I am using the following regex to find strings of type
${a-z.a-z}
where the .a-z part is optional
Here's the regex I am using -
\$\{[A-Za-z]+\.*[A-Z][a-z]*\}
This will match strings like
${users}
${user.firstName}
I wanted the above matched strings to be replaced with the following corresponding replacements
<c:out value="${users}"/>
<c:out value="${user.firstName}"/>
How to do that in Eclipse?
Note, I can write a script to replace all such occurrences, but I do NOT want to replace all occurrences, but only certain ones by looking at them manually (therefore I want to do this through an editor).
I did this using groups -
Find : (\$\{([A-Za-z]+)\.*[A-Z][a-z]*\})
Replace With: <c:out value="$1"/>
Related
I have a txt file with <i> and </i> between words that I would like to remove using Editpad
For example, I'd like to keep when it's like this:
<i>Phrases and words.</i>
And I'd like to remove the </i> and <i> tags inside the phrase, when it's like this:
<i>Phrases</i>and<i> words.</i>
<i>Phrases</i>and <i>words.</i>
I was trying to do that using regex, but I couldn't do it.
As the tag is followed by space or a word character I could find when the line has the double tag with
/ <i>|<\/i> /
but this way I can't just press replace for nothing, I have to edit line by line I search.
There's anyway to accomplish that?
* Edited *
Another example of lines found on the subtitle text
<i>- find me on the chamber.</i>
- What? <i>Go. Go, go, go!</i>
Rule number one: you can't parse html with regex.
That being said, if you know each line follows a certain pattern, you can usually hack something together to work. ;)
If I've understood correctly, it looks like you can simply remove all <i> and </i> that aren't either at the beginning or end of the lines. In that case, one method you could try is the following regex:
(?<=.)\<\/?i\>(?=.)
This will match the tags, with a lookahead and behind to make sure that we aren't at the end/start of a line (by checking if another character exists in front/behind. (Note that typically matched characters in a lookahead/behind won't be replaced when you search/replace.)
Disclaimer: this works on regex101, but notepad++ may have some differences to the pcre regex style.
update to work with Editpad
EDIT: since this question is actually wanting to know how to do this in Editpad, below is a modified alternative:
Try searching for the regex: (.)\<\/?i\>(.). This will match (and capture) exactly one character before and after the <i> tags.
When replacing, use backreferences to replace the entire match with the two captured characters - a replacement string of \1\2 should work.
I would like to use regular expression for replacing a certain pattern in the Kettle. For example, AAAA >5< BBBB, I want to replace this with AAAA 555 BBBB. I know how to find the pattern, but I am not sure how to replace that with new string. The one thing I have to keep is that I have to find pattern together ><, not separately like > or < because there is another pattern <5>.
You can use the "Replace in String" step in a transformation.
Set use RegEx to "Y", type your regex on the Search box, with capturing groups if necessary, and the replacement string in the replacement box, referring to capture groups as $1, $2, ...
It'll replace all occurrences of the regex in the original string.
If the Out Stream field is ommitted, it'll overwrite the In stream field.
If you want the pattern >\d< replaced by a triple of the found digit, you can use Replace-In-String in regex mode:
Search: (.*)(>(\d)<)(.*)
Replace: $1$3$3$3$4
If you want all such patterns treated the same:
Search: (>(\d)<)
Replace: $2$2$2
EDIT due to your improved requirement
Since you intend to convert your "simple" markup to a more HTML-like markup, you better use a User-Defined-Java-Expression. Also, you must avoid to reintroduce simple markup when replacing repeatedly.
I would like replace a standard string in a file, with another that is a result of a regular expression. The standard text looks like:
<xsl:variable name="ServiceCode" select="###"/>
I would like to replace ### with a servicecode, that I can find later in the same file, from this URL:
<a href="/Services/xyz" target="_self">
The regular expression (?<=\/Services\/)(.*)(?=\" )
returns the required service code "xyz".
So, I opened Notepad++, added "###" to the "Find what" and this RegEx to the "Replace with" section, and expected that the ### text will be replaced by xyz.
But I got this result:
<xsl:variable name="ServiceCode" select="?<=/Services/.*?=" "/>
I am new to RegEx, do I need to use different syntax in the replace section than I use to find a string? Can someone give me a hint how to achieve the required result? The goal is to standardize tons of files with similar structure as now all servicecodes are hardcoded in several places in the file. Thanks.
You could use a lookahead for capturing the part ahead.
Search for: (?s)###(?=.*/Services/([^"]+)") and replace with: $1
(?s) makes the dot also match newlines (there is also a checkbox available in np++)
[^"] matches a character that is not "
The replacement $1 corresponds to capture of first parenthesized subpattern.
I am no expert at RegEx but I think I may be able to help. It looks like you might be going at this the wrong way. The regex search that you are using would normally work like this:
The parenthesis () in RegEx allow you to select part of your search and use that in the replace section.
You place (?<=\/Services\/)(.*)(?=\" ) into the "Find what" section in Notepad++.
Then in the "Replace with" section you could use \1 or \2 or \3 to replace the contents of your search with what was found in the (?<=\/Services\/) or (.*) or (?=\" ) searches respectively.
Depending on the structure of your files, you would need to use a RegEx search that selects both lines of code (and the specific parts you need), then use a combination of \1\2\3 etc. to replace everything exactly how it was, except for the ### which you could replace with the \number associated with xyz.
See http://docs.notepad-plus-plus.org/index.php/Regular_Expressions for more info.
I am modifying a json file for Rust with the Oxide mod and I am unfamiliar with programming in general.
I need to replace several instances of "stackable": 64.0, with "stackable": 15000.0,
The value is not always "64.0". There are instances of "1000.0" and "1.0" etc...
What I am currently doing is:
Search for: "stackable":/*/*/*/*
Replace with: "stackable": 15000.0,
This replaces the text that I want but essentially the end result is it simply enters the text in before the default value. So a line will end up looking like
"stackable": 15000.0, 64.,
How can I replace using wildcards but also remove the default value?
First match "stackable" and then match spaces if any and then match numbers with . in them, using a character class.
Find: (stackable)\s*[\d.]+
Replace: $1: 15000.0
I need to replace following lines in XML file:
hashName="'Miecz Nieb. Wojownika+5IMiecz Nieb. Wojownika+5" name="Miecz Nieb. Wojownika+5"
As the above line is not correct, I want it to be replaced like this:
hashName="'Miecz Nieb. Wojownika+5'" name="Miecz Nieb. Wojownika+5"
(It should take the item name from the name="" attr!).
This is what I got at the moment, its not working as expected since it does remove my name="..." attribute.
Search for:
hashName="(')(.*)"(.)name="(.*)"(.)/
Replace with:
hashName="'\4'" name="\4"
For this simple example this is working
Search for
hashName="[^"]*"\s*name="([^"]*)"
and replace with
hashName="'\1'" name="\1"
If you don't want to capture or group characters, don't put brackets around it, therefor I removed most of them.
To avoid that too much is matched, e.g. if you have two "name" attributes in one row, I used [^"]* to do a non greedy matching.
This should work
Search for: hashName=\".+\" name=\"(.+)\"
Replace with: hashName="'\1'" name="\1"