Simple question on $VAR = num

Shell Programming and Scripting
1

Hello,
I am a begginer on shell scripting and I need a simple advice.
I have the following script.
The output of DATE is gathered from Oracle and it will be 101115 (for todays date)
Therefore the DAY will be sed-ed , and it contains 15

#!/bin/sh
DATE=`${LIB}/dt_YYMMDD 0`
DAY=`echo $DATE| sed 's/....\(.*\)/\1/'`
if [ $DAY == 15 ];
then
  echo is the same
fi

However when I am trying to execute the script, I get:

./sedtest2.sh[6]: ==: A test command parameter is not valid.

I believe it is something wrong in my comparison statement.

Any help will be much appreciated.
Thank you !

2 
#!/bin/sh
DATE=`${LIB}/dt_YYMMDD 0`
DAY=`echo $DATE| sed 's/^....//'
if [ $DAY == "15" ];
then
  echo is the same
fi

By the way, you can run command date +%d to get the day directly

1 Like
3

I tried that, but I get the same error :frowning:

4

hmm try..

if [ "${DAY}" = "15" ]
1 Like
5

Thank you all :slight_smile:

6

You can find out how comparisons work in the man page of "test" ("man test").

The opening square bracket ("[") is in fact just an alias or link for "/usr/bin/test", to make shell code more readable:

if [ "$x" = "a" ] ; then

and

 if /usr/bin/test "$x" = "a" ; then

is basically the same, the former is just "more natural" to read than the latter. "if" just takes the return code of the following command and "test" returns "0" when the comparison is true, otherwise it returns "1". The following would also work:

if 0 ; then

or

if 1 ; then

The first one would always execute the "then"-branch, the second one always the "else"-branch..

I hope this helps.

bakunin