This question already has answers here:
How to replace a string pattern with increasing integers using regex
(2 answers)
Closed 1 year ago.
When you create multiple cursors in VSCode, how can you type a number at each cursor such that each number is 1 greater than the previous number, starting at 0?
You can use the extension Regex Text Generator
For match expression use: .*
For generator expression use: {{=i}}
Related
This question already has answers here:
Assured 6 digit random number
(8 answers)
Closed 1 year ago.
I am new to Gatling and Scala Environment.
How to generate a Random Numeric String having 6 digits, in Gatling?
For example, I need to feed "123456" for an attribute in Gatling.
Thanks for the help.
You could start with what you've got and make the following change...
...filter(_.isDigit).take(6).mkString
...but I'd be more inclined toward the following.
(Random.nextInt(900000)+100000).toString
Note: For the 1st proposed solution, leading zeros are possible, "010203" for example. That's not the case for the 2nd. Not sure which is preferred.
Scala 2.13.x option:
util.Random.between(100000, 1000000)
//minimum (inclusive)--^ ^--maximum (exclusive)
This question already has answers here:
Regular Expression To Match String Not Starting With Or Ending With Spaces
(2 answers)
Regex for password: repetitive characters
(4 answers)
Closed 4 years ago.
static let password = "^(?=.*\\d)(?=.*[a-z])(?=.*[A-Z]).{8,}$"
I have this regex to validate password
at least one character
at least one Capital character
at least one number
length more than 8
I need to update it to
prevent any repeated sequence of characters or numbers
to prevent "111111" or "aaaaaaa" for example
to prevent starting and ending with space character
how to update my regex to match those requirements ?
This question already has an answer here:
Learning Regular Expressions [closed]
(1 answer)
Closed 4 years ago.
i need some help to write correct regex validation. I want password with no spaces, min. 6 symbols, doesn't matter numbers or letters or symbols. Alphabet a-zA-Z and а-яА-Я(RU). How i can do that?
"^(?=.*[A-Za-z])(?=.*\\d)[A-Za-z\\d]{6,}$"
You can take a look at this link
This question already has answers here:
Swift string formula into a real calculation
(3 answers)
Swift - Resolving a math operation in a string
(4 answers)
Closed 6 years ago.
How can I eval a string in Swift?
My strings are composed by numbers and logic operators.
Some examples are:
"2+2"
"2%2"
"2+2"
"2=3"
"2>3"
"2/5"
I've thought, if is too complex to write a function to do this, to parse a string and execute that as external process in bin/bash process than read a result. Any ideas?
Other questions, like this, are solution if there are only arithmetics operators. In my cases appears bool operators that returns error in NSExpression eval.
This question already has answers here:
How can I convert ereg expressions to preg in PHP?
(4 answers)
Closed 8 years ago.
eregi_replace('[0-9]+\.+[0-9]','',$cart['unit']);
How to change it to preg_replace?
I get an error: Warning: preg_replace() [function.preg-replace]: Unknown modifier '+' in ---
You can use your existing Regex almost unchanged in a preg_replace(). Just add delimiters and a case-insensitive modifier. You get
preg_replace('#[0-9]+\.+[0-9]#i','',$cart['unit']);
In fact, the case-sensitivity is irrelevant since your pattern only matches 0-9 and .