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# how can I use two txtbox to enter aphanumeric value and change it

 P: 12 how can I use two txtbox to enter aphanumeric value and taking the first value of the textbox and placing it in the second txtbox with the numbers value being change to the actual wording (1,2 changes to one, two). Nov 11 '06 #1
18 Replies

 100+ P: 1,646 how can I use two txtbox to enter aphanumeric value and taking the first value of the textbox and placing it in the second txtbox with the numbers value being change to the actual wording (1,2 changes to one, two). Hi. What limits did you have in mind for these numbers? For instance, do you want to have "one million five hundred thousand and forty two" Nov 11 '06 #2

 P: 12 Hi. What limits did you have in mind for these numbers? For instance, do you want to have "one million five hundred thousand and forty two" up to 10 is fine Nov 11 '06 #3

 P: 12 up to 10 is fine can someone please help me I having such a hard time. Nov 11 '06 #4

 100+ P: 1,646 can someone please help me I having such a hard time. You can do it this way using an array and the Split() function This works though there are no checks on the user input Expand|Select|Wrap|Line Numbers     Dim stAr(10) As String     Dim arInput As Variant     Dim stOutput As String     Dim intLoop As Integer       stAr(0) = "zero"     stAr(1) = "one"     stAr(2) = "two"     stAr(3) = "three"     stAr(4) = "four"     stAr(5) = "five"     stAr(6) = "six"     stAr(7) = "seven"     stAr(8) = "eight"     stAr(9) = "nine"     stAr(10) = "ten"       arInput = Split(Me.txtInput.Text, ",")     For intLoop = 0 To UBound(arInput) - 1         stOutput = stOutput & stAr(CInt(arInput(intLoop))) & ", "     Next intLoop     stOutput = stOutput & stAr(CInt(arInput(intLoop)))     Me.txtOutput.Text = stOutput   Nov 11 '06 #5

 P: 12 You can do it this way using an array and the Split() function This works though there are no checks on the user input Expand|Select|Wrap|Line Numbers     Dim stAr(10) As String     Dim arInput As Variant     Dim stOutput As String     Dim intLoop As Integer       stAr(0) = "zero"     stAr(1) = "one"     stAr(2) = "two"     stAr(3) = "three"     stAr(4) = "four"     stAr(5) = "five"     stAr(6) = "six"     stAr(7) = "seven"     stAr(8) = "eight"     stAr(9) = "nine"     stAr(10) = "ten"       arInput = Split(Me.txtInput.Text, ",")     For intLoop = 0 To UBound(arInput) - 1         stOutput = stOutput & stAr(CInt(arInput(intLoop))) & ", "     Next intLoop     stOutput = stOutput & stAr(CInt(arInput(intLoop)))     Me.txtOutput.Text = stOutput   Thanks but can I actually use this with vb I am actually trying to run it in VB 2005 express edition Nov 11 '06 #6

 P: 12 Thanks but can I actually use this with vb I am actually trying to run it in VB 2005 express edition I really appreciate your help Thanks again Nov 11 '06 #7

 100+ P: 1,646 Thanks but can I actually use this with vb I am actually trying to run it in VB 2005 express edition I have no idea if this runs on .NET or not. Why not test it and see. Put a couple of textboxes on a form with a button and paste this code into the button_click event. I tested the code with VB6 Let us know if it works. Good luck Nov 11 '06 #8

 P: 12 I have no idea if this runs on .NET or not. Why not test it and see. Put a couple of textboxes on a form with a button and paste this code into the button_click event. I tested the code with VB6 Let us know if it works. Good luck Thanks for your reply, everything seems to be fine however, I get an error "variable 'stOutput' is used before it has been assigned a value. A null refference exception could result at runtime" and is this vb coding? I think it is. Nov 13 '06 #9

 100+ P: 1,646 Thanks for your reply, everything seems to be fine however, I get an error "variable 'stOutput' is used before it has been assigned a value. A null refference exception could result at runtime" and is this vb coding? I think it is. Try this just after the last Dim statement stOutput = "" Nov 13 '06 #10

 P: 12 Try this just after the last Dim statement stOutput = "" Now I get A different error for this line stOutput = stOutput & stAr(CInt(arInput(intLoop))) error: Conversion from string "jkhkjh" to type 'Integer' is not valid. sorry about the stupid question. Yeah it is vb Nov 13 '06 #11

 P: 12 Try this just after the last Dim statement stOutput = "" Now I get a different error for this line stOutput = stOutput & stAr(CInt(arInput(intLoop))) error: Conversion from string "juji67" to type 'Integer' is not valid. Nov 13 '06 #12

 P: 12 Now I get a different error for this line stOutput = stOutput & stAr(CInt(arInput(intLoop))) error: Conversion from string "juji67" to type 'Integer' is not valid. oh this is page two I was looking for the last post I thought it was on page 1 Nov 13 '06 #13

 100+ P: 1,646 oh this is page two I was looking for the last post I thought it was on page 1 Did you copy and paste this code or did you type it in. There is no error when run under vb6 What is being typed into the first textbox. this should be numbers separated by a comma. Nov 13 '06 #14

 100+ P: 1,646 Now I get A different error for this line stOutput = stOutput & stAr(CInt(arInput(intLoop))) error: Conversion from string "jkhkjh" to type 'Integer' is not valid. sorry about the stupid question. Yeah it is vb This includes a check for numeric entries Expand|Select|Wrap|Line Numbers     Dim stAr(10) As String     Dim arInput As Variant     Dim stOutput As String     Dim intLoop As Integer       stOutput = ""       stAr(0) = "zero"     stAr(1) = "one"     stAr(2) = "two"     stAr(3) = "three"     stAr(4) = "four"     stAr(5) = "five"     stAr(6) = "six"     stAr(7) = "seven"     stAr(8) = "eight"     stAr(9) = "nine"     stAr(10) = "ten"       arInput = Split(Me.txtInput.Text, ",")     For intLoop = 0 To UBound(arInput) - 1         If IsNumeric(arInput(intLoop) Then             stOutput = stOutput & stAr(CInt(arInput(intLoop))) & ", "         Else             MsgBox "Please enter single numbers separated by a comma"             Me.txtInput.Text = ""             Me.txtInput.SetFocus             Exit Sub         End If     Next intLoop     stOutput = stOutput & stAr(CInt(arInput(intLoop)))     Me.txtOutput.Text = stOutput   Nov 13 '06 #15

 P: 12 This includes a check for numeric entries Expand|Select|Wrap|Line Numbers     Dim stAr(10) As String     Dim arInput As Variant     Dim stOutput As String     Dim intLoop As Integer       stOutput = ""       stAr(0) = "zero"     stAr(1) = "one"     stAr(2) = "two"     stAr(3) = "three"     stAr(4) = "four"     stAr(5) = "five"     stAr(6) = "six"     stAr(7) = "seven"     stAr(8) = "eight"     stAr(9) = "nine"     stAr(10) = "ten"       arInput = Split(Me.txtInput.Text, ",")     For intLoop = 0 To UBound(arInput) - 1         If IsNumeric(arInput(intLoop) Then             stOutput = stOutput & stAr(CInt(arInput(intLoop))) & ", "         Else             MsgBox "Please enter single numbers separated by a comma"             Me.txtInput.Text = ""             Me.txtInput.SetFocus             Exit Sub         End If     Next intLoop     stOutput = stOutput & stAr(CInt(arInput(intLoop)))     Me.txtOutput.Text = stOutput   ok it does work however I was hoping to enter a words/letters and numbers without commans and convert the numbers from numeric to the actual world. ex. txtInput: hello123 click button txtOutput: helloonetwothree and if you don't mind can you then try and reverse it txtInput: OneTwoThreehello Nov 13 '06 #16

 P: 12 ok it does work however I was hoping to enter a words/letters and numbers without commans and convert the numbers from numeric to the actual world. ex. txtInput: hello123 click button txtOutput: helloonetwothree and if you don't mind can you then try and reverse it txtInput: OneTwoThreehello sorry I mean txtOutput: OneTwoThreehello Nov 13 '06 #17

 100+ P: 1,646 ok it does work however I was hoping to enter a words/letters and numbers without commans and convert the numbers from numeric to the actual world. ex. txtInput: hello123 click button txtOutput: helloonetwothree and if you don't mind can you then try and reverse it txtInput: OneTwoThreehello Great so you now have a working model that you can improve upon and learn from. Happy hunting :) Nov 13 '06 #18

 P: 12 Great so you now have a working model that you can improve upon and learn from. Happy hunting :) Thanks for all your help. I appriciated Nov 13 '06 #19 