Bash - setting a string variable and printf

Question

I am setting a variable as follows and when I print it out I get something unexpected.

sql="values ('$yy-$mm-$dd $time','$tz', $load)"
printf "%s\n" "$sql"

)alues ('2011-01-01 23:55:00','EST', 5081.2
)alues ('2011-01-01 23:55:00','EST', 475.8
)alues ('2011-01-01 23:55:00','EST', 1574.9

Somehow the closing parenthesis is at the beginning of the line?! I check $load to make sure there is no newline character in it.

I am not sure what to try.

I suspect one of your variables ($load?) contains a line feed. — Cyrus
– Cyrus, Commented Mar 22, 2022 at 0:00
Your load variable is probably coming from a source that uses DOS/Windows line endings (which have a carriage return in addition to the linefeed that unix programs expect. See "Are shell scripts sensitive to encoding and line endings?" Note that both carriage return and linefeed are nonprinting characters, so checking a variable for them is nontrivial. — Gordon Davisson
– Gordon Davisson, Commented Mar 22, 2022 at 0:34

Philippe · Accepted Answer · 2022-03-22 00:39:34Z

1

You can try this command to check $load :

printf "%q\n" "$load"

You might see :

$'5081.2\r'

where \r is the problem.

Update

In your case, this check is even better :

printf "%q\n" "$sql"

answered Mar 22, 2022 at 0:17

Philippe

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1 Comment

Thanks .. yes I do have a "\r" in the $load variable!