curl command has ! character results auth failure
0
I am getting authentication failure error while using curl command in bash shell which password has ! char.
Below is curl command i am trying.
curl -D- -u 'some_user:somename@2019!' -X POST --data @data.txt -H 'Content-Type: application/json' https://help.myjira.com/rest/api/2/issue
I have also tried escaping ! char using , as below, but no luck.
curl -D- -u 'some_user:somename@2019''!' -X POST --data @data.txt -H 'Content-Type: application/json' https://help.myjira.com/rest/api/2/issue
can any one suggest.
bash shell curl sh
edited Nov 20 at 8:16
muru
1
asked Nov 20 at 7:55
Mateen Syed
11
add a comment |
0
I am getting authentication failure error while using curl command in bash shell which password has ! char.
Below is curl command i am trying.
curl -D- -u 'some_user:somename@2019!' -X POST --data @data.txt -H 'Content-Type: application/json' https://help.myjira.com/rest/api/2/issue
I have also tried escaping ! char using , as below, but no luck.
curl -D- -u 'some_user:somename@2019''!' -X POST --data @data.txt -H 'Content-Type: application/json' https://help.myjira.com/rest/api/2/issue
can any one suggest.
bash shell curl sh
edited Nov 20 at 8:16
muru
1
asked Nov 20 at 7:55
Mateen Syed
11
For debugging, replace your longcurlcommandline withecho '!'. Note how this prints an!character and does not invoke history substitution. Therefore I guess something else is going on.
– Micha Wiedenmann
Nov 20 at 8:28
thanks @MichaWiedenmann, tried you suggested, seems like this user does not have permissions to create jira. thanks for help.
– Mateen Syed
Nov 21 at 9:10
add a comment |
0
0
0
I am getting authentication failure error while using curl command in bash shell which password has ! char.
Below is curl command i am trying.
curl -D- -u 'some_user:somename@2019!' -X POST --data @data.txt -H 'Content-Type: application/json' https://help.myjira.com/rest/api/2/issue
I have also tried escaping ! char using , as below, but no luck.
curl -D- -u 'some_user:somename@2019''!' -X POST --data @data.txt -H 'Content-Type: application/json' https://help.myjira.com/rest/api/2/issue
can any one suggest.
bash shell curl sh
edited Nov 20 at 8:16
muru
1
asked Nov 20 at 7:55
Mateen Syed
11
I am getting authentication failure error while using curl command in bash shell which password has ! char.
Below is curl command i am trying.
curl -D- -u 'some_user:somename@2019!' -X POST --data @data.txt -H 'Content-Type: application/json' https://help.myjira.com/rest/api/2/issue
I have also tried escaping ! char using , as below, but no luck.
curl -D- -u 'some_user:somename@2019''!' -X POST --data @data.txt -H 'Content-Type: application/json' https://help.myjira.com/rest/api/2/issue
can any one suggest.
bash shell curl sh
bash shell curl sh
edited Nov 20 at 8:16
muru
1
asked Nov 20 at 7:55
Mateen Syed
11
edited Nov 20 at 8:16
muru
1
asked Nov 20 at 7:55
Mateen Syed
11
edited Nov 20 at 8:16
muru
1
edited Nov 20 at 8:16
muru
1
edited Nov 20 at 8:16
muru
1
1
asked Nov 20 at 7:55
Mateen Syed
11
asked Nov 20 at 7:55
Mateen Syed
11
asked Nov 20 at 7:55
Mateen Syed
11
11
For debugging, replace your longcurlcommandline withecho '!'. Note how this prints an!character and does not invoke history substitution. Therefore I guess something else is going on.
– Micha Wiedenmann
Nov 20 at 8:28
thanks @MichaWiedenmann, tried you suggested, seems like this user does not have permissions to create jira. thanks for help.
– Mateen Syed
Nov 21 at 9:10
add a comment |
For debugging, replace your longcurlcommandline withecho '!'. Note how this prints an!character and does not invoke history substitution. Therefore I guess something else is going on.
– Micha Wiedenmann
Nov 20 at 8:28
thanks @MichaWiedenmann, tried you suggested, seems like this user does not have permissions to create jira. thanks for help.
– Mateen Syed
Nov 21 at 9:10
For debugging, replace your long
curl commandline with echo '!'. Note how this prints an ! character and does not invoke history substitution. Therefore I guess something else is going on.– Micha Wiedenmann
Nov 20 at 8:28
For debugging, replace your long
curl commandline with echo '!'. Note how this prints an ! character and does not invoke history substitution. Therefore I guess something else is going on.– Micha Wiedenmann
Nov 20 at 8:28
thanks @MichaWiedenmann, tried you suggested, seems like this user does not have permissions to create jira. thanks for help.
– Mateen Syed
Nov 21 at 9:10
thanks @MichaWiedenmann, tried you suggested, seems like this user does not have permissions to create jira. thanks for help.
– Mateen Syed
Nov 21 at 9:10
add a comment |
1 Answer
1
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oldest
votes
-1
You can try with entity code for ! = %21 (and @ = %40). So:
curl -D- -u 'some_user:somename%402019%21' -X POST --data @data.txt -H 'Content-Type: application/json' https://help.myjira.com/rest/api/2/issue
answered Nov 20 at 14:06
strano
719
add a comment |
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1 Answer
1
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oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
-1
You can try with entity code for ! = %21 (and @ = %40). So:
curl -D- -u 'some_user:somename%402019%21' -X POST --data @data.txt -H 'Content-Type: application/json' https://help.myjira.com/rest/api/2/issue
answered Nov 20 at 14:06
strano
719
add a comment |
-1
You can try with entity code for ! = %21 (and @ = %40). So:
curl -D- -u 'some_user:somename%402019%21' -X POST --data @data.txt -H 'Content-Type: application/json' https://help.myjira.com/rest/api/2/issue
answered Nov 20 at 14:06
strano
719
add a comment |
-1
-1
-1
You can try with entity code for ! = %21 (and @ = %40). So:
curl -D- -u 'some_user:somename%402019%21' -X POST --data @data.txt -H 'Content-Type: application/json' https://help.myjira.com/rest/api/2/issue
answered Nov 20 at 14:06
strano
719
You can try with entity code for ! = %21 (and @ = %40). So:
curl -D- -u 'some_user:somename%402019%21' -X POST --data @data.txt -H 'Content-Type: application/json' https://help.myjira.com/rest/api/2/issue
answered Nov 20 at 14:06
strano
719
answered Nov 20 at 14:06
strano
719
answered Nov 20 at 14:06
strano
719
answered Nov 20 at 14:06
strano
719
719
add a comment |
add a comment |
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For debugging, replace your long
curlcommandline withecho '!'. Note how this prints an!character and does not invoke history substitution. Therefore I guess something else is going on.– Micha Wiedenmann
Nov 20 at 8:28
thanks @MichaWiedenmann, tried you suggested, seems like this user does not have permissions to create jira. thanks for help.
– Mateen Syed
Nov 21 at 9:10