I don't mean that this question for UNIX only, but I work on Solaris, and I didn't try it on any other OS.
I confused between the extended regular expression:
first:
[[ "str" == ?(str|STR) ]] && echo "matched"
this work correct, but when:
[[ "str str" == ?(str|STR)(.*) ]] && echo "matched"
it doesn't work, does it mean that I can only compare one pattern.
Second:
[[ "str" =~ ?(str|STR) ]] && echo "matched"
I can't use this form here why?, but when:
[[ "str" == (str|STR)? ]] && echo "matched"
it works correctly.
It looks like you are trying to combine
extended globs
with
extended regular expressions. I would say this is A Bad Thing.
$ set '(str|STR)'
$ [[ 'str' =~ $1 ]] && echo matches
matches
$ [[ 'str str' =~ $1 ]] && echo matches
matches
Related
For some reason, the following regular expression match doesn't seem to be working.
string="#Hello world";
[[ "$string" =~ 'ello' ]] && echo "matches";
[[ "$string" =~ 'el.o' ]] && echo "matches";
The first command succeeds (as expected), but the second one does not.
Shouldn't that period be treated by the regular expression as a single character?
Quoting the period causes it to be treated as a literal character, not a regular-expression metacharacter. Best practice if you want to quote the entire regular expression is to do so in a variable, where regular expression matching rules aren't in effect, then expand the parameter unquoted (which is safe to do inside [[ ... ]]).
regex='el.o'
[[ "$string" =~ $regex ]] && echo "matches"
string="#Hello world";
[[ "$string" =~ ello ]] && echo "matches";
[[ "$string" =~ el.o ]] && echo "matches";
Test
$ string="hh elxo fj"
$ [[ "$string" =~ el.o ]] && echo "matches";
matches
I have the following function
checkFormat()
{
local funcUserName=$1
if [[ "$funcUserName" != [a-zA-Z0-9][a-zA-Z0-9][a-zA-Z0-9][a-zA-Z0-9] ]];then
echo "1"
else
echo "0"
fi
}
if [[ $string != [a-zA-Z0-9]* ]]
Only returns true if the first character is not [a-zA-Z0-9]
if [[ $string != [a-zA-Z0-9]{5} ]]
Never returns true.
if [[ $string != [a-zA-Z0-9][a-zA-Z0-9][a-zA-Z0-9][a-zA-Z0-9][a-zA-Z0-9] ]]
Returns as I want it to.
Why is this?
The code is to check that a username is 5 characters long and alphanumeric i.e.
Joe12 or 12345 but not %$134.
bash version 4.2.37
I suggest to replace
if [[ $string != [a-zA-Z0-9]{5} ]]
by
if [[ ! $string =~ ^[a-zA-Z0-9]{5}*$ ]]
to match a regex.
i want to match my 2 strings index and index1 in if condition of shell programming
i tried doing this by following
if [[ $1 == [iI][nN][dD][eE][xX][1]? ]]; then
echo "matched"
but it is not working, here basically i want to say in my regular expression that 1 should occur either 0 or 1 time.
Thanks in advance!
You need to use =~ operator to match regex and make sure to use anchors ^ and $ to avoid matching unwanted text:
[[ 'index1' =~ ^[iI][nN][dD][eE][xX]1?$ ]] && echo "ok" || echo "nope"
ok
[[ 'index' =~ ^[iI][nN][dD][eE][xX]1?$ ]] && echo "ok" || echo "nope"
ok
I am doing bash , i try to test if the substring "world" in the given variable x. I have part of code working. But the other one not working. I want to figure out why
First one is working
x=helloworldfirsttime
world=world
if [[ "$x" == *$world* ]];then
echo matching helloworld
Second one is not working
x=helloworldfirsttime
if [[ "$x" == "*world*" ]];then
echo matching helloworld
How to make second one work without using variable like the 1st method
Can someone fix the second one for me.. thanks
Just remove the quotes:
x=helloworldfirsttime
if [[ "$x" == *world* ]]; then
echo matching helloworld
fi
Note that this isn't regex (a regex for this would look something like .*world.*). The pattern matching in bash is described here:
http://www.gnu.org/software/bash/manual/html_node/Pattern-Matching.html
x=helloworldfirsttime
$ if [[ "$x" == *world* ]]; then echo MATCHING; fi
MATCHING
This works because bash's builtin [[ operator treats the right-hand-side of an == test as a pattern:
When the == and != operators are used, the string to the right of the operator is used as a pattern and pattern matching is performed.
Next time if you want to provide patters with spaces you could just quote it around "" or '', only that you have to place the pattern characters outside:
[[ "$x" == *"hello world"* ]]
[[ "$x" == *'hello world'* ]]
[[ "$x" == *"$var_value_has_spaces"* ]]
You shold use without quotes and the =~ operator.
TEXT=helloworldfirsttime
SEARCH=world
if [[ "$TEXT" =~ .*${SEARCH}.* ]]; then echo MATCHING; else echo NOT MATCHING; fi
TEXT=hellowor_ldfirsttime
if [[ "$TEXT" =~ .*${SEARCH}.* ]]; then echo MATCHING; else echo NOT MATCHING; fi
Say I want to match the leading dot in a string ".a"
So I type
[[ ".a" =~ ^\. ]] && echo "ha"
ha
[[ "a" =~ ^\. ]] && echo "ha"
ha
Why am I getting the same result here?
You need to escape the dot it has meaning beyond just a period - it is a metacharacter in regex.
[[ "a" =~ ^\. ]] && echo "ha"
Make the change in the other example as well.
Check your bash version - you need 4.0 or higher I believe.
There's some compatibility issues with =~ between Bash versions after 3.0. The safest way to use =~ in Bash is to put the RE pattern in a var:
$ pat='^\.foo'
$ [[ .foo =~ $pat ]] && echo yes || echo no
yes
$ [[ foo =~ $pat ]] && echo yes || echo no
no
$
For more details, see E14 on the Bash FAQ page.
Probably it's because bash tries to treat "." as a \ character, like \n \r etc.
In order to tell \ & . as 2 separate characters, try
[[ "a" =~ ^\\. ]] && echo ha