I have a document and I want to replace all the paragraph return with the space.
I tried to replace it with space but unfortunately it did wrong like in the attache image it found space and A in search and it replace it with space nd.
What happened is that heaven." "
And
is replaced by heaven. nd
But I want it to be
heaven." "And
In MS Word Search and Replace, you may use groups and refer to their captured contents via backreferences like \1, \2, etc.
So, if you want to match a paragraph break that is followed with an ASCII letter, you need to use
^013([A-Za-z])
and replace with \1 (a space + \1, where \1 inserts the letter).
To find and replace * you can just use a normal replacing mode (uncheck Wildcard option). If you need to use * as a literal symbol in the wildcard pattern, put it into a []: [*].
Related
In a regular expression (notepad++), I want to search for:( )|(:)|(_)|(\.), and to insert \ before to, as above, a blank space, colon, under line and ".".
Search example: abcd:1234 jiod.8ufd_adfd
Result: abcd\:1234\ jiod\.8ufd\_adfd
Briefly, how can I refer to what was found in the replace expression?
Note that it is not \1, \2, \3 or \4 in the example, as I need to include what was found, there is no way to know which was found, is there?
You can use a single character class (instead of using the alternation with capturing groups) to match one of the listed
In the replacement use $& to refer to the matched text and prepend a backslash.
Match
[:\h._]
Replace with
\\$&
The character class matches either a colon, horizontal whitespace char, dot or underscore.
Regex demo
There's no such thing as insert, because if you think about it, inserting is just replacing the original with a new string that contains the old text as well.
Try this instead: search for ([ :_.]) (your original regex is pointlessly complicated) and replace with \\$1 (ie, slash followed by the original text).
I'm trying to do a project wide search and replace
from:
drivers[i].findElement(By.id("elementID")).click();
to:
findAndClick(driver[i], "elementID", true)
The issue is the elementID can be anything so I'm trying to wildcard search and replace with what's in the wildcard?
You'll need to use .+? instead of * here since this uses regular expressions.
In regular expressions a dot . means "any character", the plus + means "one or more times", and the question mark ? after this means "try to match this as few as possible times" - which is useful so it won't keep matching past your quote marks
edit
To be clear though, you have to make a valid regex, which means you'll need to escape your parenthesis, dots, etc.
Here's the full solution
Find: drivers\[i\]\.findElement\(By\.id\("(.+?)"\)\)\.click\(\);
replace with: findAndClick(driver[i], "$1", true)
Note the added unescaped parentheses in there around the "wildcard" (.+) this creates a capture group in a regex, which is what translates to $1 in the replacement since it's the 1st capture group.
I have a large file with content inside every bracket. This is not at the beginning of the line.
1. Atmos-phere (7800)
2. Atmospheric composition (90100)
3.Air quality (10110)
4. Atmospheric chemistry and composition (889s120)
5.Atmospheric particulates (10678130)
I need to do the following
Replace the entire content, get rid of line numbers
1.Atmosphere (10000) to plain Atmosphere
Delete the line numbers as well
1.Atmosphere (10000) to plain Atmosphere
make it a hyperlink
1.Atmosphere (10000) to plain linky study
[I added/Edit] Extract the words into a new file, where we get a simple list of key words. Can you also please explain the numbers in replace the \1\2, and escape on some characters
Each set of key words is a new line
Atmospheric
Atmospheric composition
Air quality
Each set is a on one line separated by one space and commas
Atmospheric, Atmospheric composition, Air quality
I tried find with regex like so, \(*\) it finds the brackets, but dont know how to replace this, and where to put the replace, and what variable holds the replacement value.
Here is mine exression for notepad ([0-9(). ]*)(.*)(\s\()(.*)
You need split your search in groups
([0-9. ]*) numbers, spaces and dots combination in 0 or more times
(.*) everything till next expression
(\s\() space and opening parenthesis
(.*) everything else
In replace box - for practicing if you place
\1\2\3\4 this do nothing :) just print all groups from above from 1.1 to 1.4
\2 this way you get only 1.2 group
new_thing\2new_thing adds your text before and after group
<a href=blah.com/\2.html>linky study</a> so now your text is added - spaces between words can be problematic when creating link - so another expression need to be made to replace all spaces in link to i.e. _
If you need add backslash as text (or other special sign used by regex) it must be escaped so you put \\ for backslash or \$ for dolar sign
Want more tune - <a href=blah.com/\2.html>\2</a> add again 1.2 group - or use whichever you want
On the screenshot you can see how I use it (I had found and replaced one line)
Ok and then we have case 4.2 with colon at the end so simply add colon after extracted section:
change replace from \2 to \2,
Now you need join it so simplest way is to Edit->Line Operations->Join Lines
but if you want to be real pro switch to Extended mode (just above Regular expression mode in Replace window) and Find \r\n and replace with space.
Removing line endings can differ in some cases but this is another story - for now I assume that you using windows since Notepad++ is windows tool and line endings are in windows style :)
The following regex should do the job: \d+\.\s*(.*?)\s*\(.*?\).
And the replacement: <a href=example.com\\\1.htm>\1</a>.
Explanation:
\d+ : Match a digit 0 or more times.
\. : Match a dot.
\s* : Match spaces 0 or more times.
(.*?) : Group and match everything until ( found.
\s* : Match spaces 0 or more times.
\(.*?\) : Match parenthesis and what's between it.
The replacement part is simple since \1 is referring to the matching group.
Online demo.
Try replacing ^\d+\.(.*) \(\w+\)$ with <a href=blah.com\\\1.htm>linky study</a>.
The ^\d+. removes the leading number and dot. The (.*) collects the words. Then there is a single space. The \(\w+\)$ matches the final number in brackets.
Update for the added Q4.
Regular expressions capture things written between round brackets ( and ). Brackets that are to be found in the text being searched must be escaped as \( and \). In the replacement expression the \1 and \2 etc are replaced by the corresponding capture expression. So a search expression such as Z(\d+)X([aeiou]+)Y might match Z29XeieiY then the replacement expression P\2Q\1R would insert PeieiQ29R. In the search at the top of this answer there is one capture, the (.) captures or collects the words and then the \1 inserts the captured words into the replacement text.
I need to reformat a text file a bit in my Notepad++ and I have a text of this kind:
This is some example text. This is some example text. This is some example text.
- This is some example text.
-This is some example text.
- This is some example text.
- This is some example text.
So as you can see in above text there are two types of "-" preceeding text the one with the space after "-" and ones without it I need to find only the ones without sapce and add it in between "-" and the "text"
If I ran piece of code below
-[A-Za-z0-9]
it finds dash and first letter right after it, which is not useful as when I replace the text it changes this first letter which is always different (depending on what is written) so I need to find this and select only the "-" and then replace it with "- " unless there is better way.
For demonstration purposes:
Find what: -([A-Za-z0-9])(.+)
Replace with: - \1\2
The parentheses denote a capture group. In the Replace with line, you use backslash and the number of group to add it.
That said, what you really want to match for is a NOT group, like -([^\s]) (match where a dash isn't immediately followed by a whitespace).
Search for
-([^ ])
and replace with
- \1
[^ ] is a negated character class and matches everything but a space. This character is stored in \1 because of the brackets () around the pattern.
In eclipse, is it possible to use the matched search string as part of the replace string when performing a regular expression search and replace?
Basically, I want to replace all occurrences of
variableName.someMethod()
with:
((TypeName)variableName.someMethod())
Where variableName can be any variable name at all.
In sed I could use something like:
s/[a-zA-Z]+\.someMethod\(\)/((TypeName)&)/g
That is, & represents the matched search string. Is there something similar in Eclipse?
Thanks!
Yes, ( ) captures a group. You can use it again with $i where i is the i'th capture group.
So:
search: (\w+\.someMethod\(\))
replace: ((TypeName)$1)
Hint: Ctrl + Space in the textboxes gives you all kinds of suggestions for regular expression writing.
Using ...
search = (^.*import )(.*)(\(.*\):)
replace = $1$2
...replaces ...
from checks import checklist(_list):
...with...
from checks import checklist
Blocks in regex are delineated by parenthesis (which are not preceded by a "\")
(^.*import ) finds "from checks import " and loads it to $1 (eclipse starts counting at 1)
(.*) find the next "everything" until the next encountered "(" and loads it to $2. $2 stops at the "(" because of the next part (see next line below)
(\(.*\):) says "at the first encountered "(" after starting block $2...stop block $2 and start $3. $3 gets loaded with the "('any text'):" or, in the example, the "(_list):"
Then in the replace, just put the $1$2 to replace all three blocks with just the first two.
NomeN has answered correctly, but this answer wouldn't be of much use for beginners like me because we will have another problem to solve and we wouldn't know how to use RegEx in there. So I am adding a bit of explanation to this. The answer is
search: (\w+\\.someMethod\\(\\))
replace: ((TypeName)$1)
Here:
In search:
First and last (, ) depicts a group in regex
\w depicts words (alphanumeric + underscore)
+ depicts one or more (ie one or more of alphanumeric + underscore)
. is a special character which depicts any character (ie .+ means
one or more of any character). Because this is a special character
to depict a . we should give an escape character with it, ie \.
someMethod is given as it is to be searched.
The two parenthesis (, ) are given along with escape character
because they are special character which are used to depict a group
(we will discuss about group in next point)
In replace:
It is given ((TypeName)$1), here $1 depicts the
group. That is all the characters that are enclosed within the first
and last parenthesis (, ) in the search field
Also make sure you have checked the 'Regular expression' option in
find an replace box
At least at STS (SpringSource Tool Suite) groups are numbered starting form 0, so replace string will be
replace: ((TypeName)$0)
For someone who needs an explanation and an example of how to use a regxp in Eclipse. Here is my example illustrating the problem.
I want to rename
/download.mp4^lecture_id=271
to
/271.mp4
And there can be multiple of these.
Here is how it should be done.
Then hit find/replace button