I'm using the VS2012 replace feature to replace some text in the editor.
I want to replace >OneWord</Label> with Content="OneWord" />.
And I want to replace >More than one word</Label> with Content="More than one word" />.
At the moment I have Find what filled with
>(\w+|[^\S\r\n]+)</Label>
and Replace with filled with
Content="$1" />
This works for the first case where only one word is used, but not for the second case.
If I use >(\w+|[^\S\r\n]+)+</Label> I get Content="word" /> for the secod case.
How can I define my regular expressions to work in both cases?
At the moment your are matching either a sequence of word characters \w+ or a sequence of whitespace [^\S\r\n]+.
To solve your problem, just move the quantifier and add another group:
>((\w|[^\S\r\n])+)</Label>
Your result is still in $1.
Related
I need to transform text using regex
TPI +2573<br>
NM$ +719<br>
Молоко +801<br>
Прод. жизнь +6.5<br>
Оплод-сть +3.6<br>
Л. отела 6.3/3.9<br>
Вымя +1.48<br>
Ноги +1.61<br>
to this one
<strong>TPI</strong> +2573<br>
<strong>NM$</strong> +719<br>
<strong>Молоко</strong> +801<br>
<strong>Прод. жизнь</strong> +6.5<br>
<strong>Оплод-сть</strong> +3.6<br>
<strong>Л. отела</strong> 6.3/3.9<br>
<strong>Вымя</strong> +1.48<br>
<strong>Ноги</strong> +1.61<br>
Is it possible with regex in PhpStorm's search and replace dialog?
Given your text, you can use this regex,
.* +
and replace it with <strong>$0</strong> (Notice there is a space after </strong>)
We're using .* to capture everything but stop just before one (possible one or more) space because that's the point after which we want the text to remain intact. Once we capture the text, we use back-reference $0 to replace the match with <strong>$0</strong> so only matched text is placed within <strong> tags.
Regex Demo
Just in case, if this doesn't work for any of the samples you haven't included in your post, then please list the rules of replacement and I will give you a more robust solution, that will work flawlessly for your given rules.
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 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 have this dataset: (about 10k times)
<Id>HOW2SING</Id>
<PopularityRank>1</PopularityRank>
<Title><![CDATA[Superior Singing Method - Online Singing Course]]></Title>
<Description><![CDATA[High Quality Vocal Improvement Product With High Conversions. Online Singing Lessons Course Converts Like Crazy Using Content Packed Sales Video. You Make 75% On Every Sale Including Front End, Recurring, And 1-click Upsells!]]></Description>
<HasRecurringProducts>true</HasRecurringProducts>
<Gravity>45.9395</Gravity>
<PercentPerSale>74.0</PercentPerSale>
<PercentPerRebill>20.0</PercentPerRebill>
<AverageEarningsPerSale>74.9006</AverageEarningsPerSale>
<InitialEarningsPerSale>70.1943</InitialEarningsPerSale>
<TotalRebillAmt>16.1971</TotalRebillAmt>
<Referred>75.0</Referred>
<Commission>75</Commission>
<ActivateDate>2011-06-23</ActivateDate>
</Site>
I am trying to do the following:
Get the data from within the tags, and use it to create a URL, so in this example it should make
http://www.reviews.how2sing.domain.com
also, all other data has to go, i want to perform a REGEX function that will just give me a list of URLS.
I prefer to do it using notepad++ but i suck at regex, any help would be welome
To keep the regex relatively simple you can just use:
.*?<id>(.+?)</id>
Replace with:
http://www.reviews.\1.domain.com\n
That will search and replace all instances of Id tag and preceding text. You can then just remove the last manually.
Make sure matches newline is selected.
Regex is straightforward, only slightly tricky part is that it uses +? and *? which are non-greedy. This prevents the whole file from being matched. The () indicate a capture group that is used in the replacement, i.e. \1.
If you want to a regex that will include replacing the last part then use:
.*?(?:(<id>)?(.+?)</id>).+?(?:<id>|\Z)
This is a bit more tricky, it uses:
?:. A non-capturing group.
| OR
\Z end of file
Basically, the first time it will match everything up to the end of the first </id> and replace up to and including the next <id>. After that it will have replaced the starting <id> so everything before </id> goes in the group. On the last match it will match the end of file \Z.
If you only want the Id values, you can do:
'<Id>([^<]*)<\/Id>'
Then you can get the first captured group \1 which is the Id text value and then create a link from it.
Here is a demo:
http://regex101.com/r/jE9qN8
[UPDATE]
To get rid of all other lines, match this regex: '.*<Id>([^<]*)<\/Id>.*' and replace by first captured group \1. Note for the regex match, since there are multiple lines, you will need to have the DOTALL or /s flag activated to also match newlines.
Hope that helps.
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"