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Full Discussion: Slack message multi line from UNIX script
Top Forums Shell Programming and Scripting Slack message multi line from UNIX script Post 303026042 by onenessboy on Monday 19th of November 2018 12:03:15 AM
Old 11-19-2018
Quote:
Originally Posted by onenessboy
Thank you very much

Its worked like a charm. bit modified to display output like below and terminal output is like:

Code:
==============================================
Oracle Host: development_box
==============================================
orcl:
Database UP -- OK
Listener UP -- OK
Database Read&Write Mode -- OK
Instance Status -- OK
Instance Login Allowed -- OK
---------------------------------------------
orcltest:
Database UP -- OK
Listener UP -- OK
Database Read&Write Mode -- OK
Instance Status -- OK
Instance Login Allowed -- OK
---------------------------------------------

However, as you suggested I am using below line to slack...however only first two line going into slack..

the below is code currently being used for slack post:

Code:
token="xxxxxxxxxxxxxxxxxxxxx"

if [[ $token == "" ]]
then
        echo "No token specified"
        exit 1
fi

shift
channel="#are_you_listening"
if [[ $channel == "" ]]
then
        echo "No channel specified"
        exit 1
fi

shift

text="$msg"
text1="$lmsg"

if [[ $text == "" ]]
then
        echo "No text specified"
        exit 1
fi

#escapedText=$(echo $text | $text1 | sed 's/"/\"/g' | sed "s/'/\'/g" )
escapedText=$(
echo "$text
$text1" | sed 's/"/\\"/g; s/'\''/\\'\''/g'
)
json="{\"channel\": \"#$channel\", \"text\": \"$escapedText\"}"

curl -X POST --data-urlencode "payload=$json" "https://hooks.slack.com/services/xxxx/xxx/xxxxxx

On terminal I am getting all output but, on slack i get only the below values

Code:
orcltest Database UP -- OK
Listener UP -- OK

no clue why it printing only these lines.. do I need to print instead of echo ?. any suggestion pls
Is there any way to send terminal screen output same format to slack as it is?

------ Post updated at 01:13 PM ------



Hi,

Some how i am able to output all values in terminal.. Up to this point i am able to achieve...now problem is with posting it to slack..even though I mention \n its not printing line by line in slack...

Terminal output (I have stored into variable, so echo it):

Code:
echo "$text"

==============================================
Oracle Host: xxxxxxxxx
==============================================
orcl:
Database UP -- OK
Listener UP -- OK
Database Read&Write Mode -- OK
Instance Status -- OK
Instance Login Allowed -- OK
---------------------------------------------
orcltest:
Database UP -- OK
Listener UP -- OK
Database Read&Write Mode -- OK
Instance Status -- OK
Instance Login Allowed -- OK
---------------------------------------------

Now trying to read line by line from this variable with below code:

Code:
while IFS= read -r line; do
  #printf '%s\n' "$line"
  text="$text$line\n"
done <<< "$text"

escapedText=$(echo $text | sed 's/"/\\"/g; s/'\''/\\'\''/g' )
json="{\"channel\": \"#$channel\", \"text\": \"$escapedText\"}"

curl -X POST --data-urlencode "payload=$json" "https://hooks.slack.com/services/xxxxx/xxxx/xxxxx"

its showing error like:
Code:
missing_text_or_fallback_or_attachments

but in above code I guess while loop is not working properly...is that correct way of reading multi line variable and pass on to payload in curl statement
This is resolved now. Omitting loop for text variable and updating escapedtext = "$text" does the trick..

Thank you for all help and patience..
 

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