I need to format a CString that contains a phone number.
I need to remove all spaces and all characters that are not a number.
Please advice on how I can do this.
tel1 = replace(tel1,' ','')
That is how it would be done in C# but I don't have a clue how to do it in Clarion.
As Griffo mentioned, the built-in procedure DeFormat should do this for you.
There is no online help for Clarion, this is copied from relevant section:
The DEFORMAT procedure removes formatting characters from a numeric string, returning only the numbers contained in the string.
tel1 = '(02) 1234-5678'
tel1 = DeFormat(tel1)
! tel1 now contains 0212345678
tel1 = deformat(tel1)
Check out Deformat in your help
Related
Can someone please help me with how can a string value be converted into a predefined format by splitting the string?
For example:
If the
Stringvalue = "20210819"
needs to be converted as "08/19/2021"
(Splitting first four and then next two and again next two)
Thanks,
Thanks, #DaveNewton, I was able to do that as you suggested.
Stringvalue = "20210819"
valuedate=Stringvalue
valueyear=(valuedate[0..3]);
valuemonth=(valuedate[4..5]);
valueday=(valuedate[6..7]);
println (valuemonth+"/"+valueday+"/"+valueyear);
I am not a professional programmer, just FYI.
I am having trouble splitting a string. The resulting array has a length of 3, with the second spot (index 1) being completely empty.
I could manipulate the array to work the way I'd like, but I would rather understand why it is acting this way and code it properly from the beginning.
Dim defaultSetting() As String
Dim curSetting as String = "MENU_ITEM_ON_OPEN;;OPTIONAL_LEAVE"
defaultSetting = curSetting.Split(";;")
MsgBox(defaultSetting.Length) 'this is 3
MsgBox(defaultSetting(0)) 'this is as expected "MENU_ITEM_ON_OPEN"
MsgBox(defaultSetting(1)) 'this is empty and I do not know why
MsgBox(defaultSetting(2)) 'this is "OPTIONAL_LEAVE" and should be stored in defaultSetting(1)
Any help would be appreciated, thank you.
The problem here is that Option Strict is set to Off.
The overload of Split which is used expects a ParamArray of Char.
Because of this, the string ";;" is "silently" converted to a single char.
You can check this with following code:
Dim x As Char = ";;"
MsgBox(x)
You want to split by a string, which means you have to use another overload:
defaultSetting = curSetting.Split({";;"}, StringSplitOptions.None)
Thanks to a comment made by dbasnett I was able to find code that worked the way I was expecting, although I am not really sure why if anyone care to explain. But if not, this question has been answered, thanks.
defaultSetting = testString.Split(CType(";;", Char()), StringSplitOptions.RemoveEmptyEntries)
Hi I would like to extract dynamically the numbers from string in Excel.
I have the following strings and I would like to have only the numbers before ". pdf". taken out of the string into the next column.
As you can see the number of characters varies from line to line.
I have invented something like this:
=MID(M20;SEARCH("_";M20);20)
But this takes out only the numbers after "_" and .pdf after this....
How to make it the way I like?
D:\Users\xxxx\Desktop\1610\ts25b_4462.pdf
D:\Users\xxx\Desktop\1610\ts02b_39522.pdf
D:\Users\xxxxx\Desktop\1610\ts02b_except_39511.pdf
D:\Users\xxxx\Desktop\1610\ts02b_except_39555.pdf
D:\Users\xxxx\Desktop\1610\ts22b_6118.pdf
So that I have just :
4462
39522
39511
39555
6118
and so on...
Thank you!!!
With VBA, try to do it like this:
Public Function splitThings(strInput As String) As String
splitThings = Split(Split(strInput, "_")(1), ".")(0)
End Function
Concerning your formula, try to use =LEFT(MID(M20;SEARCH("_";M20);20),K), where K is the difference of the length of ts22b_6118.pdf and 4 (.pdf). 4 is the length of .pdf.
Something like this should do the work:
=LEFT(MID(I3,SEARCH("_",I3)+1,LEN(I3)),LEN(MID(I3,SEARCH("_",I3),LEN(I3)))-5)
You should do it using Excel formula. For example:
=SUBSTITUTE(LEFT(A1,FIND(".pdf",A1)-1),LEFT(A1,FIND("_",A1)),"")
Using the first line as an example, with LEFT(A1,FIND(".pdf",A1)-1) you will have D:\Users\xxxx\Desktop\1610\ts25b_4462 and with the LEFT(A1,FIND("_",A1)) D:\Users\xxxx\Desktop\1610\ts25b_, if you SUBSTITUTE the first part by "" you will have 4462.
Hope this can help.
With this formula, you should be able to get the numbers you want:
=MID(A1,FIND("|",SUBSTITUTE(A1,"_","|",LEN(A1)-LEN(SUBSTITUTE(A1,"_",""))))+1,FIND(".",A1)-FIND("|",SUBSTITUTE(A1,"_","|",LEN(A1)-LEN(SUBSTITUTE(A1,"_",""))))-1)
Basically, this is the initial fomula:
=MID(A1,FIND("_",A1)+1,FIND(".",A1)-FIND("_",A1)-1)
But since there may be two _ in the string so this is the one to find the 2nd _:
=SUBSTITUTE(A1,"_","|",LEN(A1)-LEN(SUBSTITUTE(A1,"_","")))
Now just replace this SUBSTITUTE with A1 above and you get that long formula. Hope this helps.
This will return the number you want regardless of extension (could be .pdf, could be .xlsx, etc) and regardless of the number of underscores present in the filename and/or filepath:
=TRIM(LEFT(RIGHT(SUBSTITUTE(SUBSTITUTE(M20,".",REPT(" ",LEN(M20))),"_",REPT(" ",LEN(M20))),LEN(M20)*2),LEN(M20)))
Something like that is coming in:
str="Hello;this;is;a;text"
What I do want as result is this:
result="72:101:108:108:111;116:104:105:115;..."
which should be the Text in ASCII.
You could use string matching to get each word separated by ; and then convert, concat:
local str = "Hello;this;is;a;text"
for word in str:gmatch("[^;]+") do
ascii = table.pack(word:byte(1, -1))
local converted = table.concat(ascii, ":")
print(converted)
end
The output of the above code is:
72:101:108:108:111
116:104:105:115
105:115
97
116:101:120:116
I'll leave the rest of work to you. Hint: use table.concat.
Here is another approach, which exploits that fact that gsub accepts a table where it reads replacements:
T={}
for c=0,255 do
T[string.char(c)]=c..":"
end
T[";"]=";"
str="Hello;this;is;a;text"
result=str:gsub(".",T):gsub(":;",";")
print(result)
Another possibility:
function convert(s)
return (s:gsub('.',function (s)
if s == ';' then return s end
return s:byte()..':'
end)
:gsub(':;',';')
:gsub(':$',''))
end
print(convert 'Hello;this;is;a;text')
Finding certain character or string (such as ";") can be done by using string.find - https://www.lua.org/pil/20.1.html
Converting character to its ASCII code can be done by string.byte - https://www.lua.org/pil/20.html
What you need to do is build a new string using two functions mentioned above. If you need more string-based functions please visit official Lua site: https://www.lua.org/pil/contents.html
Okay...I got way further, but I can't find how to return a string made up of two seperate strings like
str=str1&" "&str2
I am currently working on a project that dynamically displays DB content into table.
To edit the table contents i am want to use the dynamically created "string"+id value.
Is there any way to retrieve the appended int value from the whole string in javaScript?
Any suggestions would be appreciative...
Thanks!!!
If you know that the string part is only going to consist of letters or non-numeric characters, you could use a regular expression:
var str = "something123"
var id = str.replace(/^[^\d]+/i, "");
If it can consist of numbers as well, then things get complicated unless you can ensure that string always ends with a non-numeric character. In which case, you can do something like this:
var str = "something123"
var id = str.match(/\d+$/) ? str.match(/\d+$/)[0] : "";
(''+string.match(/\d+/) || '')
Explanation: match all digits in the variable string, and make a string from it (''+).
If there is no match, it would return null, but thanks to || '', it will always be a string.
You might try using the regex:
/\d+$/
to retrieve the appended number