Comparision in shell script

Question

I am doing comparision in shell script

              if [ $# -lt 10  ]
              then
              echo "PROCEED"
              else 
              echo "STOP"

I run the script as ./test.sh 1 2 3 4 5 6 7 8 9 10 11 then output is "STOP" Instead of using "-lt" if I use "<"

              if [ $# < 10  ]
              then
              echo "PROCEED"
              else 
              echo "STOP"

then output is "PROCEED".

My question is in case of shell script if I compare the no greater than 10 cause this problem. Please guide me on this

Answer 1

As seen in Advanced Bash Shell Scripting Guide - 7.3. Other Comparison Operators:

string comparison

<

is less than, in ASCII alphabetical order

if [[ "$a" < "$b" ]]

if [ "$a" \< "$b" ]

Note that the "<" needs to be escaped within a [ ] construct.

So the way to do it is with -lt or, quoting again:

integer comparison

-lt is less than

if [ "$a" -lt "$b" ]

< is less than (within double parentheses)

(("$a" < "$b"))

Answer 2

You need to use < rather than -lt when comparing numbers, since -lt does STRING comparisons.

Answer 3

Please be aware, Bash uses the buildin "test" implementation , man test -> link

This is also eplained here -> Stackoverflow Link

The angle brackets are commonly used for command redirections! link

If you are comparing integers use -lt for lower than.