhava
Nov 14 '18 at 17:18
2
Your output seems to be a
grepor whatever on a JSON file. You might be interested in usingjqon your full file. But for this, we should need to see how this thing looks like.– kvantour
Nov 14 '18 at 17:27
1
Use some JSON parser, whether it be
jqor a module in your language of choice. Don't use line-oriented tools likegreporawkto try to parse JSON.– chepner
Nov 14 '18 at 17:27