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Umbraco 7 regex error: Value is invalid, it does not match the correct pattern


Umbraco 7 regex error: Value is invalid, it does not match the correct pattern

By : user2954270
Date : November 22 2020, 01:01 AM
like below fixes the issue By providing a regular expression for the field you in turn make it mandatory. If you wish for the field to be optional then you must either omit your regular expression or alter it so that it also accepts an empty value.
code :


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Regex help: My regex pattern will match invalid strings

Regex help: My regex pattern will match invalid strings


By : user1654395
Date : March 29 2020, 07:55 AM
like below fixes the issue Use ^ to anchor the start and $ to anchor the end. E.g.: ^(abc)*$, this matches zero or more repetitions of the group ("abc" in this example) and that must start at the start of the input string and end at the end of it.
^(?:[(?[^,][}' ]+?,[S|D],\d{1})])+$—using an ungreedy +? doesn't matter, as you require it to match until the end anyway. However, your regex has a few issues.
code :
import re
def find_segments(input_string):
  results = []
  regex = re.compile(r"\[([^],]+),([SD]),(\d)\]")
  start = 0
  while True:
    m = regex.match(input_string, start)
    if not m: # no match
      return None # whole string didn't match, do another action as appropriate
    results.append(m.group(1, 2, 3))
    start = m.end(0)
    if start == len(input_string):
      break
  return results

print find_segments("[A-Z,S,3][klm,D,4][0-9,S,1]")
# output:
#[('A-Z', 'S', '3'), ('klm', 'D', '4'), ('0-9', 'S', '1')]
Regex help: My regex pattern will match invalid Dictionary

Regex help: My regex pattern will match invalid Dictionary


By : Patrik Barta
Date : March 29 2020, 07:55 AM
I wish this helpful for you Here is a complete rework of the regex in C#.
Assumptions : (tell me if one of them is false or all are false)
code :

const string dataFileScr = @"
Start 0
{
    Next = 1
    Author = rk
    Date = 2011-03-10
/*  Description = simple */
}

PZ 11
{
IA_return()
}

GDC 7
{
    Message = 6
    Message = 7
        Message = 8
        Message = 8
    RepeatCount = 2
    ErrorMessage = 10
    ErrorMessage = 11
    onKey[5] = 6
    onKey[6] = 4
    onKey[9] = 11
}

[Section1]
key1 = value1
key2 = value2

[Section2]
key1 = value1
key2 = value2
";

const string patternFileScr = @"
(?<Section>                                                              (?# Start of a non ini file section)
  (?<SectionName>[\w ]+)\s*                                              (?# Capture section name)
     {                                                                   (?# Match but don't capture beginning of section)
        (?<SectionBody>                                                  (?# Capture section body. Section body can be empty)
         (?<SectionLine>\s*                                              (?# Capture zero or more line(s) in the section body)
         (?:                                                             (?# A line can be either a key/value pair, a comment or a function call)
            (?<KeyValuePair>(?<Key>[\w\[\]]+)\s*=\s*(?<Value>[\w-]*))    (?# Capture key/value pair. Key and value are sub-captured separately)
            |
            (?<Comment>/\*.+?\*/)                                        (?# Capture comment)
            |
            (?<FunctionCall>[\w]+\(\))                                   (?# Capture function call. A function can't have parameters though)
         )\s*                                                            (?# Match but don't capture white characters)
         )*                                                              (?# Zero or more line(s), previously mentionned in comments)
        )
     }                                                                   (?# Match but don't capture beginning of section)
)
|
(?<Section>                                                              (?# Start of an ini file section)
  \[(?<SectionName>[\w ]+)\]                                             (?# Capture section name)
  (?<SectionBody>                                                        (?# Capture section body. Section body can be empty)
     (?<SectionLine>                                                     (?# Capture zero or more line(s) in the section body. Only key/value pair allowed.)
        \s*(?<KeyValuePair>(?<Key>[\w\[\]]+)\s*=\s*(?<Value>[\w-]+))\s*  (?# Capture key/value pair. Key and value are sub-captured separately)
     )*                                                                  (?# Zero or more line(s), previously mentionned in comments)
  )
)
";
Using RegEX to match URL pattern, invalid quantifier?

Using RegEX to match URL pattern, invalid quantifier?


By : user3290722
Date : March 29 2020, 07:55 AM
help you fix your problem I am trying to use RegEX to match a URL pattern. I found an answer here: Check if a Javascript string is a url , You need to escape the first \ in here:
code :
'(\?[;&a-z\d%_.~+=-]*)?'
'(\\?[;&a-z\d%_.~+=-]*)?'
Python Regex Multiple Pattern Match and Anti Pattern Match needed

Python Regex Multiple Pattern Match and Anti Pattern Match needed


By : user2875762
Date : March 29 2020, 07:55 AM
hope this fix your issue Although your requirements are a bit fuzzy, a reasonable shot given your particular input string seems to be to split on any space that is preceded by a literal . and followed by the literal (letter) pattern.
code :
import re

s = "(a) be given by XXX for its election per scenario 3.3(a). (b) Second statement has a section 4.2(b) which might elect for scenario 2.4(a) potentially"

print(re.split(r"(?<=\.) (?=\([a-z]\))", s))
['(a) be given by XXX for its election per scenario 3.3(a).', 
 '(b) Second statement has a section 4.2(b) which might elect for scenario 2.4(a) potentially']
import re
from string import ascii_lowercase

s = "(a) be given by XXX for its election per scenario 3.3(a). (b) Second statement has a section 4.2(b) which might elect for scenario 2.4(a) potentially. (c) blah blah  (c) blah blah (d) asd ad(a) (b) (e) ee (b) (a) (d) (f) (f) fff f ff (g) (a) gggg (h) hhhh (b) (i) iii i i (i) i (j) jjj (k) k (l) ll (a) (b) (x) (m) mm (n) nn (o) oo) () () (p) ppp (A) (B) (Q) (q) qq (r) rr (s) ss (t) tt( u ) (u) uu (v) vvv (ww) (w) ww (x) xx (y) yy (z) zzz"

pattern = "".join([f"((?: |^)\({l}\) .+)" for l in ascii_lowercase])

for result in re.findall(pattern, s)[0]:
    print(result.strip())
(a) be given by XXX for its election per scenario 3.3(a).
(b) Second statement has a section 4.2(b) which might elect for scenario 2.4(a) potentially. (c) blah blah
(c) blah blah
(d) asd ad(a) (b)
(e) ee (b) (a) (d) (f)
(f) fff f ff
(g) (a) gggg
(h) hhhh (b) (i) iii i i
(i) i
(j) jjj
(k) k
(l) ll (a) (b) (x)
(m) mm
(n) nn
(o) oo) () ()
(p) ppp (A) (B) (Q)
(q) qq
(r) rr
(s) ss
(t) tt( u )
(u) uu
(v) vvv (ww)
(w) ww
(x) xx
(y) yy
(z) zzz
Ruby look behind regex error: invalid pattern in look-behind

Ruby look behind regex error: invalid pattern in look-behind


By : user2897275
Date : March 29 2020, 07:55 AM
it should still fix some issue The challenge calls to capture all characters including the second "." , You may use
code :
s = s.gsub(/\A([^.]*\.[^.]*)\..*/, '\1')
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