how the keyword “if” test the value true of false?

Question

In bash script

if [ 1 ]
then
   echo "Yes"
else
   echo "No"
fi

Output: Yes It represent that '1' is treated as true value.

But in code:

word = Linux
letter= nuxi
if echo "$word" | grep -q "$letter"
then
    echo "Yes"
else
    echo "No"
fi

Output:No But echo "$word" | grep -q "$letter" will return 1, why the result is No.

how the keyword “if” test  the value  returned by the command after "if"?

Answer 1

The return value of a command is checked. [ 1 ] has a return value of 0 (true). Any other return value (like 1) indicates an error.

You can display the return value of the last executed command using the $? variable:

true
echo $?
# returned 0
false
echo $?
# returned 1
echo $?
# returned 0 as the last executed command is 'echo', and not 'false'

Answer 2

This is because the grep failed to find the $letter in $word, hence the exit code is 1. Whenever a process in linux return a code other than 0 then it means it failed. 0 means exited successfully. You can verify this by echo "Linux" | grep -d "nuxi"; echo $?

On the other hand in scripting world 0 means false and 1 mean true. So the grep failed to find the word and send 1 as an exit code to if, which took it as a true value.

Answer 3

In unix land, 0 is true and 1 is false.

For your first example:

if [ 1 ]
then
   echo "Yes"
else
   echo "No"
fi

"If" checks the exit code of the given command for true/false (i.e. zero/non-zero).

The square brackets actually invoke the "test" command (see "man test" for more information) and give the exit code to if.

"test 1" (or indeed "test any_string") returns true (0) so "Yes" is output.

For your second example, this outputs "No" because "nuxi" isn't found in "Linux", if you change "nuxi" to "nux" (perhaps this was a typo?) and remove the spaces around the = then you will get the behaviour you expect. e.g.

word=Linux
letter=nux
if echo "$word" | grep -q "$letter"
then
    echo "Yes"
else
    echo "No"
fi