I am trying to find replace thousands of strings in code - form.controls['property'] with form.get('property') where property is a variable.
So far I got find expression .controls\['\w+'\] with replace expression .get\('\w+'\) but VSCode is replacing .controls['products'] with .get\('\w+'\) for string productForm.controls['products'].controls.length. I need to replace it with productForm.get('products').controls.length Any ideas how to fix it? Thanks!
Parentheses () capture a group. Money sign $ accesses a group by index.
.controls\['(\w+)'\]
.get('$1')
This is the result in VS Code:
See also: What flavor of Regex does Visual Studio Code use?
Related
I have phone numbers on my file and I would like to surround each of them with braces, so if I have this:
"+49-1111"
"+49-2222"
I would like to run a replace operation and to get this:
["+49-1111"]
["+49-2222"]
Is that possible with the current vs code find/replace action?
If it is a dup sorry but I could find any explanation anywhere.
Thanks
This is more relevant to regular expression.
Find ("\+\d{2}-\d{4}"), replace the occurrences with [$1]
Be sure to turn the "Use Regular Expression" switch on.
I'm getting a syntax error and invalid quantifier error in dreamweaver when I try to use a regexp in the source code.
The purpose is to find spaces in front of numbers on table cells and delete them.
(?<=>)\s+(?=\d)
this expression works on notepad++ but not in dreamweaver.
Can this be a Dreamweaver bug or the syntax is wrong?
Of course I can make a text search looking for >\s and replace by > but then I cant catch more spaces than the ones specified in the search string
thanks in advance
PS: Would be nice also to have a multisearch option in the dreamweaver search screen, to run multiple search and replace in one operation, like code clean up. An extension maybe?
I don't use DW but, since I have read several posts about lookarounds problems with DW, I assume that DW doesn't support these regex features.
You can use capturing groups instead (if DW supports it!):
search : (>)\s+(\d)
replace: $1$2
or
replace: \1\2
To append to the previous answer, when formulating a replace statement in DreamWeaver, use the format of $1 rather than ^1 for the variables.
I receive a similar "invalid quantifier" response in DreamWeaver CC 2015.1 when using a negative lookbehind:
(?<!somephrase)
I have the following string 3}HFB}4AF4}1 -M}1.
I have searched for this string using the regex :
([0-9])(\})([A-Z]{3})(\})([0-9][A-Z]{2}[0-9])(\})([0-9])(\s\-)([A-Z])(\})([0-9]).
I want to replace the } with 0. The Result I am looking for is 30HFB04AF401-M01, any assistance is appriciated. The tool I am using is Regex Buddy
A possible solution
Problem solved? In JavaScript at least :-)
"3}HFB}4AF4}1 -M}1".replace(/\}/g, "0");
// "30HFB04AF401 -M01"
I'm missing the point, right?
Assuming the language is JavaScript, we can write something like
"dfghj456783}HFB}4AF4}1 -M}1fghjkl8765".replace(/(?:[\d\w\s]+)([0-9]}[A-Z]{3}}[0-9][A-Z]{2}[0-9]}[0-9] -[A-Z]}[0-9])(?:[\d\w\s]+)/g, function () {
return arguments[1].replace(/}/g, "0");
});
What's possible in other languages though may be a different story.
Try the home of RegexBuddy for details.
So you've already got an expression to find instances of the string. Now you can either use groups to replace the characters, or you can use a separate regular expression over the string you found, simply replacing the } character within group(0) (which is the entire matched part of the input). I would certainly prefer the latter.
Fred seems to have created the replacement method for you already, so I won't repeat it here.
I have managed to find a solution to the formating in the JGSoft Lanugage used by Regex Buddy, thanks to all that provided suggestions that helped me channel my thoughts in the right direction.
Solution(I am still a beginner with Regex hence the syntax might not be efficent, but it does the job!!)
Using Group Names instead of Regex assiging groups with backreference and $ syntax.
Hence to replace 0 for } in the string 3}HFB}4AF4}1 -M}1 or any similar string. I used the following search and replacement syntax
Search : (?<Gp1>([0-9]))(?:})(?<Gp2>([A-Z]){3})(?:})(?<Gp3>([0-9])([A-Z]{2})([0-9]))(?:})(?<Gp4>([0-9]))(?:\s-)(?<Gp5>([A-Z]))(?:})(?<Gp6>[0-9])
Replace : ${Gp1}0${Gp2}0${Gp3}0${Gp4}-${Gp5}0${Gp6}
Result : 30HFB04AF401-M01
I want to replace C# attributes with VB.NET, which means, [Serializable] should become <Serializable>.
The pattern (\[)(.+)(\]) does find the results but I don't know how to replace the first and the last groups with the appropriate parenthesis.
I read this page, but I didn't understand how to use the curly braces for F&R, I tried to wrap the groups with it but it didn't work.
If you are using the Productivity Power Tools extension from Microsoft that support normal .NET regexes, what you would put in the textbox for the replacement given your regular expression above is:
<$2>
where $2 refers to the second capture group in your regex, i.e. the text between the brackets.
Note that this only works with the Quick Find from Productivity Power Tools though. The normal find/replace in Visual Studio use another syntax altogether.
Find what: \[{Serializable}\]
Replace with: <\1>
I have a large solution with a lot of lines that I need to replace.
In Visual Studio, you can search and replace with the aid of regular expressions.
I want to replace lines like:
rst.Fields("CustomerName").Value
rst.Fields("Address").Value
rst.Fields("Invoice").Value
To:
row("CustomerName").ToString()
row("Address").ToString()
row("Invoice").ToString()
Thus keeping the dynamic text part, which can vary.
Is this possible and how?
Update, solution:
Search: rst.Fields{\(.*\)}\.Value
Replace: rst\1.ToString()
Thanks JaredPar!
Try the following
Search Expression: ASpecificCommand(\(.*\))\.ASpecificProperty
Replace Expression: ATotallyDifferentCommand\1.ATotallyDifferentProperty
Note: This is not a perfect solution. Since there are (s involved and hence matching of nested parens, a regex won't ever be a perfect solution. However it should get the job done for the specific pattern you posted
The answer and solution provided helpful in doing a find-replace on messageboxes.
This worked in Visual Studio 2008 (VB .NET):
Example:
MessageBox.Show("Invalid Entry","Error")
Find What:
MessageBox.Show{(.*,*)}
Replace WIth:
Error.ShowError\1\2
Results in:
Error.ShowError("Invalid Entry","Error")
Looks like you have it nailed. It's what is called a "tagged expression" and you can see another example here:
http://blogs.msdn.com/b/zainnab/archive/2010/09/12/replace-in-files-tagged-expressions-vstipfind0016.aspx