I'm trying to write a regex that checks if string contains 6 or more signs including 1 or more special sign [^0-9a-zA-Z\s] and 1 or more [0-9a-zA-Z].
Spent like 2h and not getting any closer :/
maybe this is of some help:
^(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?!.*\s).{6,13}$
Password expresion that requires one lower case letter, one upper case letter, one digit, 6-13 length, and no spaces.
Matches:
1agdH*$# | 1agdC*$# | 1agdB*$#
Non-Matches:
wyrn%#*&$# f | mbndkfh782 | BNfhjdhfjd&*)%#$)
This is based on the Regex Lib entry here
Taking the style of Hasson's answer . . .
grep -P '^(?=.*[^a-zA-Z0-9\s])(?=.*[a-zA-Z0-9])(?!.*\s).{6}'
6 or more chars (regexp not ended with $)
1 or more special char (?=.*[^0-9a-zA-Z\s])
1 or more (?=.*[0-9a-zA-Z])
no whitespace (?!.*\s)
Some test data, NO match:
password
pa5sword
pa5sWord
pa5sWord
password
test
1agdA
1agd
wyrn%#*&$# f
mbndkfh782
t1*$
Some test data, YES match:
pa5*Word
pa5*Word
pa5*Word1
pa5*Wor
1agdA*
1agdA*$
1agdA*$#
1agdA*$#1
1agdA*$#12
1agdA*$#123
1agdA*$#a
1agdA*$#ab
1agdA*$#abc
1agdA*$#abcd
BNfhjdhfjd&*)%#$)
Related
I'm a regex newbie and I've got a valid regex for SSNs:
/^(\d{3}(\s|-)?\d{2}(\s|-)?\d{4})|[\d{9}]*$/
But I now need to expand it to accept either an SSN or another alphanumeric ID of 7 characters, like this:
/^[a-zA-Z0-9]{7}$/
I thought it'd be as simple as grouping the SSN and adding an OR | but my tests are still failing. This is what I've got now:
/^((\d{3}(\s|-)?\d{2}(\s|-)?\d{4})|[\d{9}])|[a-zA-Z0-9]{7}$/
What am I doing wrong? And is there a more elegant way to say either SSN or my other ID?
Thanks for any helpful tips.
Valid SSNs:
123-45-6789
123456789
123 45 6789
Valid ID: aCe8999
I have modified your first regex also a bit, below is demo program. This is as per my understanding of the problem. Let me know if any modification is needed.
my #ids = (
'123-45-6789',
'123456789',
'123 45 6789',
'1234567893434', # invalid
'123456789wwsd', # invalid
'aCe8999',
'aCe8999asa' # invalid
);
for (#ids) {
say "match = $&" if $_ =~ /^ (?:\d{3} ([ \-])? \d{2} \1? \d{4})$ | ^[a-zA-Z0-9]{7}$/x ;
}
Output:
match = 123-45-6789
match = 123456789
match = 123 45 6789
match = aCe8999
Your first regex got some problems. The important thing about it is that it accepts {{{{}}}}} which means you have built a wrong character class. Also it matches 123-45 6789 (notice the mixture of space and dash).
To mean OR in regular expressions you need to use pipe | and remember that each symbol belongs to the side that it resides. So for example ^1|2$ checks for strings beginning with 1 or ending with 2 not only two individual input strings 1 and 2.
To apply the exact match you need to do ^1$|^2$ or ^(1|2)$.
With the second regex ^[a-zA-Z0-9]{7}$ you are not saying alphanumeric ID of 7 characters but you are saying numeric, alphabetic or alphanumeric. So it matches 1234567 too. If this is not a problem, the following regex is the solution by eliminating the said issues:
^\d{3}([ -]?)\d\d\1\d{4}$|^[a-zA-Z0-9]{7}$
I am trying to select some codes from a PostgreSQl table.
I only want the codes that have numbers in them e.g
GD123
GD564
I don't want to pick any codes like `GDTG GDCNB
Here's my query so far:
select regexp_matches(no_, '[a-zA-Z0-9]*$')
from myschema.mytable
which of course doesn't work.
Any help appreciated.
The pattern to match a string that has at least 1 letter followed by at least 1 number is '[A-Za-z]+[0-9]+'.
Now, if the valid patterns had to start with two letters, and then have 3 digits after as your examples show, then replace the + with {2} & {4} respectively, and enclose the pattern in ^$, like this: '^[A-Za-z]{2}[0-9]{3}$'
The regex match operator is ~ which you can use in the where clause:
SELECT no_
FROM myschema.mytable
WHERE no_ ~ '[A-Za-z]+[0-9]+'
You may use
CREATE TABLE tb1
(s character varying)
;
INSERT INTO tb1
(s)
VALUES
('GD123'),
('12345'),
('GDFGH')
;
SELECT * FROM tb1 WHERE s ~ '^(?![A-Za-z]+$)[a-zA-Z0-9]+$';
Result:
Details
^ - start of string
(?![A-Za-z]+$) - a negative lookahead that fails the match if there are only letters to the end of the string
[a-zA-Z0-9]+ - 1 or more alphanumeric chars
$ - end of string.
If you want to avoid matching 12345, use
'^(?![A-Za-z]+$)(?![0-9]+$)[a-zA-Z0-9]+$'
Here, (?![0-9]+$) will similarly fail the match if, from the string start, all chars up to the end of the string are digits. Result:
smth like:
so=# with c(v) as (values('GD123'),('12345'),('GD ERT'))
select v ~ '[A-Z]{1,}[0-9]+', v from c;
?column? | v
----------+--------
t | GD123
f | 12345
f | GD ERT
(3 rows)
?..
If the format of the data you want to obtain is a set of characters follewd by a set of digits (i.e., GD123) you can use the regex:
[a-zA-Z0-9]+[0-9]
This captures every digit and letter which is in front of the digits:
([A-z]+\d+)
The general problem
I am trying to understand how to prevent the existence of some pattern before or after a sought-out pattern when writing regex's!
A more specific example
I'm looking for a regex that will match dates in the format YYMMDD ((([0-9]{2})(0[1-9]|1[0-2])(0[1-9]|[1-2][0-9]|3[0-1]))) inside a long string while ignoring longer numeric sequences
it should be able to match:
text151124moretext
123text151124moretext
text151124
text151124moretext1944
151124
but should ignore:
text15112412moretext
(reason: it has 8 numbers instead of 6)
151324
(reason: it is not a valid date YYMMDD - there is no 13th month)
how can I make sure that if a number has more than these 6 digits, it won't picked up as a date inside one single regex (meaning, that I would rather avoid preprocessing the string)
I've thought of \D((19|20)([0-9]{2})(0[1-9]|1[0-2])(0[1-9]|[1-2][0-9]|3[0-1]))\D but doesn't this mean that there has to be some character before and after?
I'm using bash 3.2 (ERE)
thanks!
Try:
#!/usr/bin/env bash
extract_date() {
local string="$1"
local _date=`echo "$string" | sed -E 's/.*[^0-9]([0-9]{6})[^0-9].*/\1/'`
#date -d $_date &> /dev/null # for Linux
date -jf '%y%m%d' $_date &> /dev/null # for MacOS
if [ $? -eq 0 ]; then
echo $_date
else
return 1
fi
}
extract_date text15111224moretext # ignore n_digits > 6
extract_date text151125moretext # take
extract_date text151132 # # ignore day 32
extract_date text151324moretext1944 # ignore month 13
extract_date text150931moretext1944 # ignore 31 Sept
extract_date 151126 # take
Output:
151125
151126
If your tokens are line-separated (i.e. there is only one token per line):
^[\D]*[\d]{6}([\D]*|[\D]+[\d]{1,6})$
Basically, this regex looks for:
Any number of non-digits at the beginning of the string;
Exactly 6 digits
Any number of non-digits until the end OR at least one non-digit and at least one digit (up to 6) to the end of the string
This regex passes all of your given sample inputs.
You could use non-capturing groups to define non-digits either side of your date Regex. I had success with this expression and your same test data.
(?:\D)([0-9]{2})(0[1-9]|1[0-2])(0[1-9]|[1-2][0-9]|3[0-1])(?:\D)
Why does the following literal string
1998-${year}
..match against the grep command:
grep "[0-9 ]*-[ 0-9]*" filename.txt ?
What I need is a regex to match any of the following strings containing either a year range or one value of year only.
sdkfmslf 1998-2008
asdassdadsa 1998 - 2008
mkklml mklsmdf 2006
..but NOT this one:
asdsad a s 1998-${year}
* means "match zero or more". You want + which means "one or more."
grep "[0-9 ]+-[0-9]+" filename.txt
Try [0-9]{4}(\s*-\s*[0-9]{4})?. This will match a 4 digit number, or if it is followed by (optional white space)-(optional whitespace) then that must be followed by another 4 digit number.
Your string "asdsad a s 1998-${year}" would still match, since it has a single 4 digit value in it.
I don't like answering my own question, but none of the above worked. Here is what I found by experimenting. I'm sure there could be more elegant solutions, but here is a working version:
grep "[0-9][0-9][0-9][0-9][ ]*[\-]*[ ]*[0-9]*" filename.txt
I am working on a Rails 3 application that needs to validate the password based on the following criteria: must be at least 6 characters and include one number and one letter.
Here is my Regex:
validates :password, :format => {:with => /^[([a-z]|[A-Z])0-9_-]{6,40}$/, message: "must be at least 6 characters and include one number and one letter."}
Right now if I put in a password of (for ex: dogfood) it will pass. But what I need it to do is to pass the criteria above.
I am not all that great at regex, so any and all help is greatly appreciated!
Use lookahead assertions:
/^(?=.*[a-zA-Z])(?=.*[0-9]).{6,}$/
| | |
| | |
| | Ensure there are at least 6 characters.
| |
| Look ahead for an arbitrary string followed by a number.
|
Look ahead for an arbitrary string followed by a letter.
Technically in this case you don't need the anchors, but it's good habit to use them.