How can i convert pascal string to double?
Like this:
Val(YourString, YourDouble, ErrorCode);
ErrorCode (which is Integer) is needed to know if the conversion was successful.
Hope it helps :)
Related
This question already has answers here:
How to convert an int value to string in Go?
(10 answers)
Closed last year.
this is my day 1 using goLang, I am currently trying to consume a data but I encounter an error, and that's converting integer to string
func createLink(title string, page int) string {
url := url.URL{
Scheme: "https",
Host: "jsonmock.hackerrank.com",
Path: "/api/movies/search/",
}
query := url.Query()
query.Set("page", string(page))
query.Set("title", title)
url.RawQuery = query.Encode()
return url.String()
}
you can try that code, and the result is
actual result :
https://jsonmock.hackerrank.com/api/movies/search/?page=%01&title=spiderman
expected result :
https://jsonmock.hackerrank.com/api/movies/search/?page=1&title=spiderman
There's %01 , an that's something that I do not want. i believe that I made a mistake in converting an integer to string
You should use strconv.Itoa() method to format your integers as strings. This is better explained in the linked answer. For the sake of completeness, here's how you end up with %01 in your result:
first, int 1 gets "plain-converted" to string by following this conversion rule:
Converting a signed or unsigned integer value to a string type yields
a string containing the UTF-8 representation of the integer. Values
outside the range of valid Unicode code points are converted to
"\uFFFD".
then the resulting string (with character of unicode code point equal to 1) gets URL-encoded, ending up with %01 as its representation.
As a sidenote, you're warned about this if you run go vet over your code:
hello.go:19:20: conversion from int to string yields a string of one
rune, not a string of digits (did you mean fmt.Sprint(x)?)
While this doesn't always give you absolutely the best advice on how to fix your error, it at least pushed you into the right direction. And it's strongly recommended to get used to the idea of running this (or similar) kind of checks from day 1 of learning language.
I have an ASCII decimal that i want to convert to a character in vb 6. Is it possible? if yes, how do i do that? In other words, how do i convert vb 6 ASCII decimal to vb 6 character?
Dim myChar As Char
Dim myBtye As byte
myChar = ? myByte
The Chr() function is the slow Variant version of the deprecated Chr$() function.
Use ChrW$() instead, which was introduced over 15 years ago and is far faster.
Dim myByte As byte
Dim myChar As Char
myChar = Chr(myByte)
Chr is returning the string value of the given argument.
btw if you want to convert it back to ascii, you can use the Asc function.
I need correct sample of using decimal format and decimal rounding to closest integer.
Sample1: I have number 123.345,51 result need to be 123.346,00
Sample2: I have number 123.345,49 result need to be 123.345,00
I found samples but they all use format with comma first 123,345.00 I need in first place point like 123.345,00
Tried with culture info but did not success ..
Sample code:
var amount = 123.345,77;
var cultureInfo = CultureInfo.GetCultureInfo("da-DK");
var formattedAmount = String.Format(cultureInfo, "{0:C}", amount);
When I need to convert formattedAmount to decimal its breaks... I am converting for Rounding values after ,
Solution is to enable users to input in format they want, like 123123,55 then use .Replace(",", ".") to replace , colon character with dot, after that made Decimal.Round and all problem are solved!
I have the double value
3,64171229302883E-02
I would like to convert it so that the E-02 is gone and zeros are used instead.
How would I do that in VB6, please?
Assuming its in a string cast to Decimal ("," is locale specific, for me its ".")
?cdec("3.64171229302883E-02")
0.0364171229302883
I have some picture named like this . (like "2.jpeg", "1234.gif" etc.)
how can I get the integer from the string?
I mean, is there something like the C function:
sscanf(myString,"%d", myInteger);
thanks
Alessandro
You will need to remove the non-numeric portion and then call TryParse.
Dim numericPortion As String
Dim result As Integer
numericPortion = myString.Substring(0, myString.IndexOf('.'))
If (Not Int32.TryParse(numericPortion, ByRef result)) Then
''// Handle Error
Else
''// Use Result
End If