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How to print variable value in next line using echo command


How to use “quickly run” and pass command-line arguments?Problem in unmounting a disk image with a script… but not manually!Mysterious behavior of echo commandecho #? is not printing any kind of valueEscape command in echoecho command variationsecho command echo the blank output instead of Variable value in Ubuntu 18.04






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;









6


















I'm getting a curl command output as below 



curl -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all | jq -r '.[]|(.number|tostring)+" "+.user.login+" "+.created_at+" "+.merged_at'


5 test 2019-09-27T11:06:23Z 2019-09-27T11:09:28Z
4 test1 2019-09-26T16:56:40Z 2019-09-26T16:57:02Z
3 test2 2019-09-26T16:54:25Z 2019-09-26T16:54:55Z
2 test3 2019-09-26T16:52:59Z 2019-09-26T16:55:19Z
1 test4 2019-09-26T16:46:52Z 2019-09-26T16:47:25Z


and I'm storing it in a variable and trying to pass echo the variable
to pass the value to another command to parse the output. But when I'm trying to echo the variable it is printing all the lines as a single line.



prlist=$(curl -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all | jq -r '.[]|(.number|tostring)+" "+.user.login+" "+.created_at+" "+.merged_at')
echo $prlist


5 test 2019-09-27T11:06:23Z 2019-09-27T11:09:28Z 4 test1 2019-09-26T16:56:40Z 2019-09-26T16:57:02Z 3 test2 2019-09-26T16:54:25Z 2019-09-26T16:54:55Z 2 test3 2019-09-26T16:52:59Z 2019-09-26T16:55:19Z1 test4 2019-09-26T16:46:52Z 2019-09-26T16:47:25Z


How to avoid this printing in different line? Please help.






command-line bash printing echo







share|improve this question






















  • 1





    Please edit your question and show us the exact commands you are using. We need to see the curl command and the echo command. You're probably simply not quoting correctly, but you almost certainly don't need to echo at all. If you give us the full commands, we can give you a better solution.

    – terdon
    Sep 28 at 12:47











  • @terdon I have updated the question with the curl command I'm using.

    – harsha
    Sep 28 at 12:52











  • Thank you, but as I said before, we need all commands, including the echo one. In fact, especially the echo. How are you saving the curl output in a variable? How are you then printing that variable? And what parsing are you trying to do?

    – terdon
    Sep 28 at 12:55












  • @dessert I have updated it

    – harsha
    Sep 28 at 12:58

















6


















I'm getting a curl command output as below 



curl -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all | jq -r '.[]|(.number|tostring)+" "+.user.login+" "+.created_at+" "+.merged_at'


5 test 2019-09-27T11:06:23Z 2019-09-27T11:09:28Z
4 test1 2019-09-26T16:56:40Z 2019-09-26T16:57:02Z
3 test2 2019-09-26T16:54:25Z 2019-09-26T16:54:55Z
2 test3 2019-09-26T16:52:59Z 2019-09-26T16:55:19Z
1 test4 2019-09-26T16:46:52Z 2019-09-26T16:47:25Z


and I'm storing it in a variable and trying to pass echo the variable
to pass the value to another command to parse the output. But when I'm trying to echo the variable it is printing all the lines as a single line.



prlist=$(curl -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all | jq -r '.[]|(.number|tostring)+" "+.user.login+" "+.created_at+" "+.merged_at')
echo $prlist


5 test 2019-09-27T11:06:23Z 2019-09-27T11:09:28Z 4 test1 2019-09-26T16:56:40Z 2019-09-26T16:57:02Z 3 test2 2019-09-26T16:54:25Z 2019-09-26T16:54:55Z 2 test3 2019-09-26T16:52:59Z 2019-09-26T16:55:19Z1 test4 2019-09-26T16:46:52Z 2019-09-26T16:47:25Z


How to avoid this printing in different line? Please help.






command-line bash printing echo







share|improve this question






















  • 1





    Please edit your question and show us the exact commands you are using. We need to see the curl command and the echo command. You're probably simply not quoting correctly, but you almost certainly don't need to echo at all. If you give us the full commands, we can give you a better solution.

    – terdon
    Sep 28 at 12:47











  • @terdon I have updated the question with the curl command I'm using.

    – harsha
    Sep 28 at 12:52











  • Thank you, but as I said before, we need all commands, including the echo one. In fact, especially the echo. How are you saving the curl output in a variable? How are you then printing that variable? And what parsing are you trying to do?

    – terdon
    Sep 28 at 12:55












  • @dessert I have updated it

    – harsha
    Sep 28 at 12:58













6













6









6








I'm getting a curl command output as below 



curl -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all | jq -r '.[]|(.number|tostring)+" "+.user.login+" "+.created_at+" "+.merged_at'


5 test 2019-09-27T11:06:23Z 2019-09-27T11:09:28Z
4 test1 2019-09-26T16:56:40Z 2019-09-26T16:57:02Z
3 test2 2019-09-26T16:54:25Z 2019-09-26T16:54:55Z
2 test3 2019-09-26T16:52:59Z 2019-09-26T16:55:19Z
1 test4 2019-09-26T16:46:52Z 2019-09-26T16:47:25Z


and I'm storing it in a variable and trying to pass echo the variable
to pass the value to another command to parse the output. But when I'm trying to echo the variable it is printing all the lines as a single line.



prlist=$(curl -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all | jq -r '.[]|(.number|tostring)+" "+.user.login+" "+.created_at+" "+.merged_at')
echo $prlist


5 test 2019-09-27T11:06:23Z 2019-09-27T11:09:28Z 4 test1 2019-09-26T16:56:40Z 2019-09-26T16:57:02Z 3 test2 2019-09-26T16:54:25Z 2019-09-26T16:54:55Z 2 test3 2019-09-26T16:52:59Z 2019-09-26T16:55:19Z1 test4 2019-09-26T16:46:52Z 2019-09-26T16:47:25Z


How to avoid this printing in different line? Please help.






command-line bash printing echo







share|improve this question
















I'm getting a curl command output as below 



curl -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all | jq -r '.[]|(.number|tostring)+" "+.user.login+" "+.created_at+" "+.merged_at'


5 test 2019-09-27T11:06:23Z 2019-09-27T11:09:28Z
4 test1 2019-09-26T16:56:40Z 2019-09-26T16:57:02Z
3 test2 2019-09-26T16:54:25Z 2019-09-26T16:54:55Z
2 test3 2019-09-26T16:52:59Z 2019-09-26T16:55:19Z
1 test4 2019-09-26T16:46:52Z 2019-09-26T16:47:25Z


and I'm storing it in a variable and trying to pass echo the variable
to pass the value to another command to parse the output. But when I'm trying to echo the variable it is printing all the lines as a single line.



prlist=$(curl -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all | jq -r '.[]|(.number|tostring)+" "+.user.login+" "+.created_at+" "+.merged_at')
echo $prlist


5 test 2019-09-27T11:06:23Z 2019-09-27T11:09:28Z 4 test1 2019-09-26T16:56:40Z 2019-09-26T16:57:02Z 3 test2 2019-09-26T16:54:25Z 2019-09-26T16:54:55Z 2 test3 2019-09-26T16:52:59Z 2019-09-26T16:55:19Z1 test4 2019-09-26T16:46:52Z 2019-09-26T16:47:25Z


How to avoid this printing in different line? Please help.






command-line bash printing echo




command-line bash printing echo






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Sep 28 at 12:56







harsha

















asked Sep 28 at 12:41









harshaharsha

18313 bronze badges




18313 bronze badges










  • 1





    Please edit your question and show us the exact commands you are using. We need to see the curl command and the echo command. You're probably simply not quoting correctly, but you almost certainly don't need to echo at all. If you give us the full commands, we can give you a better solution.

    – terdon
    Sep 28 at 12:47











  • @terdon I have updated the question with the curl command I'm using.

    – harsha
    Sep 28 at 12:52











  • Thank you, but as I said before, we need all commands, including the echo one. In fact, especially the echo. How are you saving the curl output in a variable? How are you then printing that variable? And what parsing are you trying to do?

    – terdon
    Sep 28 at 12:55












  • @dessert I have updated it

    – harsha
    Sep 28 at 12:58












  • 1





    Please edit your question and show us the exact commands you are using. We need to see the curl command and the echo command. You're probably simply not quoting correctly, but you almost certainly don't need to echo at all. If you give us the full commands, we can give you a better solution.

    – terdon
    Sep 28 at 12:47











  • @terdon I have updated the question with the curl command I'm using.

    – harsha
    Sep 28 at 12:52











  • Thank you, but as I said before, we need all commands, including the echo one. In fact, especially the echo. How are you saving the curl output in a variable? How are you then printing that variable? And what parsing are you trying to do?

    – terdon
    Sep 28 at 12:55












  • @dessert I have updated it

    – harsha
    Sep 28 at 12:58







1




1





Please edit your question and show us the exact commands you are using. We need to see the curl command and the echo command. You're probably simply not quoting correctly, but you almost certainly don't need to echo at all. If you give us the full commands, we can give you a better solution.

– terdon
Sep 28 at 12:47





Please edit your question and show us the exact commands you are using. We need to see the curl command and the echo command. You're probably simply not quoting correctly, but you almost certainly don't need to echo at all. If you give us the full commands, we can give you a better solution.

– terdon
Sep 28 at 12:47













@terdon I have updated the question with the curl command I'm using.

– harsha
Sep 28 at 12:52





@terdon I have updated the question with the curl command I'm using.

– harsha
Sep 28 at 12:52













Thank you, but as I said before, we need all commands, including the echo one. In fact, especially the echo. How are you saving the curl output in a variable? How are you then printing that variable? And what parsing are you trying to do?

– terdon
Sep 28 at 12:55






Thank you, but as I said before, we need all commands, including the echo one. In fact, especially the echo. How are you saving the curl output in a variable? How are you then printing that variable? And what parsing are you trying to do?

– terdon
Sep 28 at 12:55














@dessert I have updated it

– harsha
Sep 28 at 12:58





@dessert I have updated it

– harsha
Sep 28 at 12:58










1 Answer
1






active

oldest

votes


















8



















What you describe is the standard behavior of echoing an unquoted variable:



$ prlist=$(curl -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all | 
 jq -r '.[]|(.number|tostring)+" "+.user.login+" "+.created_at+" "+.merged_at')
$ echo $prlist
5 SMYALTAMASH 2019-09-27T11:06:23Z 2019-09-27T11:09:28Z 4 ganesh-28 2019-09-26T16:56:40Z 2019-09-26T16:57:02Z 3 ganesh-28 2019-09-26T16:54:25Z 2019-09-26T16:54:55Z 2 ganesh-28 2019-09-26T16:52:59Z 2019-09-26T16:55:19Z 1 ganesh-28 2019-09-26T16:46:52Z 2019-09-26T16:47:25Z


That's because $prlist is not quoted. Compare with what happens if you quote it:



$ echo "$prlist"
5 SMYALTAMASH 2019-09-27T11:06:23Z 2019-09-27T11:09:28Z
4 ganesh-28 2019-09-26T16:56:40Z 2019-09-26T16:57:02Z
3 ganesh-28 2019-09-26T16:54:25Z 2019-09-26T16:54:55Z
2 ganesh-28 2019-09-26T16:52:59Z 2019-09-26T16:55:19Z
1 ganesh-28 2019-09-26T16:46:52Z 2019-09-26T16:47:25Z


But why use a variable at all? You can just parse the output of curl directly:



curl ... | jq ... | someOtherCommand


Like this:



curl -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all | 
 jq -r '.[]|(.number|tostring)+" 
 "+.user.login+" "+.created_at+" "+.merged_at'curl 
 -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all |
 someOtherCommand





share|improve this answer






















  • 1





    Thanks @terdon, adding quotes worked.

    – harsha
    Sep 28 at 13:01











  • I'm storing it in a variable because I need to pass same output as input to multiple api's down the script. @terdon

    – harsha
    Sep 28 at 17:13












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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









8



















What you describe is the standard behavior of echoing an unquoted variable:



$ prlist=$(curl -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all | 
 jq -r '.[]|(.number|tostring)+" "+.user.login+" "+.created_at+" "+.merged_at')
$ echo $prlist
5 SMYALTAMASH 2019-09-27T11:06:23Z 2019-09-27T11:09:28Z 4 ganesh-28 2019-09-26T16:56:40Z 2019-09-26T16:57:02Z 3 ganesh-28 2019-09-26T16:54:25Z 2019-09-26T16:54:55Z 2 ganesh-28 2019-09-26T16:52:59Z 2019-09-26T16:55:19Z 1 ganesh-28 2019-09-26T16:46:52Z 2019-09-26T16:47:25Z


That's because $prlist is not quoted. Compare with what happens if you quote it:



$ echo "$prlist"
5 SMYALTAMASH 2019-09-27T11:06:23Z 2019-09-27T11:09:28Z
4 ganesh-28 2019-09-26T16:56:40Z 2019-09-26T16:57:02Z
3 ganesh-28 2019-09-26T16:54:25Z 2019-09-26T16:54:55Z
2 ganesh-28 2019-09-26T16:52:59Z 2019-09-26T16:55:19Z
1 ganesh-28 2019-09-26T16:46:52Z 2019-09-26T16:47:25Z


But why use a variable at all? You can just parse the output of curl directly:



curl ... | jq ... | someOtherCommand


Like this:



curl -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all | 
 jq -r '.[]|(.number|tostring)+" 
 "+.user.login+" "+.created_at+" "+.merged_at'curl 
 -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all |
 someOtherCommand





share|improve this answer






















  • 1





    Thanks @terdon, adding quotes worked.

    – harsha
    Sep 28 at 13:01











  • I'm storing it in a variable because I need to pass same output as input to multiple api's down the script. @terdon

    – harsha
    Sep 28 at 17:13















8



















What you describe is the standard behavior of echoing an unquoted variable:



$ prlist=$(curl -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all | 
 jq -r '.[]|(.number|tostring)+" "+.user.login+" "+.created_at+" "+.merged_at')
$ echo $prlist
5 SMYALTAMASH 2019-09-27T11:06:23Z 2019-09-27T11:09:28Z 4 ganesh-28 2019-09-26T16:56:40Z 2019-09-26T16:57:02Z 3 ganesh-28 2019-09-26T16:54:25Z 2019-09-26T16:54:55Z 2 ganesh-28 2019-09-26T16:52:59Z 2019-09-26T16:55:19Z 1 ganesh-28 2019-09-26T16:46:52Z 2019-09-26T16:47:25Z


That's because $prlist is not quoted. Compare with what happens if you quote it:



$ echo "$prlist"
5 SMYALTAMASH 2019-09-27T11:06:23Z 2019-09-27T11:09:28Z
4 ganesh-28 2019-09-26T16:56:40Z 2019-09-26T16:57:02Z
3 ganesh-28 2019-09-26T16:54:25Z 2019-09-26T16:54:55Z
2 ganesh-28 2019-09-26T16:52:59Z 2019-09-26T16:55:19Z
1 ganesh-28 2019-09-26T16:46:52Z 2019-09-26T16:47:25Z


But why use a variable at all? You can just parse the output of curl directly:



curl ... | jq ... | someOtherCommand


Like this:



curl -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all | 
 jq -r '.[]|(.number|tostring)+" 
 "+.user.login+" "+.created_at+" "+.merged_at'curl 
 -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all |
 someOtherCommand





share|improve this answer






















  • 1





    Thanks @terdon, adding quotes worked.

    – harsha
    Sep 28 at 13:01











  • I'm storing it in a variable because I need to pass same output as input to multiple api's down the script. @terdon

    – harsha
    Sep 28 at 17:13













8















8











8









What you describe is the standard behavior of echoing an unquoted variable:



$ prlist=$(curl -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all | 
 jq -r '.[]|(.number|tostring)+" "+.user.login+" "+.created_at+" "+.merged_at')
$ echo $prlist
5 SMYALTAMASH 2019-09-27T11:06:23Z 2019-09-27T11:09:28Z 4 ganesh-28 2019-09-26T16:56:40Z 2019-09-26T16:57:02Z 3 ganesh-28 2019-09-26T16:54:25Z 2019-09-26T16:54:55Z 2 ganesh-28 2019-09-26T16:52:59Z 2019-09-26T16:55:19Z 1 ganesh-28 2019-09-26T16:46:52Z 2019-09-26T16:47:25Z


That's because $prlist is not quoted. Compare with what happens if you quote it:



$ echo "$prlist"
5 SMYALTAMASH 2019-09-27T11:06:23Z 2019-09-27T11:09:28Z
4 ganesh-28 2019-09-26T16:56:40Z 2019-09-26T16:57:02Z
3 ganesh-28 2019-09-26T16:54:25Z 2019-09-26T16:54:55Z
2 ganesh-28 2019-09-26T16:52:59Z 2019-09-26T16:55:19Z
1 ganesh-28 2019-09-26T16:46:52Z 2019-09-26T16:47:25Z


But why use a variable at all? You can just parse the output of curl directly:



curl ... | jq ... | someOtherCommand


Like this:



curl -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all | 
 jq -r '.[]|(.number|tostring)+" 
 "+.user.login+" "+.created_at+" "+.merged_at'curl 
 -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all |
 someOtherCommand





share|improve this answer
















What you describe is the standard behavior of echoing an unquoted variable:



$ prlist=$(curl -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all | 
 jq -r '.[]|(.number|tostring)+" "+.user.login+" "+.created_at+" "+.merged_at')
$ echo $prlist
5 SMYALTAMASH 2019-09-27T11:06:23Z 2019-09-27T11:09:28Z 4 ganesh-28 2019-09-26T16:56:40Z 2019-09-26T16:57:02Z 3 ganesh-28 2019-09-26T16:54:25Z 2019-09-26T16:54:55Z 2 ganesh-28 2019-09-26T16:52:59Z 2019-09-26T16:55:19Z 1 ganesh-28 2019-09-26T16:46:52Z 2019-09-26T16:47:25Z


That's because $prlist is not quoted. Compare with what happens if you quote it:



$ echo "$prlist"
5 SMYALTAMASH 2019-09-27T11:06:23Z 2019-09-27T11:09:28Z
4 ganesh-28 2019-09-26T16:56:40Z 2019-09-26T16:57:02Z
3 ganesh-28 2019-09-26T16:54:25Z 2019-09-26T16:54:55Z
2 ganesh-28 2019-09-26T16:52:59Z 2019-09-26T16:55:19Z
1 ganesh-28 2019-09-26T16:46:52Z 2019-09-26T16:47:25Z


But why use a variable at all? You can just parse the output of curl directly:



curl ... | jq ... | someOtherCommand


Like this:



curl -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all | 
 jq -r '.[]|(.number|tostring)+" 
 "+.user.login+" "+.created_at+" "+.merged_at'curl 
 -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all |
 someOtherCommand






share|improve this answer















share|improve this answer




share|improve this answer








edited Sep 28 at 13:03

























answered Sep 28 at 12:58









terdonterdon

76.2k14 gold badges153 silver badges239 bronze badges




76.2k14 gold badges153 silver badges239 bronze badges










  • 1





    Thanks @terdon, adding quotes worked.

    – harsha
    Sep 28 at 13:01











  • I'm storing it in a variable because I need to pass same output as input to multiple api's down the script. @terdon

    – harsha
    Sep 28 at 17:13












  • 1





    Thanks @terdon, adding quotes worked.

    – harsha
    Sep 28 at 13:01











  • I'm storing it in a variable because I need to pass same output as input to multiple api's down the script. @terdon

    – harsha
    Sep 28 at 17:13







1




1





Thanks @terdon, adding quotes worked.

– harsha
Sep 28 at 13:01





Thanks @terdon, adding quotes worked.

– harsha
Sep 28 at 13:01













I'm storing it in a variable because I need to pass same output as input to multiple api's down the script. @terdon

– harsha
Sep 28 at 17:13





I'm storing it in a variable because I need to pass same output as input to multiple api's down the script. @terdon

– harsha
Sep 28 at 17:13


















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