Is it some function which can return me text between qoutes.? Text before and between quotes is variable length. I find a function mid but in specify is length. I would like to get from this string text between both quots(APP_STATUS_RUNNING) and (PRIMARY):
string: Error gim_icon_cfg_1 Application with DBID 736 has status "APP_STATUS_RUNNING", runmode "PRIMARY"
Thank you
EDIT: I try to get output to the label but show me error:BC30452: Operator '&' is not defined for types 'String' and '1-dimensional array of String'.
Dim output As String = myProcess.StandardOutput.ReadToEnd()
Dim StandardError As String = myProcess.StandardError.ReadToEnd()
Dim Splitted() As String
Splitted = Split(output, """")
Label1.text="Ahoj " & Splitted & "Error " & StandardError
You can split your string by quotes beeing the delimiter Split(InputString, """") the odd numbers of the output array then are the strings between quotes, the even numbers are the rest.
Option Explicit
Public Sub Example()
Dim InputString As String
InputString = "Error gim_icon_cfg_1 Application with DBID 736 has status ""APP_STATUS_RUNNING"", runmode ""PRIMARY"""
Dim Splitted() As String
Splitted = Split(InputString, """")
' between quotes (all odd numbers)
Debug.Print Splitted(1) ' APP_STATUS_RUNNING
Debug.Print Splitted(3) ' PRIMARY
' rest (all even numbers)
Debug.Print Splitted(0) ' Error gim_icon_cfg_1 Application with DBID 736 has status
Debug.Print Splitted(2) ' , runmode
End Sub
Or:
The follow code gives you what you are trying to have.
In practice you can search patterns that are between quotes then get from each element the first element sliced by quote (which means quote = 0 element = 1 other quote = 2)
Dim s As String = "Error gim_icon_cfg_1 Application With DBID 736 has status ""APP_STATUS_RUNNING"", runmode ""PRIMARY"""
Dim parts() As String = s.Split(" "c).Where(Function(el) el Like "*""*""*").Select(Function(el) el.Split(""""c)(1)).ToArray
Related
I have a variable with a string...and I want to know if it contains any value other than single quote, comma and a space ("', ") I'm using vba in excel.
for example, i have a varible strA = "'test', 'player'"
I want to check to see if strA has any characters other than "', " (single quote, comma and space).
Thanks
Here is a strategy based on Count occurrences of a character in a string
I don't have vba handy, but this should work. The idea is to remove all these characters and see if anything is left. text represents your string that is being tested.
Dim TempS As String
TempS = Replace(text, " " , "")
TempS = Replace(TempS, "," , "")
TempS = Replace(TempS, "'" , "")
and your result is Len(TempS>0)
Another approach is to use recursion by having a base case of false if the string is empty, if the first character is one of the three call ourselves on the rest of the string, or if not the value is true. Here is the code
function hasOtherChars(s As String) As Boolean
hasOtherChars=false
if (len(s)=0) then
exit function
end if
Dim asciiSpace As Integer
asciiSpace = Asc(" ")
Dim asciiComma As Integer
asciiComma= Asc(",")
Dim asciiApostrophe As Integer
asciiApostrophe = Asc("'")
Dim c as Integer
c = Asc(Mid$(s, 1, 1))
if ((c=asciiSpace) or (c=asciiComma) or (c=asciiApostrophe)) then
hasOtherChars = hasOtherChars(Mid$(s,2))
else
hasOtherChars=true
end if
End function
Again I am borrowing from the other thread.
I have an XML file with about 30000 characters I want to put into a String variable. In that variable, I want to replace certain characters using Replace function.
In the below code neither Replace nor Instr detect the characters I'm looking for. When I shorten the XML to around 3000 characters, the code works.
Strings can be up to 2 billion characters long, so what's going on here?
The only thing that comes to mind is that str is a String and the Instr/Replace require a String Expression as an argument. Is there a limit on the length of String Expression these functions can handle?
Dim str As String
Dim arrInvalidChars() As String
Dim arrValidChars() As String
Dim i As Integer
arrInvalidChars = Split(Expression:="ä, ü, ö, ß, é, ç", Delimiter:=", ")
arrValidChars = Split(Expression:="ae, ue, oe, ss, e, c", Delimiter:=", ")
For i = LBound(arrInvalidChars) To UBound(arrInvalidChars)
On Error Resume Next
Debug.Print InStr(1, str, arrInvalidChars(i), vbTextCompare)
str = Replace(str, arrInvalidChars(i), arrValidChars(i), , , vbTextCompare)
On Error GoTo 0
Next i
I'm a programing student, so I've started with vb.net as my first language and I need some help.
I need to know how I delete excess white spaces between words in a sentence, only using these string functions: Trim, instr, char, mid, val and len.
I made a part of the code but it doesn't work, Thanks.
enter image description here
Knocked up a quick routine for you.
Public Function RemoveMyExcessSpaces(str As String) As String
Dim r As String = ""
If str IsNot Nothing AndAlso Len(str) > 0 Then
Dim spacefound As Boolean = False
For i As Integer = 1 To Len(str)
If Mid(str, i, 1) = " " Then
If Not spacefound Then
spacefound = True
End If
Else
If spacefound Then
spacefound = False
r += " "
End If
r += Mid(str, i, 1)
End If
Next
End If
Return r
End Function
I think it meets your criteria.
Hope that helps.
Unless using those VB6 methods is a requirement, here's a one-line solution:
TextBox2.Text = String.Join(" ", TextBox1.Text.Split(New Char() {" "c}, StringSplitOptions.RemoveEmptyEntries))
Online test: http://ideone.com/gBbi55
String.Split() splits a string on a specific character or substring (in this case a space) and creates an array of the string parts in-between. I.e: "Hello There" -> {"Hello", "There"}
StringSplitOptions.RemoveEmptyEntries removes any empty strings from the resulting split array. Double spaces will create empty strings when split, thus you'll get rid of them using this option.
String.Join() will create a string from an array and separate each array entry with the specified string (in this case a single space).
There is a very simple answer to this question, there is a string method that allows you to remove those "White Spaces" within a string.
Dim text_with_white_spaces as string = "Hey There!"
Dim text_without_white_spaces as string = text_with_white_spaces.Replace(" ", "")
'text_without_white_spaces should be equal to "HeyThere!"
Hope it helped!
I am try to get value in the middle of a string
My string is
"arely browning,68097,19
I know I can use a substring and pick a specific place but I want it to work for any line that is format like this
If the format of the string remains the same, then this will do it:
Dim MyString As String = "arely browning,68097,19"
Dim StringPart As String = Mid(MyString, InStr(MyString, " "), InStr(MyString, ",") - InStr(MyString, " "))
I have three values which need to be sorted from highest to lowest value. I use the following code which works like a charm until I want to use periods "." and commas ",". If I type "1,3" it displays as I like, but if I type "1.3" it changes to 13. My end users need to be able to use both commas and periods.
How can I fix this?
Dim IntArr(2) As Decimal
IntArr(0) = TextBox1.Text
IntArr(1) = TextBox2.Text
IntArr(2) = TextBox3.Text
Array.Sort(IntArr)
Dim highestNum As Decimal
Dim Midelnum As Decimal
Dim lowestNum As Decimal
lowestNum = IntArr(0)
Midelnum = IntArr(1)
highestNum = IntArr(2)
MsgBox("Highest " & highestNum)
MsgBox("lowest " & lowestNum)
MsgBox("middel " & Midelnum)
The problem is that it's based on culture. I say this because if I enter numbers as you described, I get the opposite effect ("1,3" -> "13", etc).
Here's a quick way to change the values to match the current culture.
At the top of your class, put this:
Imports System.Globalization
Then, you can do this:
Dim IntArr(2) As Decimal
Dim nfi As NumberFormatInfo = CultureInfo.CurrentCulture.NumberFormat
Dim sep1 As String = nfi.NumberDecimalSeparator
Dim sep2 As String = If(sep1.Equals("."), ",", ".")
Dim useComma As Boolean = (TextBox1.Text.Contains(",") Or TextBox2.Text.Contains(",") Or TextBox3.Text.Contains(","))
'Replace the separator to match the current culture for parsing
Decimal.TryParse(TextBox1.Text.Replace(sep2, sep1), IntArr(0))
Decimal.TryParse(TextBox2.Text.Replace(sep2, sep1), IntArr(1))
Decimal.TryParse(TextBox3.Text.Replace(sep2, sep1), IntArr(2))
Array.Sort(IntArr)
sep1 = If(useComma, ",", ".")
sep2 = If(useComma, ".", ",")
'Reformat the results to match the user's input
Dim lowestNum As String = IntArr(0).ToString().Replace(sep2, sep1)
Dim middleNum As String = IntArr(1).ToString().Replace(sep2, sep1)
Dim highestNum As String = IntArr(2).ToString().Replace(sep2, sep1)
Dim msg As String = "Highest: {0}" & Environment.NewLine & _
"Lowest: {1}" & Environment.NewLine & _
"Middle: {2}"
msg = String.Format(msg, highestNum, lowestNum, middleNum)
MessageBox.Show(msg)
Also, since you are using .NET, you may want to skip the VB6 way of doing things. Refer to my example to see what I've used.
You could use the hack of altering the string before saving it:
TextBox.Text.Replace(".",",")
But if you want to show the original input you could have a variable to detect the entered character:
Dim isDot As Boolean = False
Dim number As String = TextBox.Text
If number.Contains(".") Then
isDot = True
End If
And in the end replace it just for purposes of displaying
If isDot Then
number.Replace(",",".")
End If
The accepted answer uses too much unnecessary string manipulation. You can use the CultureInfo object to get what you need:
Sub Main
Dim DecArr(2) As Decimal
'Select the input culture (German in this case)
Dim inputCulture As CultureInfo = CultureInfo.GetCultureInfo("de-DE")
Dim text1 As String = "1,2"
Dim text2 As String = "5,8"
Dim text3 As String = "4,567"
'Use the input culture to parse the strings.
'Side Note: It is best practice to check the return value from TryParse
' to make sure the parsing actually succeeded.
Decimal.TryParse(text1, NumberStyles.Number, inputCulture, DecArr(0))
Decimal.TryParse(text2, NumberStyles.Number, inputCulture, DecArr(1))
Decimal.TryParse(text3, NumberStyles.Number, inputCulture, DecArr(2))
Array.Sort(DecArr)
Dim format As String = "Highest: {0}" & Environment.NewLine & _
"Lowest: {1}" & Environment.NewLine & _
"Middle: {2}"
'Select the output culture (US english in this case)
Dim ouputCulture As CultureInfo = CultureInfo.GetCultureInfo("en-US")
Dim msg As String = String.Format(ouputCulture, format, DecArr(2), DecArr(1), DecArr(0))
Console.WriteLine(msg)
End Sub
This code outputs:
Highest: 5.8
Lowest: 4.567
Middle: 1.2