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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;
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 $prlist
5 test 2019-09-27T11:06:23Z 2019-09-27T11:09:28Z 4 test1 2019-09-26T16:56:40Z 2019-09-26T16:57:02Z 3 test2 2019-09-26T16:54:25Z 2019-09-26T16:54:55Z 2 test3 2019-09-26T16:52:59Z 2019-09-26T16:55:19Z1 test4 2019-09-26T16:46:52Z 2019-09-26T16:47:25Z
How to avoid this printing in different line? Please help.
command-line bash printing echo
asked 10 hours ago
harshaharsha
9812 bronze badges
add a comment
|
I'm getting a curl command output as below
curl -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all | jq -r '.[]|(.number|tostring)+" "+.user.login+" "+.created_at+" "+.merged_at'
5 test 2019-09-27T11:06:23Z 2019-09-27T11:09:28Z
4 test1 2019-09-26T16:56:40Z 2019-09-26T16:57:02Z
3 test2 2019-09-26T16:54:25Z 2019-09-26T16:54:55Z
2 test3 2019-09-26T16:52:59Z 2019-09-26T16:55:19Z
1 test4 2019-09-26T16:46:52Z 2019-09-26T16:47:25Z
and I'm storing it in a variable and trying to pass echo the variable
to pass the value to another command to parse the output. But when I'm trying to echo the variable it is printing all the lines as a single line.
prlist=$(curl -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all | jq -r '.[]|(.number|tostring)+" "+.user.login+" "+.created_at+" "+.merged_at')
echo $prlist
5 test 2019-09-27T11:06:23Z 2019-09-27T11:09:28Z 4 test1 2019-09-26T16:56:40Z 2019-09-26T16:57:02Z 3 test2 2019-09-26T16:54:25Z 2019-09-26T16:54:55Z 2 test3 2019-09-26T16:52:59Z 2019-09-26T16:55:19Z1 test4 2019-09-26T16:46:52Z 2019-09-26T16:47:25Z
How to avoid this printing in different line? Please help.
command-line bash printing echo
asked 10 hours ago
harshaharsha
9812 bronze badges
1
Please edit your question and show us the exact commands you are using. We need to see thecurlcommand and theechocommand. You're probably simply not quoting correctly, but you almost certainly don't need to echo at all. If you give us the full commands, we can give you a better solution.
– terdon♦
10 hours ago
@terdon I have updated the question with the curl command I'm using.
– harsha
10 hours ago
Thank you, but as I said before, we need all commands, including theechoone. In fact, especially theecho. How are you saving thecurloutput in a variable? How are you then printing that variable? And what parsing are you trying to do?
– terdon♦
10 hours ago
@dessert I have updated it
– harsha
10 hours ago
add a comment
|
I'm getting a curl command output as below
curl -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all | jq -r '.[]|(.number|tostring)+" "+.user.login+" "+.created_at+" "+.merged_at'
5 test 2019-09-27T11:06:23Z 2019-09-27T11:09:28Z
4 test1 2019-09-26T16:56:40Z 2019-09-26T16:57:02Z
3 test2 2019-09-26T16:54:25Z 2019-09-26T16:54:55Z
2 test3 2019-09-26T16:52:59Z 2019-09-26T16:55:19Z
1 test4 2019-09-26T16:46:52Z 2019-09-26T16:47:25Z
and I'm storing it in a variable and trying to pass echo the variable
to pass the value to another command to parse the output. But when I'm trying to echo the variable it is printing all the lines as a single line.
prlist=$(curl -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all | jq -r '.[]|(.number|tostring)+" "+.user.login+" "+.created_at+" "+.merged_at')
echo $prlist
5 test 2019-09-27T11:06:23Z 2019-09-27T11:09:28Z 4 test1 2019-09-26T16:56:40Z 2019-09-26T16:57:02Z 3 test2 2019-09-26T16:54:25Z 2019-09-26T16:54:55Z 2 test3 2019-09-26T16:52:59Z 2019-09-26T16:55:19Z1 test4 2019-09-26T16:46:52Z 2019-09-26T16:47:25Z
How to avoid this printing in different line? Please help.
command-line bash printing echo
asked 10 hours ago
harshaharsha
9812 bronze badges
I'm getting a curl command output as below
curl -s https://api.github.com/repos/harshavardhanc/dockerfile-ansible/pulls?state=all | jq -r '.[]|(.number|tostring)+" "+.user.login+" "+.created_at+" "+.merged_at'
5 test 2019-09-27T11:06:23Z 2019-09-27T11:09:28Z
4 test1 2019-09-26T16:56:40Z 2019-09-26T16:57:02Z
3 test2 2019-09-26T16:54:25Z 2019-09-26T16:54:55Z
2 test3 2019-09-26T16:52:59Z 2019-09-26T16:55:19Z
1 test4 2019-09-26T16:46:52Z 2019-09-26T16:47:25Z
and I'm storing it in a variable and trying to pass echo the variable
to pass the value to another command to parse the output. But when I'm trying to echo the variable it is printing all the lines as a single line.
prlist=$(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
asked 10 hours ago
harshaharsha
9812 bronze badges
asked 10 hours ago
harshaharsha
9812 bronze badges
edited 10 hours ago
harsha
asked 10 hours ago
harshaharsha
9812 bronze badges
asked 10 hours ago
harshaharsha
9812 bronze badges
asked 10 hours ago
harshaharsha
9812 bronze badges
9812 bronze badges
1
Please edit your question and show us the exact commands you are using. We need to see thecurlcommand and theechocommand. 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♦
10 hours ago
@terdon I have updated the question with the curl command I'm using.
– harsha
10 hours ago
Thank you, but as I said before, we need all commands, including theechoone. In fact, especially theecho. How are you saving thecurloutput in a variable? How are you then printing that variable? And what parsing are you trying to do?
– terdon♦
10 hours ago
@dessert I have updated it
– harsha
10 hours ago
add a comment
|
1
Please edit your question and show us the exact commands you are using. We need to see thecurlcommand and theechocommand. 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♦
10 hours ago
@terdon I have updated the question with the curl command I'm using.
– harsha
10 hours ago
Thank you, but as I said before, we need all commands, including theechoone. In fact, especially theecho. How are you saving thecurloutput in a variable? How are you then printing that variable? And what parsing are you trying to do?
– terdon♦
10 hours ago
@dessert I have updated it
– harsha
10 hours ago
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♦
10 hours ago
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♦
10 hours ago
@terdon I have updated the question with the curl command I'm using.
– harsha
10 hours ago
@terdon I have updated the question with the curl command I'm using.
– harsha
10 hours ago
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♦
10 hours ago
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♦
10 hours ago
@dessert I have updated it
– harsha
10 hours ago
@dessert I have updated it
– harsha
10 hours ago
add a comment
|
1 Answer
1
active
oldest
votes
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
answered 10 hours ago
terdon♦terdon
74.5k14 gold badges151 silver badges235 bronze badges
Thanks @terdon, adding quotes worked.
– harsha
10 hours ago
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
6 hours ago
add a comment
|
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
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
answered 10 hours ago
terdon♦terdon
74.5k14 gold badges151 silver badges235 bronze badges
Thanks @terdon, adding quotes worked.
– harsha
10 hours ago
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
6 hours ago
add a comment
|
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
answered 10 hours ago
terdon♦terdon
74.5k14 gold badges151 silver badges235 bronze badges
Thanks @terdon, adding quotes worked.
– harsha
10 hours ago
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
6 hours ago
add a comment
|
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
answered 10 hours ago
terdon♦terdon
74.5k14 gold badges151 silver badges235 bronze badges
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
answered 10 hours ago
terdon♦terdon
74.5k14 gold badges151 silver badges235 bronze badges
edited 10 hours ago
answered 10 hours ago
terdon♦terdon
74.5k14 gold badges151 silver badges235 bronze badges
answered 10 hours ago
terdon♦terdon
74.5k14 gold badges151 silver badges235 bronze badges
answered 10 hours ago
terdon♦terdon
74.5k14 gold badges151 silver badges235 bronze badges
74.5k14 gold badges151 silver badges235 bronze badges
Thanks @terdon, adding quotes worked.
– harsha
10 hours ago
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
6 hours ago
add a comment
|
Thanks @terdon, adding quotes worked.
– harsha
10 hours ago
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
6 hours ago
Thanks @terdon, adding quotes worked.
– harsha
10 hours ago
Thanks @terdon, adding quotes worked.
– harsha
10 hours ago
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
6 hours ago
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
6 hours ago
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1
Please edit your question and show us the exact commands you are using. We need to see the
curlcommand and theechocommand. 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♦
10 hours ago
@terdon I have updated the question with the curl command I'm using.
– harsha
10 hours ago
Thank you, but as I said before, we need all commands, including the
echoone. In fact, especially theecho. How are you saving thecurloutput in a variable? How are you then printing that variable? And what parsing are you trying to do?– terdon♦
10 hours ago
@dessert I have updated it
– harsha
10 hours ago