The FAQ is somewhat misleading, starting from its use of “should” followed by the inconsistent use of “requirement” about the same thing. The Unicode Standard itself (cited in the FAQ) is more accurate. Basically, you should not expect programs to treat canonically equivalent strings as different, but neither should you expect all programs to treat … Read more
You’re looking for Integer/parseInt. user=> (map #(Integer/parseInt %) [“1” “2” “3” “4”]) (1 2 3 4) You have to wrap Integer/parseInt in an anonymous function because Java methods aren’t functions. read-string would also work in this case: user=> (map read-string [“1” “2” “3” “4”]) (1 2 3 4) read-string reads any object from a string, … Read more
This is part of the rules of ~-expansion. It is clearly stated in the Bash manual that this expansion is not performed when the ~ is quoted. Workaround 1 Don’t quote the ~. file=~/path/to/file If you need to quote the rest of the filename: file=~/”path with spaces/to/file” (This is perfectly legal in a garden-variety shell.) … Read more
VB6 was designed to be easy Use If str = “” Then ‘ uninitialised, null or empty “” Strings are automatically initialized to [edit] a null string. The null string is vbNullString. But don’t worry about null strings. A VB6 null string is indistinguishable from an empty string “” for (almost) any string manipulation.
Another way to do this is with operator -replace. $TestString = “test=keep this, but not this.” $NewString = $TestString -replace “.*=” -replace “,.*” .*= means any number of characters up to and including an equals sign. ,.* means a comma followed by any number of characters. Since you are basically deleting those two parts of … Read more
Razor helpers return HelperResult objects. You can get the raw HTML by calling ToString(). For more information, see my blog post.
try the LastDelimiter function which is part of the SysUtils unit.
If it is acceptable that newline (and other whitespace) characters are removed from both ends of the string then you can use let string = “\n\nBLA\nblub” let trimmed = string.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet()) // In Swift 1.2 (Xcode 6.3): let trimmed = (string as NSString).stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet()) To remove leading newline/whitespace characters only you can (for example) use a regular … Read more
Also awk – use slash as separator and print last field echo “/gas/string” | awk -F/ ‘{print $NF}’ Or cut – but that will only work if you have same number of directories to strip echo “/gasg/string” |cut -d/ -f 3
I just answered a similar question on SO and thought I can provide a more concise solution: Swift 2 func split(str: String, _ count: Int) -> [String] { return 0.stride(to: str.characters.count, by: count).map { i -> String in let startIndex = str.startIndex.advancedBy(i) let endIndex = startIndex.advancedBy(count, limit: str.endIndex) return str[startIndex..<endIndex] } } Swift 3 func … Read more