Usage of Regular Expression Extractor JMeter? - testing

Using Regular Extractor in JMeter, I need to get the value of "fullBkupUNIXTime" from the below response,
{"fullBackupTimeString":["Mon 10 Apr 2017 14:14:36"],"fullBkupUNIXTime":["1491833676"],"fullBackupDirName":["10_04_2017_0636"]}
I tried with Ref Name as time and
Regular Expression: "fullBkupUNIXTime": "([0-9])" and "(.+?)"
and pass them as input for 2nd request ${time}
The above 2 two doesn't work out for me.
Please Help me out of this.

First of all: why not just use this thing?
Then, if you firm with your RegExp adventure to get happen.
First expression is not going to work because you've defined it to match exactly one [0-9] charcter.
Add the appropriate repetition character, like "fullBkupUNIXTime": "([0-9]+)".
And basically it make sense to tell the engine to stop at first narrowest match too: "fullBkupUNIXTime": "([0-9]+?)"
Next, make sure you're handling space chars between key and value and colon mark properly. Better mark them explicitly, if any, with \s
And last but not least: make sure you're properly handle multiple lines (if appropriate, of course). Add the (?m) modifier to your expression.
And/or (?im) to be not case-sensitive, in addition.

[ is a reserve character in regex, you need to escape it, in your case use:
Regular Expression fullBkupUNIXTime":\["(\d+)
Template: $1$
Match No.: 1

Related

Input character validation using word validation regular expression

Let's say, I have a regular expression that checks the validation of the input value as a whole. For example, it is an email input box and when user hits enter, I check it against ^[A-Z0-9._%+-]+#[A-Z0-9.-]+\.[A-Z]{2,4}$ to see if it is a valid email address.
What I want to achieve is, I want to intercept the character input too, and check every single input character to see if that character is also a valid character. I can do this by adding an extra regular expression, e.g. [A-Z0-9._%+-] but that is not what I want.
Is there a way to extract the widest possible range of acceptable characters from a given regular expression? So in the example above, can I extract all the valid characters that are defined by the original regular expression (i.e. ^[A-Z0-9._%+-]+#[A-Z0-9.-]+\.[A-Z]{2,4}$) programmatically?
I would appreciate any help or hint.
P.S. This is project for iOS written in Objective-C.
If you don't mind writing half a regex parser, certainly. You would have to be able to distinguish literals from meta-characters and to unroll/merge all character classes (including negated character classes, and nested negated character classes, if you regex flavor supports them).
If NSRegularExpressions doesn't come with some convenience method, I cannot imagine how it would be possible otherwise. Just think about ^. When it is outside of a character class, it's a meta-character that you can ignore. If it is inside a character class, it's a meta-character, that negates the character class unless it is not the first character. - is a meta-character inside character classes, unless it is the first character, the last character, or right after another character range (depending on regex flavor). And I'm not even speaking about escaped characters.
I don't know about NSRegularExpressions, but some flavors also support nested character classes (like [a-z[^aeiou]] for all consonants). I think you get where I am going with this.

IP Address/Hostname match regex

I need to match two ipaddress/hostname with a regular expression:
Like 20.20.20.20
should match with 20.20.20.20
should match with [http://20.20.20.20/abcd]
should not match with 20.20.20.200
should not match with [http://20.20.20.200/abcd]
should not match with [http://120.20.20.20/abcd]
should match with AB_20.20.20.20
should match with 20.20.20.20_AB
At present i am using something like this regular expression: "(.*[^(\w)]|^)20.20.20.20([^(\w)].*|$)"
But it is not working for the last two cases. As the "\w" is equal to [a-zA-Z0-9_]. Here I also want to eliminate the "_" underscore. I tried different combination but not able to succeed. Please help me with this regular expression.
(.*[_]|[^(\w)]|^)10.10.10.10([_]|[^(\w)].*|$)
I spent some more time on this.This regular expression seems to work.
I don't know which language you're using, but with Perl-like regular expressions you could use the following, shorter expression:
(?:\b|\D)20\.20\.20\.20(?:\b|\D)
This effectively says:
Match word boundary (\b, here: the start of the word) or a non-digit (\D).
Match IP address.
Match word boundary (\b, here: the end of the word) or a non-digit (\D).
Note 1: ?: causes the grouping (\b|\D) not to create a backreference, i.e. to store what it has found. You probably don't need the word boundaries/non-digits to be stored. If you actually need them stored, just remove the two ?:s.
Note 2: This might be nit-picking, but you need to escape the dots in the IP address part of the regular expression, otherwise you'd also match any other character at those positions. Using 20.20.20.20 instead of 20\.20\.20\.20, you might for example match a line carrying a timestamp when you're searching through a log file...
2012-07-18 20:20:20,20 INFO Application startup successful, IP=20.20.20.200
...even though you're looking for IP addresses and that particular one (20.20.20.200) explicitly shouldn't match, according to your question. Admittedly though, this example is quite an edge case.

SQL to return results for the following regex

I have the following regular expression:
WHERE A.srvc_call_id = '40750564' AND REGEXP_LIKE (A.SRVC_CALL_DN, '[^TEST]')
The row that contains 40750564 has "TEST CALL" in the column SRVC_CALL_DN and REGEXP_LIKE doesn't seem to be filtering it out. Whenever I run the query it returns the row when it shouldn't.
Is my regex pattern wrong? Or does SQL not accept [^whatever]?
The carat anchors the expression to the start of a string. By enclosing the letters T, E, S & T in square brackets you're searching, as barsju suggests for any of these characters, not for the string TEST.
You say that SRVC_CALL_DN contains the string 'TEST CALL', but you don't say where in the string. You also say that you're looking for where this string doesn't match. This implies that you want to use not regexp_like(...
Putting all this together I think you need:
AND NOT REGEXP_LIKE (A.SRVC_CALL_DN, '^TEST[[:space:]]CALL')
This excludes every match from your query where the string starts with 'TEST CALL'. However, if this string may be in any position in the column you need to remove the carat - ^.
This also assumes that the string is always in upper case. If it's in mixed case or lower, then you need to change it again. Something like the following:
AND NOT REGEXP_LIKE (upper(A.SRVC_CALL_DN), '^TEST[[:space:]]CALL')
By upper-casing SRV_CALL_DN you ensure that you're always going to match but ensure that your query may not use an index on this column. I wouldn't worry about this particular point as regular expressions queries can be fairly poor at using indexes anyway and it appears as though SRVC_CALL_ID is indexed.
Also if it may not include 'CALL' you will have to remove this. It is best when using regular expressions to make your match pattern as explicit as possible; so include 'CALL' if you can.
Try with '^TEST' or '^TEST.*'
Your regexp means any string not starting with any of the characters: T,E,S,T.
But your case is so simple, starts with TEST. Why not use a simple like:
LIKE 'TEST%'

Is it possible to ignore characters in a string when matching with a regular expression

I'd like to create a regular expression such that when I compare the a string against an array of strings, matches are returned with the regex ignoring certain characters.
Here's one example. Consider the following array of names:
{
"Andy O'Brien",
"Bob O'Brian",
"Jim OBrien",
"Larry Oberlin"
}
If a user enters "ob", I'd like the app to apply a regex predicate to the array and all of the names in the above array would match (e.g. the ' is ignored).
I know I can run the match twice, first against each name and second against each name with the ignored chars stripped from the string. I'd rather this by done by a single regex so I don't need two passes.
Is this possible? This is for an iOS app and I'm using NSPredicate.
EDIT: clarification on use
From the initial answers I realized I wasn't clear. The example above is a specific one. I need a general solution where the array of names is a large array with diverse names and the string I am matching against is entered by the user. So I can't hard code the regex like [o]'?[b].
Also, I know how to do case-insensitive searches so don't need the answer to focus on that. Just need a solution to ignore the chars I don't want to match against.
Since you have discarded all the answers showing the ways it can be done, you are left with the answer:
NO, this cannot be done. Regex does not have an option to 'ignore' characters. Your only options are to modify the regex to match them, or to do a pass on your source text to get rid of the characters you want to ignore and then match against that. (Of course, then you may have the problem of correlating your 'cleaned' text with the actual source text.)
If I understand correctly, you want a way to match the characters "ob" 1) regardless of capitalization, and 2) regardless of whether there is an apostrophe in between them. That should be easy enough.
1) Use a case-insensitivity modifier, or use a regexp that specifies that the capital and lowercase version of the letter are both acceptable: [Oo][Bb]
2) Use the ? modifier to indicate that a character may be present either one or zero times. o'?b will match both "o'b" and "ob". If you want to include other characters that may or may not be present, you can group them with the apostrophe. For example, o['-~]?b will match "ob", "o'b", "o-b", and "o~b".
So the complete answer would be [Oo]'?[Bb].
Update: The OP asked for a solution that would cause the given character to be ignored in an arbitrary search string. You can do this by inserting '? after every character of the search string. For example, if you were given the search string oleary, you'd transform it into o'?l'?e'?a'?r'?y'?. Foolproof, though probably not optimal for performance. Note that this would match "o'leary" but also "o'lea'r'y'" if that's a concern.
In this particular case, just throw the set of characters into the middle of the regex as optional. This works specifically because you have only two characters in your match string, otherwise the regex might get a bit verbose. For example, match case-insensitive against:
o[']*b
You can add more characters to that character class in the middle to ignore them. Note that the * matches any number of characters (so O'''Brien will match) - for a single instance, change to ?:
o[']?b
You can make particular characters optional with a question mark, which means that it will match whether they're there or not, e.g:
/o\'?b/
Would match all of the above, add .+ to either side to match all other characters, and a space to denote the start of the surname:
/.+? o\'?b.+/
And use the case-insensitivity modifier to make it match regardless of capitalisation.

Change Url using Regex

I have url, for example:
http://i.myhost.com/myimage.jpg
I want to change this url to
http://i.myhost.com/myimageD.jpg.
(Add D after image name and before point)
i.e I want add some words after image name and before point using regex.
What is the best way do it using regex?
Try using ^(.*)\.([a-zA-Z]{3,5}) and replacing with \1D\2. I'm assuming the extension is 3-5 alphanumeric numbers but you can modify it to suit. E.g. if it's just jpg images then you can put that instead of the [a-zA-Z]{3,5}.
Sounds like a homework question given the solution must use a regex, on that assumption here is an outline to get you going.
If all you have is a URL then #mathematical.coffee's solution will suit. However if you have a chunk of text within which is one or more URLs and you have to locate and change just those then you'll need something a little more involved.
Look at the structure of a URL: {protocol}{address}{item}; where
{protocol} is "http://", "ftp://" etc.;
{address} is a name, e.g. "www.google.com", or a number, e.g. "74.125.237.116" - there will always be at least one dot in the address; and
{item} is "/name" where name is quite flexible - there will be zero or more items, you can think of them as directories and a file but this isn't strictly true. Also the sequence of items can end in a "/" (including when there are zero of them).
To make a regex which matches a URL start by matching each part. In the case of the items you'll want to match the last in the sequence separately - you'll have zero or more "directories" and one "file", the latter must be of the form "name.extension".
Once you have regexes for each part you just concatenate them to produce a regex for the whole. To form the replacement pattern you can surround parts of your regex with parentheses and refer to those parts using \number in the replacement string - see #mathematical.coffee's solution for an example.
The best way to learn regexs is to use an editor which supports them and just experiment. The exact syntax may not be the same as NSRegularExpression but they are mostly pretty similar for the basic stuff and you can translate from one to another easily.