Is there a way to print a double dagger symbol in SQR language?
I am trying to see if we can use unicode to print the same in a SQR report
Any help in this regard will be helpful.
Use the unicode function.
LET $str = unicode('U+2021')
print $str ()
Related
How can i convert string into XPATH, below is the code
let $ti := "item/title"
let $tiValue := "Welcome to America"
return db:open('test')/*[ $tiValue = $ti]/base-uri()
Here is one way to solve it:
let $ti := "item/title"
let $tiValue := "Welcome to America"
let $input := db:open('test')
let $steps := tokenize($ti, '/')
let $process-step := function($input, $step) { $input/*[name() = $step] }
let $output := fold-left($input, $steps, $process-step)
let $test := $output[. = $tiValue]
return $test/base-uri()
The path string is split into single steps (item, title). With fold-left, all child nodes of the current input (initially db:open('test')) will be matched against the current step (initially, item). The result will be used as new input and matched against the next step (title), and so on. Finally, only those nodes with $tiValue as text value will be returned.
Your question is very unclear - the basic problem is that you've shown us some code that doesn't do what you want, and you're asking us to work out what you want by guessing what was going on in your head when you wrote the incorrect code.
I suspect -- I may be wrong -- that you were hoping this might somehow give you the result of
db:open('test')/*[item/title = $ti]/base-uri()
and presumably $ti might hold different path expressions on different occasions.
XQuery 3.0/3.1 doesn't have any standard way to evaluate an XPath expression supplied dynamically as a string (unless you count the rather devious approach of using fn:transform() to invoke an XSLT transformation that uses the xsl:evaluate instruction).
BaseX however has an query:eval() function that will do the job for you. See https://docs.basex.org/wiki/XQuery_Module
I'm trying to learn F# at the moment and have come up on a problem I can't solve and can't find any answers for on google.
Initially I wanted a log function that would work like the printf family of functions whereby I could provide a format string and a number of arguments (statically checked) but which would add a little metadata before printing it out. With googling, I found this was possible using a function like the following:
let LogToConsole level (format:Printf.TextWriterFormat<'T>) =
let extendedFormat = (Printf.TextWriterFormat<string->string->'T> ("%s %s: " + format.Value))
let date = DateTime.UtcNow.ToString "yyyy-MM-dd HH:mm:ss.fff"
let lvl = string level
printfn extendedFormat date lvl
having the printfn function as the last line of this function allows the varargs-like magic of the printf syntax whereby the partially-applied printfn method is returned to allow the caller to finish applying arguments.
However, if I have multiple such functions with the same signature, say LogToConsole, LogToFile and others, how could I write a function that would call them all keeping this partial-application magic?
Essential I'm looking for how I could implement a function MultiLog
that would allow me to call multiple printf-like functions from a single function call Such as in the ResultIWant function below:
type LogFunction<'T> = LogLevel -> Printf.TextWriterFormat<'T> -> 'T
let MultiLog<'T> (loggers:LogFunction<'T>[]) level (format:Printf.TextWriterFormat<'T>) :'T =
loggers
|> Seq.map (fun f -> f level format)
|> ?????????
let TheResultIWant =
let MyLog = MultiLog [LogToConsole; LogToFile]
MyLog INFO "Text written to %i outputs" 2
Perhaps the essence of this question can be caught more succintly: given a list of functions of the same signature how can I partially apply them all with the same arguments?
type ThreeArg = string -> int -> bool -> unit
let funcs: ThreeArg seq = [func1; func2; func3]
let MagicFunction = ?????
// I'd like this to be valid
let partiallyApplied = MagicFunction funcs "string"
// I'd also like this to be valid
let partiallyApplied = MagicFunction funcs "string" 255
// and this (fullyApplied will be `unit`)
let fullyApplied = MagicFunction funcs "string" 255 true
To answer the specific part of the question regarding string formatting, there is a useful function Printf.kprintf which lets you do what you need in a very simple way - the first parameter of the function is a continuation that gets called with the formatted string as an argument. In this continuation, you can just take the formatted string and write it to all the loggers you want. Here is a basic example:
let Loggers = [printfn "%s"]
let LogEverywhere level format =
Printf.kprintf (fun s ->
let date = DateTime.UtcNow.ToString "yyyy-MM-dd HH:mm:ss.fff"
let lvl = string level
for logger in Loggers do logger (sprintf "%s %s %s" date lvl s)) format
LogEverywhere "BAD" "hi %d" 42
I don't think there is a nice and simple way to do what you wanted to do in the more general case - I suspect you might be able to use some reflection or static member constraints magic, but fortunately, you don't need to in this case!
There is almost nothing to add to a perfect #TomasPetricek answer as he is basically a "semi-god" in F#. Another alternative, which comes to mind, is to use a computation expression (see, for example: https://fsharpforfunandprofit.com/series/computation-expressions.html). When used properly it does look like magic :) However, I have a feeling that it is a little bit too heavy for the problem, which you described.
I'am new to the Julia language and below is a code that I encode using the Jupyter Notebook, but there is no output but when i tried the same code using the REPL, there is an output. Please help me with this.
NOTE: the value of the variable is set to either 'S' or 's' and the input is a function that i copied from Ismael Venegas Castelló (Julia request user input from script). Thank you by the way Mr. Castelló.
if choose == 'S' || choose == 's'
str = input("Please input a String.");
che = input("please input a character to be search");
search(str, che);
end
Worked totally fine for me this way in JuliaPro(0.5.1.1).
julia> choose='s'
's'
julia> function input(prompt::AbstractString="")
print(prompt)
return chomp(readline())
end
input (generic function with 2 methods)
julia> if choose == 'S' || choose == 's'
str = input("Please input a String.");
che = input("please input a character to be search");
search(str, che);
end
Please input a String.It is working.
please input a character to be searchk
10:10
How do I request user input from a running script in Julia? In MATLAB, I would do:
result = input(prompt)
Thanks
The easiest thing to do is readline(stdin). Is that what you're looking for?
I like to define it like this:
julia> #doc """
input(prompt::AbstractString="")::String
Read a string from STDIN. The trailing newline is stripped.
The prompt string, if given, is printed to standard output without a
trailing newline before reading input.
""" ->
function input(prompt::AbstractString="")::String
print(prompt)
return chomp(readline())
end
input (generic function with 2 methods)
julia> x = parse(Int, input());
42
julia> typeof(ans)
Int64
julia> name = input("What is your name? ");
What is your name? Ismael
julia> typeof(name)
String
help?> input
search: input
input(prompt::AbstractString="")::String
Read a string from STDIN. The trailing newline is stripped.
The prompt string, if given, is printed to standard output without a trailing newline before reading input.
julia>
A function that checks that the answer provided matches the expected Type:
Function definition:
function getUserInput(T=String,msg="")
print("$msg ")
if T == String
return readline()
else
try
return parse(T,readline())
catch
println("Sorry, I could not interpret your answer. Please try again")
getUserInput(T,msg)
end
end
end
Function call (usage):
sentence = getUserInput(String,"Write a sentence:");
n = getUserInput(Int64,"Write a number:");
Now in Julia 1.6.1, it's as simple as typing:
num = readline()
Yea! without any arguments since the default value for the IO positional argument of the readline() function is "stdin". So in the above example Julia will read the input from the user and store it in the variable "num".
First I ran
Pkg.add("Dates")
then
using Dates
println()
print("enter year "); year = int(readline(STDIN))
print("enter month "); month = int(readline(STDIN))
print("enter day "); day = int(readline(STDIN))
date = Date(year, month, day)
println(date)
I'm trying to replace two backslashes with a single one within Oracle Service Bus xquery transformation with the replace function:
let $str := replace($srcStr, "\\\\", "\\"), where $srcStr holds the value "^\\d{1,4}$"
But for some reason this does not work. The result is stil "^\\d{1,4}$"
When I'm calling the same function in e.g. Altova XmlSpy this works fine: replace("^\\d{1,4}$", "\\\\", "\\") results in ^\d{1,4}
Does anybody have an idea why OSB does not match the backslashes in the source string? What could be a workaround?
This is a bug.
You can write custom regexp to workaround this bug.
declare function xf:replace_test($e as element()) as xs:string {
let $str := replace("junk (\)\ junk", ".*\\.*", "\$1")
return $str
};
declare variable $e as element() external;
xf:replace_test($e)`