I want to ensure valid input on my text field where user can key in simple math expression for numbers up to 2 d.p such as "1.10 + 3.21 x 0.07". I am doing this through the adding regex to the input formatter constructor in the TextField class.
Following the regex example for 2 d.p here, I modified the code for my text field input formatter to include operators:
String dp = (decimalRange != null && decimalRange > 0)
? "([.][0-9]{0,$decimalRange}){0,1}"
: "";
String num = "($dp)|([0-9]{1,4}$dp)";
_exp = new RegExp(
"^($num){0,1}[-+x/]{0,1}($num){0,1}[-+x/]{0,1}($num){0,1}\$");
I am able to achieve "1.10 + 3.21 x 0.07", however, the user can also type invalid value into the textfield such as "1...10", "1.10 + 3..21". Any advice to improve the Regex above would be greatly appreciated!
Note that I also limit the user to key in a maximum of 3 decimal numbers. so "(2d.p)(operator)(2d.p)(operator)(2d.p)is the maximum limit.
Related
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How to find sum of integers in a string using JavaScript
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Closed 3 years ago.
I am getting a string back "1+2" and would like to remove the "+" and then add the numbers together.
Is this possible using Regex? So far I have:
let matches = pattern.exec(this.expression);
matches.input.replace(/[^a-zA-Z ]/g, "")
I am now left with two numbers. How would I add together?
"this.a + this.b"
Assuming the string returned only has '+' operation how about:
const sum = str.split('+').reduce((sumSoFar, strNum) => sumSoFar + parseInt(strNum), 0);
You cannot add two numbers using regex.
If what you have is a string of the form "1+2", why not simply split the string on the + symbol, and parseInt the numbers before adding them?
var str = "1+2";
var parts = str.split("+"); //gives us ["1", "2"]
console.log(parseInt(parts[0]) + parseInt(parts[1]));
If you don't always know what the delimiter between the two numbers is going to be you could use regex to get your array of numbers, and then reduce or whatever from there.
var myString = '1+2 and 441 with 9978';
var result = myString.match(/\d+/g).reduce((a,n)=> a+parseInt(n),0);
console.log(result); // 1 + 2 + 441 + 9978 = 10422
*Edit: If you actually want to parse the math operation contained in the string, there are a couple of options. First, if the string is from a trusted source, you could use a Function constructor. But this can be almost as dangerous as using eval, so it should be used with great caution. You should NEVER use this if you are dealing with a string entered by a user through the web page.
var myFormula = '1+2 * 441 - 9978';
var fn = new Function('return ' + myFormula);
var output = fn();
console.log(myFormula, ' = ', output); //1+2 * 441 - 9978 = -9095
A safer (but more difficult) course would be to write your own math parser which would detect math symbols and numbers, but would prevent someone from injecting other random commands that could affect global scope variables and such.
I have thousands of lines of text that I need to work through and the lines I am interested with lines that look like the following:
01/04/2019 09:35:41 - Test user (Additional Comments)
I am currently using this code to filter out all the other rows:
If InStr(FullCell(i), " - ") <> 0 And InStr(FullCell(i), ":") <> 0 And InStr(FullCell(i), "(") <> 0 Then
FullCell is the array that I am working through.
which I know is not the best way to do it. Is there a way to check that there is a date at the beginning of the string in the format dd/mm/yyyy and then extract the user name inbetween the '-' and the '(' symbol.
I had a play with regex to see if that could help but i'm limited in skills to be able to pull off both VBA and regex in the same code.
Whats the best way to do this.
Assuming Fullcell(i) contains the string,
If Left(Fullcell(i), 10) Like "##/##/####"
Will return True if you have a date (note that it will not differentiate between dd/mm/yyyy and mm/dd/yyyy.
And
Mid(Fullcell(i), InStr(Fullcell(i), " - ") + 2, InStr(Fullcell(i), " (") - InStr(Fullcell(i), " - ") - 2)
Will return the username
I'm sure there is a more efficient way to do this, but I've used the following solution quite a few times:
This will select the date:
x = 1
Do While Mid(FullCell,1,x) <> " "
x = x + 1
Loop
strDate = Left(FullCell,x)
This will find the character number of the hyphen, the username starts 2 characters after.
x = 1
Do While Mid(FullCell,x,1) <> "-"
x = x + 1
Loop
Then we will find the end of the username
y = x + 2
Do While Mid(FullCell,y,1) <> " "
y = y + 1
Loop
The username should now be characters (x+2 to y-1)
strUsername = Mid(FullCell, x + 2, y - (x + 2) - 1)
Here's how I would do it
Dim your variables
Dim ring as Range
Dim dat as variant
Dim FullCell() as string
Dim User as string
Dim I as long
Set your range
Set rng = ` any way you choose
Dat = rng.value2
Loop dat
For i = 1 to UBound(dat, 1)
Split the data
FullCell = Trim(Split(FullCell, "-"))
Test if it split
If UBound(FullCell) > 0 Then
Test if it matches
If IsDate(FullCell(0)) Then
i = Instr(FullCell(1), "(")-1)
If i then
User = left$(FullCell(1), i)
' Found a user
End If
End If
End If
Next
Abstraction is your friend, it's always helpful to break these into their own private functions whenever you can. You could put your code in a function and call it something like ExtractUsername.
Below I did an example of this, and I decided to go with the RegExp approach (late binding), but you could use string functions like the examples above as well.
This function returns the username if it finds the pattern you mentioned above, otherwise, it returns an empty string.
Private Function ExtractUsername(ByVal SourceString As String) As String
Dim RegEx As Object
Set RegEx = CreateObject("vbscript.regexp")
'(FIRST GROUP FINDS THE DATE FORMATTED AS DD/MM/YYY, AS WELL AS THE FORWARD SLASH)
'(SECOND GROUP FINDS THE USERNAME) THIS WILL BE SUBMATCH 1
With RegEx
.Pattern = "(^\d{2}\/\d{2}\/\d{4}.*-)(.+)(\()"
.Global = True
End With
Dim Match As Object
Set Match = RegEx.Execute(SourceString)
'ONLY RETURN IF A MATCH WAS FOUND
If Match.Count > 0 Then
ExtractUsername = Trim(Match(0).SubMatches(1))
End If
Set RegEx = Nothing
End Function
The regex pattern is grouped into three parts, the date (and slash), username, and opening parentheses. What you are interested in is the username, which in the SubMatch would be number 1.
Regexr is a helpful site for practicing regular expressions and can show you a bit more of what the pattern I went with is doing.
Please note that using regular expressions might give you performance issues and you should test it against regular string functions to see what works best for your situation.
I have an excel sheet where i use the follwoing command to get numbers from a cell that contains a form text:
=MID(D2;SEARCH("number";D2)+6;13)
It searches for the string "number" and gets the next 13 characters that comes after it. But some times the results get more than the number due to the fact these texts within the cells do not have a pattern, like the example below:
62999999990
21999999990
11999999990
6299999993) (
17999999999)
21914714753)
58741236714 P
18888888820
How do i avoid taking anything but numbers OR how do i remove everything but numbers from what i get?
You can user this User Defined Function (UDF) that will get only the numbers inside a specific cell.
Code:
Function only_numbers(strSearch As String) As String
Dim i As Integer, tempVal As String
For i = 1 To Len(strSearch)
If IsNumeric(Mid(strSearch, i, 1)) Then
tempVal = tempVal + Mid(strSearch, i, 1)
End If
Next
only_numbers = tempVal
End Function
To use it, you must:
Press ALT + F11
Insert new Module
Paste code inside Module window
Now you can use the formula =only_numbers(A1) at your spreadsheet, by changing A1 to your data location.
Example Images:
Inserting code at module window:
Executing the function
Ps.: if you want to delimit the number of digits to 13, you can change the last line of code from:
only_numbers = tempVal
to
only_numbers = Left(tempVal, 13)
Alternatively you can take a look a this topic to understand how to achieve this using formulas.
If you are going to go to a User Defined Function (aka UDF) then perform all of the actions; don't rely on the preliminary worksheet formula to pass a stripped number and possible suffix text to the UDF.
In a standard code module as,
Function udfJustNumber(str As String, _
Optional delim As String = "number", _
Optional startat As Long = 1, _
Optional digits As Long = 13, _
Optional bCaseSensitive As Boolean = False, _
Optional bNumericReturn As Boolean = True)
Dim c As Long
udfJustNumber = vbNullString
str = Trim(Mid(str, InStr(startat, str, delim, IIf(bCaseSensitive, vbBinaryCompare, vbTextCompare)) + Len(delim), digits))
For c = 1 To Len(str)
Select Case Asc(Mid(str, c, 1))
Case 32
'do nothing- skip over
Case 48 To 57
If bNumericReturn Then
udfJustNumber = Val(udfJustNumber & Mid(str, c, 1))
Else
udfJustNumber = udfJustNumber & Mid(str, c, 1)
End If
Case Else
Exit For
End Select
Next c
End Function
I've used your narrative to add several optional parameters. You can change these if your circumstances change. Most notable is whether to return a true number or text-that-looks-like-a-number with the bNumericReturn option. Note that the returned values are right-aligned as true numbers should be in the following supplied image.
By supplying FALSE to the sixth parameter, the returned content is text-that-looks-like-a-number and is now left-aligned in the worksheet cell.
If you don't want VBA and would like to use Excel Formulas only, try this one:
=SUMPRODUCT(MID(0&MID(D2,SEARCH("number",D2)+6,13),LARGE(INDEX(ISNUMBER(--MID(MID(D2,SEARCH("number",D2)+6,13),ROW($1:$13),1))* ROW($1:$13),0),ROW($1:$13))+1,1)*10^ROW($1:$13)/10)
At the moment I am saving a set of variables to a text file. I am doing following to check if my code works, but whenever I use a two-digit numbers such as 10 it would not print this number as the max number.
If my text file looked like this.
tom:5
tom:10
tom:1
It would output 5 as the max number.
name = input('name')
score = 4
if name == 'tom':
fo= open('tom.txt','a')
fo.write('Tom: ')
fo.write(str(score ))
fo.write("\n")
fo.close()
if name == 'wood':
fo= open('wood.txt','a')
fo.write('Wood: ')
fo.write(str(score ))
fo.write("\n")
fo.close()
tomL2 = []
woodL2 = []
fo = open('tom.txt','r')
tomL = fo.readlines()
tomLi = tomL2 + tomL
fo.close
tomLL=max(tomLi)
print(tomLL)
fo = open('wood.txt','r')
woodL = fo.readlines()
woodLi = woodL2 + woodL
fo.close
woodLL=max(woodLi)
print(woodLL)
You are comparing strings, not numbers. You need to convert them into numbers before using max. For example, you have:
tomL = fo.readlines()
This contains a list of strings:
['tom:5\n', 'tom:10\n', 'tom:1\n']
Strings are ordered lexicographically (much like how words would be ordered in an English dictionary). If you want to compare numbers, you need to turn them into numbers first:
tomL_scores = [int(s.split(':')[1]) for s in tomL]
The parsing is done in the following way:
….split(':') separates the string into parts using a colon as the delimiter:
'tom:5\n' becomes ['tom', '5\n']
…[1] chooses the second element from the list:
['tom', '5\n'] becomes '5\n'
int(…) converts a string into an integer:
'5\n' becomes 5
The list comprehension [… for s in tomL] applies this sequence of operations to every element of the list.
Note that int (or similarly float) are rather picky about what it accepts: it must be in the form of a valid numeric literal or it will be rejected with an error (although preceding and trailing whitespace is allowed). This is why you need ….split(':')[1] to massage the string into a form that it's willing to accept.
This will yield:
[5, 10, 1]
Now, you can apply max to obtain the largest score.
As a side-note, the statement
fo.close
will not close a file, since it doesn't actually call the function. To call the function you must enclose the arguments in parentheses, even if there are none:
fo.close()
I want to accomplish the following requirements using Regex only (no C# code can be used )
• BTN length is 12 and BTN starts with 0[123456789] then it should remove one digit from left and one digit from right.
WORKING CORRECTLY
• BTN length is 12 and it’s not the case stated above then it should always return 10 right digits by removing 2 from the start. (e.g. 491234567891 should be changed to 1234567891)
NOT WORKING CORRECTLY
• BTN length is 11 and it should remove one digit from left. WORKING CORRECTLY
for length <=10 BTNs , nothing is required to be done , they would remain as it is or Regex may get failed too on them , thats acceptable .
USING SQL this can be achieved like this
case when len(BTN) = 12 and BTN like '0[123456789]%' then SUBSTRING(BTN,2,10) else RIGHT(BTN,10) end
but how to do this using Regex .
So far I have used and able to get some result correct using this regex
[0*|\d\d]*(.{10}) but by this regex I am not able to correctly remove 1st and last character of a BTN like this 015732888810 to 1573288881 as this regex returns me this 5732888810 which is wrong
code is
string s = "111112573288881,0573288881000,057328888105,005732888810,15732888815,344956345335,004171511326,01777203102,1772576210,015732888810,494956345335";
string[] arr = s.Split(',');
foreach (string ss in arr)
{
// Match mm = Regex.Match(ss, #"\b(?:00(\d{10})|0(\d{10})\d?|(\d{10}))\b");
// Match mm = Regex.Match(ss, "0*(.{10})");
// ([0*|\\d\\d]*(.{10}))|
Match mm = Regex.Match(ss, "[0*|\\d\\d]*(.{10})");
// Match mm = Regex.Match(ss, "(?(^\\d{12}$)(.^{12}$)|(.^{10}$))");
// Match mm = Regex.Match(ss, "(info)[0*|\\d\\d]*(.{10}) (?(1)[0*|\\d\\d]*(.{10})|[0*|\\d\\d]*(.{10}))");
string m = mm.Groups[1].Value;
Console.WriteLine("Original BTN :"+ ss + "\t\tModified::" + m);
}
This should work:
(0(\d{10})0|\d\d(\d{10}))
UPDATE:
(0(\d{10})0|\d{1,2}(\d{10}))
1st alternate will match 12-digits with 0 on left and 0 on right and give you only 10 in between.
2nd alternate will match 11 or 12 digits and give you the right 10.
EDIT:
The regex matches the spec, but your code doesn't read the results correctly. Try this:
Match mm = Regex.Match(ss, "(0(\\d{10})0|\\d{1,2}(\\d{10}))");
string m = mm.Groups[2].Value;
if (string.IsNullOrEmpty(m))
m = mm.Groups[3].Value;
Groups are as follows:
index 0: returns full string
index 1: returns everything inside the outer closure
index 2: returns only what matches in the closure inside the first alternate
index 3: returns only what matches in the closure inside the second alternate
NOTE: This does not deal with anything greater than 12 digits or less than 11. Those entries will either fail or return 10 digits from somewhere. If you want results for those use this:
"(0(\\d{10})0|\\d*(\\d{10}))"
You'll get rightmost 10 digits for more than 12 digits, 10 digits for 10 digits, nothing for less than 10 digits.
EDIT:
This one should cover your additional requirements from the comments:
"^(?:0|\\d*)(\\d{10})0?$"
The (?:) makes a grouping excluded from the Groups returned.
EDIT:
This one might work:
"^(?:0?|\\d*)(\\d{10})\\d?$"
(?(^\d{12}$)(?(^0[1-9])0?(?<digit>.{10})|\d*(?<digit>.{10}))|\d*(?<digit>.{10}))
which does the exact same thing as sql query + giving result in Group[1] all the time so i didn't had to change the code a bit :)