Check whether string starts with "!" in POSIX sh - sh

Question

Trying to create a condition based on whether the line starts with an "!".
Note: this is sh not bash
#!/bin/sh
if [ $line = "!*" ] ;
then
echo "$0: $line has !"
else
echo "$0: $line has no !"
fi

Answer 1

In POSIX test, = performs exact string comparisons only.
Use a case statement instead.
case $line in
"!"*) echo "$line starts with an exclamation mark" ;;
*) echo "$line does not start with an exclamation mark" ;;
esac
If you really want to put this in an if, you can do that:
if case $string in "!"*) true;; *) false;; esac; then
echo "$line starts with an exclamation mark"
else
echo "$line does not start with an exclamation mark"
fi

Answer 2

You can use the standard POSIX parameter substitution syntax.
${var#repl} will remove the repl from the beginning of the content of the $var variable.
So, you'll have:
$ var=test
$ echo ${var#t}
est
$ echo ${var#X}
test
So, in order to have a simple if statement to check if a variable starts with a string (! in your case), you can have:
#!/bin/sh
if test "$line" = "${line#!}"; then
echo "$0: $line has no !"
else
echo "$0: $line has !"
fi
PS: test ... is equivalent to [ ... ], so the above script is exactly the same as
#!/bin/sh
if [ "$line" = "${line#!}" ]; then
echo "$0: $line has no !"
else
echo "$0: $line has !"
fi

Answer 3

Could be as simple as echo "$STR" | cut -c -1.
#!/bin/sh
STR="!abcd"
if [ "!" = $(echo "$STR" | cut -c -1) ]; then
echo "true"
fi