I am writing a program to be able to input a password and take that password and display a new password in a label. the new password will display the results as follows. Any vowels in the old password will be displayed as an x in the new password and numbers will be displayed as x's, and the last part it will display in reverse order so if I enter in timmy1 it should display zxmmxt. I do not know how to write the code to be able to reverse the string to be able to display the password backwards.
Option Explicit On Option Strict On Option Infer Off Public Class frmMain
For each character of this string I want a new character out of the string and then remove the character from the list of characters that still maybe used for other characters. It may not get the same character, you could basically just call this encryption, but it's not what I am making. I don't want to waste my time doing this one hour while VB can do this for me in <1 second.
I am writing a hangman type game and I am displaying the word to the user in a label as all *'s, but I cannot figure out how to have just one of the *'s changed in the label to the correct letter when the user inputs the correct letter into the text box and clicks the check letter button.Everything else in the program works perfectly, except for this part.[code]When I use the .Replace it changes all of the *'s to the correct selected letter.
I've been working with the substring command and after coding up all the things I needed it to do, I saw a post on here where the "For Each" statement was used basically to do the same thing.Lets say we just want to take a string apart one character at a time and add each character to a label. Which would be more efficient?I made a cheap example to show ...
Code: ABinary = "0110 1100 0001 1011" For x = 0 To Len(ABinary) - 1
I was told to use bitblt to crop my image. But i am new to vb and am really confused as to how to do it. I am making a board game. The player chooses its colour of player (a pawn),this image is then placed on the board. I want just the pawn to show and not the box around it.
I don't believe this is possible by conventional methods, but something like this verbose code: For Each s As String In myStringList Step -1 //' Do stuff here Next I will probably have to invert the myString object before a conventional For..Each Loop, correct?
Does anyone know of any reason why my FilmID field (which is an autonumber) is going backwards? When I add a new record, it is going -1, -2, -3, -4, instead of carrying on from the last number. If you need the code, just ask, but I was thinking it might be something in properties or something?
How do I add a space between every character in a string of text?For example:Change abcdefg" To "a b c d e f g"I have two text boxes on my form, one for input, the other for output of the re-formatted string.
I would like to know how i can count the character from a string.dim mystring as string = "myfilename_employee--2010-11-23-45-00--empid200"i need to see if this string have "--" two if it is less or more than two i need to alert to the user.
I have been making a VB app that adds two text boxes together. This all adds up fine but I was wondering if there was any code to be able to turn the answer into a value such as 10.54 instead of 10.54653
I have been making a VB app that adds two text boxes together. This all adds up fine but I was wondering if there was any code to be able to turn the answer into a value such as 10.54 instead of 10.54653Basically turning the answer into a money value?
I want to take each character in a string and multiply it by 7 then 3 then 1. And then loop back to 7 etc. This is what I got but it doesn't work For Each ch As Char In row("SCAN2").ToString product = Convert.ToInt64(ch) * 7 Next Lets say the first character is 8. It should be 8 * 7 = 56 but I get 392.
I have a multi line string consisting of links, some of which will have a # and an id number following it. I need to be able to remove the # and everything after it on each line.I have been looking at using InStr and Split(), but I have had not had any success.