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Thread: Help with if logic - was JAVA easy

  1. #1
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    Post Help with if logic - was JAVA easy

    So I cant seem to remember how to do this so hopoe fully someone on here will
    import java.io.*;
    import java.util.*;
    public class relearn{
    public static void main( String args[] )
    {
    Scanner kbReader = new Scanner(System.in);
     
    System.out.println("Enter the nucleotide sequence below: "); //Enter One Two
    String AA = kbReader.nextLine( ); 
    int AL = AA.length();
    int x;
     
    for( x=0; x<AL; x++){
    if(AA.equals("a")||AA.equals("A")){
    	System.out.print("U");
    }
    else if(AA.equalsIgnoreCase("T")){
    	System.out.print("A");
    }
    else if(AA.equalsIgnoreCase("C")){
    	System.out.print("G");
    }
    else if(AA.equalsIgnoreCase("G")){
    	System.out.print("C");
    }
     
    }
     
    }
    }

    heres my code what I need it to do is when ever I put in " ATCGTC" I need it to output "UAGCAG" but when I type in more that one letter is just ends the program


  2. #2
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    Default re: Help with if logic - was JAVA easy

    Those are strange and confusing variable names. Which variable holds the user's input? One with a name like: usersInput would make the code easier to read.

    If the purpose of the code is to look at the letters the user has typed in one at a time, the String class has some methods for getting the letters one at a time so that they can be compared to single letters or characters. Look at the API doc for the String class for methods that will get a sub part of the String or that will get one char at a time.
    If you don't understand my answer, don't ignore it, ask a question.

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    Default re: Help with if logic - was JAVA easy

    Quote Originally Posted by Norm View Post
    Those are strange and confusing variable names. Which variable holds the user's input? One with a name like: usersInput would make the code easier to read.

    If the purpose of the code is to look at the letters the user has typed in one at a time, the String class has some methods for getting the letters one at a time so that they can be compared to single letters or characters. Look at the API doc for the String class for methods that will get a sub part of the String or that will get one char at a time.
    I didn't realy go into much back story but its for coding out DNA strands. My bio teacher challenged me to it but ive forgotten java, but he asked me to make a program that would code out Amino Acids ( String AA) and convert them to mRNA. String AA stores the data then int AL (Acid length) takes the length of AA so if the enter 50 letters 50 should be converted

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    Default re: Help with if logic - was JAVA easy

    Quote Originally Posted by nickerb2k View Post
    So I cant seem to remember how to do this so hopoe fully someone on here will
    import java.io.*;
    import java.util.*;
    public class relearn{
    public static void main( String args[] )
    {
    Scanner kbReader = new Scanner(System.in);
     
    System.out.println("Enter the nucleotide sequence below: "); //Enter One Two
    String AA = kbReader.nextLine( ); 
    int AL = AA.length();
    int x;
     
    for( x=0; x<AL; x++){
    if(AA.equals("a")||AA.equals("A")){
    	System.out.print("U");
    }
    else if(AA.equalsIgnoreCase("T")){
    	System.out.print("A");
    }
    else if(AA.equalsIgnoreCase("C")){
    	System.out.print("G");
    }
    else if(AA.equalsIgnoreCase("G")){
    	System.out.print("C");
    }
     
    }
     
    }
    }

    heres my code what I need it to do is when ever I put in " ATCGTC" I need it to output "UAGCAG" but when I type in more that one letter is just ends the program
    This is not the proper logic ..
    please think differently..
    insted of checking char with the string ..
    make that string to char array and then check each char.
    make use of toCharArray().

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