或者你可以使用流API为一个字符或从Java 1.8开始的字符串:

public class ASCIIConversion {
    public static void main(String[] args) {
        String text = "adskjfhqewrilfgherqifvehwqfjklsdbnf";
        text.chars()
                .forEach(System.out::println);
    }
}

String str = "abc";  // or anything else

// Stores strings of integer representations in sequence
StringBuilder sb = new StringBuilder();
for (char c : str.toCharArray())
    sb.append((int)c);

 // store ascii integer string array in large integer
BigInteger mInt = new BigInteger(sb.toString());
System.out.println(mInt);

我知道这个问题已经以几种形式回答了，但这里是我的一点代码，看看所有的字符。

下面是代码，从类开始

public class CheckChValue {  // Class name
public static void main(String[] args) { // class main

    String name = "admin"; // String to check it's value
    int nameLenght = name.length(); // length of the string used for the loop

    for(int i = 0; i < nameLenght ; i++){   // while counting characters if less than the length add one        
        char character = name.charAt(i); // start on the first character
        int ascii = (int) character; //convert the first character
        System.out.println(character+" = "+ ascii); // print the character and it's value in ascii
    }
}

}

为此，我们可以直接使用String类

    input.codePointAt(index);

我想再给出一个建议，以获得整个字符串转换为相应的ascii码，使用java 8 例如:“abcde”~“979899100101”。

    String input = "abcde";
    System.out.println(
            input.codePoints()
                    .mapToObj((t) -> "" + t)
                    .collect(joining()));

public class Ascii {
    public static void main(String [] args){
        String a=args[0];
        char [] z=a.toCharArray();
        for(int i=0;i<z.length;i++){ 
            System.out.println((int)z[i]);
        }
    }
}

正如@Raedwald指出的那样，Java的Unicode并不能满足所有字符获取ASCII值的需求。正确的方法(Java 1.7+)如下:

byte[] asciiBytes = "MyAscii".getBytes(StandardCharsets.US_ASCII);
String asciiString = new String(asciiBytes);
//asciiString = Arrays.toString(asciiBytes)