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I want to find all words in a string.
The user can add words in any pattern like comma seperated or ; seperated or anything.
Input: Hello, Hi, test word
Output: array("Hello","Hi","test","word");
If a word is string that contains only word characters \w, you could do:
preg_match_all('/(\w+)/', $input_string, $matches);
Or use preg_split:
$words = preg_split('/\W+/', $input_string);
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Hi I'm trying to sed replace the entire string after a match.
my scenario will be inside my file I have the below values:
"XX|TESTFILE|MATCH" "CHANGEME"
"XX|TESTFILE|MATCH1" "CHANGEME1"
I need to replace the string after the text MATCH" " so my expected output will be
"XX|TESTFILE|MATCH" "THIS IS CHANGED"
"XX|TESTFILE|MATCH1" "CHANGEME1"
Hope it is clear and someone will help.
Thank you!
sed -ri '/MATCH\"/ {s/(^\"XX.* \")(.*)(\"$)/\1THIS IS THE CHANGE\3/}'
Using sed with -r flag for regular expressions, search for entries with MATCH followed by a double quote and then when encountered, split the line into three section. Print the first section followed by "THIS IS THE CHANGE" and then the third section.
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How can I extract substring from a string using key pattern and delimiter. e.g.
mystring = 'toto=1,2,3 mynames=ralf,john,franky myhobbies=tennis,soccer,naps'
I want to extract: ralf,john,franky
The pattern here is: mynames
The delimiter is: =
You can use this regular expression:
mynames=([^\s]+)
And then, look for the first group: $1
Here is a live example in JavaScript (works also in other languages):
var regex = /mynames=([^\s]+)/;
var text = "toto=1,2,3 mynames=ralf,john,franky myhobbies=tennis,soccer,naps"
console.log(regex.exec(text)[1]);
If you are looking for a regex which does only match everything after mynames=, so that you don't need to look for the first group, you can also use a positive lookbehind:
(?<=mynames=)[^\s]+
Here is a live example: https://regex101.com/r/mem6mA/1
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a= <<EOF
Password
:
7UV1ceFQ (You will be asked to change this after logging in for the first time)
EOF
I need to extract the value "7UV1ceFQ" using regular expression, I have tried using '/Password : 7UV1ceFQ/ but it's not working, I think it's because next line character is included, Can anyone please suggest me to exact this value?
▶ a[/^\S+(?=\s\(You will be)/]
#⇒ "7UV1ceFQ"
The regular expression above reads as:
starting with a new line start ^
get all non-space symbols greedy \S+
until the positive lookahead (?=\s\(You will be)
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In Ruby, I would like to create a regular expression that matches the following:
building/liberty-green/6d
(the word building and some number somewhere after it)
Currently, I have /building/ and need to add \d (any digit) to it, but I don't know how.
You need /building\/[\w-]+\/\w+/. For example:
irb(main):001:0> /building\/[\w-]+\/\w+/.match("building/liberty-green/6d")
=> #<MatchData "building/liberty-green/6d">
That expression will match any string that:
Starts with /building/
Then follows with one or more word characters or dashes (eg. foo-bar, foo, bar-1)
Then follows with a /
Finally ends with one or more word characters (eg. foo, 6d, 12345)
Note that \w includes digits.
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How can I convert the string russ(ai)(edocn)cup to russiancodecup using Ruby?
By using gsub with a block, you can replace any match of a regular expression by the result of this block.
s = "russ(ai)(edocn)cup"
s.gsub(/\(([^)]*)\)/) {$1.reverse} # => "russiancodecup"
Here the regular expression will match any non-) character between brackets. Then it will send reverse to $1 which is gonna be the content between brackets.
$0 will be the complete match and $n, the nth "submatch". (anybody for the correct word ?)