I have instances of JSON in the following format - "firstName":"sample" and "firstName"="sample". I am trying to modify my regex for the colon that replaced the "sample" with asterisks to include cases where instead of a colon there is an equal sign.
\"firstName\"\s*:\s*\"(.*?)\"
I tried adding an OR -
\"firstName\"\s*(:|=)\s*\"(.*?)\"
However, this gives me the following output --> "firstName" * "******************" for cases with the colon as well as an equals sign. The expected output should not replace the colon or equals, only what is inside the string. Any help is appreciated.
\"firstName\"\s*[:|=]\s*\"(.*?)\"
Related
$.validator.addMethod('AZ09_', function (value) {
return /^[a-zA-Z0-9.-_]+$/.test(value);
}, 'Only letters, numbers, and _-. are allowed');
When I use somehting like test-123 it still triggers as if the hyphen is invalid. I tried \- and --
Escaping using \- should be fine, but you can also try putting it at the beginning or the end of the character class. This should work for you:
/^[a-zA-Z0-9._-]+$/
Escaping the hyphen using \- is the correct way.
I have verified that the expression /^[a-zA-Z0-9.\-_]+$/ does allow hyphens. You can also use the \w class to shorten it to /^[\w.\-]+$/.
(Putting the hyphen last in the expression actually causes it to not require escaping, as it then can't be part of a range, however you might still want to get into the habit of always escaping it.)
The \- maybe wasn't working because you passed the whole stuff from the server with a string. If that's the case, you should at first escape the \ so the server side program can handle it too.
In a server side string: \\-
On the client side: \-
In regex (covers): -
Or you can simply put at the and of the [] brackets.
Generally with hyphen (-) character in regex, its important to note the difference between escaping (\-) and not escaping (-) the hyphen because hyphen apart from being a character themselves are parsed to specify range in regex.
In the first case, with escaped hyphen (\-), regex will only match the hyphen as in example /^[+\-.]+$/
In the second case, not escaping for example /^[+-.]+$/ here since the hyphen is between plus and dot so it will match all characters with ASCII values between 43 (for plus) and 46 (for dot), so will include comma (ASCII value of 44) as a side-effect.
\- should work to escape the - in the character range. Can you quote what you tested when it didn't seem to? Because it seems to work: http://jsbin.com/odita3
A more generic way of matching hyphens is by using the character class for hyphens and dashes ("\p{Pd}" without quotes). If you are dealing with text from various cultures and sources, you might find that there are more types of hyphens out there, not just one character. You can add that inside the [] expression
I have the following two lines of code:
json_str = _cases.to_json
path += " #{USER} #{PASS} #{json_str}"
When I use the debugger, I noticed that json_str appears to be formatted as JSON:
"[["FMCE","Wiltone","Wiltone","04/10/2018","Marriage + - DOM"]]"
However, when I interpolate it into another string, the quotes are removed:
"node superuser 123456 [["FMCE","Wiltone","Wiltone","04/10/2018","Marriage + - DOM"]]"
Why does string interpolation remove the quotes from JSON string and how can I resolve this?
I did find one solution to the problem, which was manually escaping the string:
json_str = _cases.to_json.gsub('"','\"')
path += " #{USER} #{PASS} \"#{json_str}\""
So basically I escape the double quotes generated in the to_json call. Then I manually add two escaped quotes around the interpolated variable. This will produce a desired result:
node superuser 123456 "[[\"FMCE\",\"Wiltone\",\"Wiltone\",\"04/10/2018\",\"Marriage + - DOM\"]]"
Notice how the outer quotes around the collection are not escaped, but the strings inside the collection are escaped. That will enable JavaScript to parse it with JSON.parse.
It is important to note that in this part:
json_str = _cases.to_json.gsub('"','\"')
it is adding a LITERAL backslash. Not an escape sequence.
But in this part:
path += " #{USER} #{PASS} \"#{json_str}\""
The \" wrapping the interpolated variable is an escape sequence and NOT a literal backslash.
Why do you think the first and last quote marks are part of the string? They do not belong to the JSON format. Your program’s behavior looks correct to me.
(Or more precisely, your program seems to be doing exactly what you told it to. Whether your instructions are any good is a question I can’t answer without more context.)
It's hard to tell with the small sample, but it looks like you might be getting quotes from your debugger output. assuming the output of .to_json is a string (usually is), then "#{json_str}" should be exactly equal to json_str. If it isn't, that's a bug in ruby somehow (doubtful).
If you need the quotes, you need to either add them manually or escape the string using whatever escape function is appropriate for your use case. You could use .to_json as your escape function even ("#{json_str.to_json}", for example).
I am trying to extract an email address from user input text in Watson Conversation. First thing first, I need to trigger a particular node using an if condition like this:
input.text.contains('\^(([^<>()[].,;:s#\"]+(.[^<>()[].,;:s#\"]+)*)|(\".+\"))#(([[0-9]{1,3}.[0-9]{1,3}.[0-9]{1,3}.[0-9]{1,3}])|(([a-zA-Z-0-9]+.)+[a-zA-Z]{2,}))$\')
But it doesn't work, I tried a lot of regexes that I found on the internet but none of them work. Does anyone know how to write a proper regex?
I suggest using a much simpler, approximate, regex to match emails that you need to use with String.matches(string regexp) method that accepts a regex:
input.text.matches('^\\S+#\\S+\\.\\S+$')
Do not forget to double escape backslashes so as to define literal backslashes in the pattern.
Pattern details:
^ - start of string
\\S+ - one or more non-whitespace chars
# - a # symbol
\\S+ - one or more non-whitespace chars
\\. - a literal dot
\\S+ - one or more non-whitespace chars
$ - end of string.
How to write a regular expression that matches characters between first '/' and second '/'.
Example :
5/2/266/267 (Expected Output : 2)
68/567/8/56 (Expected Output : 567)
......
Similarly also how write the regex that matches characters between second '/' and third '/'.
Example :
5/2/266/267 (Expected Output : 266)
68/567/8/56 (Expected Output : 8)
I suggest to use this one:
^[^\/]*\/([^\/]+?)\/.*
Largely depends on the engine, but most should support the reluctant quantifier. This will work at least in JavaScript, PHP, ruby, python, and possibly others:
/(.*?)/
You may need to escape delimiters, e.g. \/
A pattern like this would work:
/([^/]*)/
Or possibly using lookarounds:
(?<=/)[^/]*(?=/)
Note however, that many regular expression engines use / to delimit regular expression literals. In that case, you'd need to escape the / like this:
(?<=\/)[^\/]*(?=\/)
I want to replace "\cite{foo123a}" with "[1]" and backwards. So far I was able to replace text with the following command
body.replaceText('.cite{foo}', '[1]');
but I did not manage to use
body.replaceText('\cite{foo}', '[1]');
body.replaceText('\\cite{foo}', '[1]');
Why?
The back conversion I cannot get to work at all
body.replaceText('[1]', '\\cite{foo}');
this will replace only the "1" not the [ ], this means the [] are interpreted as regex character set, escaping them will not help
body.replaceText('\[1\]', '\\cite{foo}');//no effect, still a char set
body.replaceText('/\[1\]/', '\\cite{foo}');//no matches
The documentation states
A subset of the JavaScript regular expression features are not fully supported, such as capture groups and mode modifiers.
Can I find a full description of what is supported and what not somewhere?
I'm not familiar with Google Apps Script, but this looks like ordinary regular expression troubles.
Your second conversion is not working because the string literal '\[1\]' is just the same as '[1]'. You want to quote the text \[1\] as a string literal, which means '\\[1\\]'. Slashes inside of a string literal have no relevant meaning; in that case you have written a pattern which matches the text /1/.
Your first conversion is not working because {...} denotes a quantifier, not literal braces, so you need \\\\cite\\{foo\\}. (The four backslashes are because to match a literal \ in a regular expression is \\, and to make that a string literal it is \\\\ — two escaped backslashes.)