I have one question in shell script for escape "\" with command substitution "` `". I post there to seek help to understand how it works.
My original one piece of code in script like this: This piece of code in whole script is working without errors
chk_mode=`sqlplus -s /nolog<<EOF
connect / as sysdba
set pagesize 0 feedback off heading off
select open_mode from v\\\$database;
EOF`
Usually if I want to escape $ sign from v$database, I should use "v\$database". Why is this piece of code using 3 backslash as v\\\$database? Is this because the backslash is in command substitution(backtick)? I tested in command line: only v\$database works, v\\\$database is not working. My Unix server is solaris 11.3 and SHELL is bash.
Please help me to understand why using 3 backslashes as escape. Thanks for your advice.
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Yes it is because the command is backticks. This is one of the reasons to avoid using this deprecated form of command substitution. The preferred command substitution method uses $( ... ) .
Try this instead:
chk_mode=$(sqlplus -s /nolog<<EOF
connect / as sysdba
set pagesize 0 feedback off heading off
select open_mode from v\$database;
EOF
)
