下面是字符串,例如:

"Apple"

我想加零来填充8个字符:

"000Apple"

我该怎么做呢?


当前回答

public static void main(String[] args)
{
    String stringForTest = "Apple";
    int requiredLengthAfterPadding = 8;
    int inputStringLengh = stringForTest.length();
    int diff = requiredLengthAfterPadding - inputStringLengh;
    if (inputStringLengh < requiredLengthAfterPadding)
    {
        stringForTest = new String(new char[diff]).replace("\0", "0")+ stringForTest;
    }
    System.out.println(stringForTest);
}

其他回答

你可能得处理edgecase。这是一个泛型方法。

public class Test {
    public static void main(String[] args){
        System.out.println(padCharacter("0",8,"hello"));
    }
    public static String padCharacter(String c, int num, String str){
        for(int i=0;i<=num-str.length()+1;i++){str = c+str;}
        return str;
    }
}
String input = "Apple";
StringBuffer buf = new StringBuffer(input);

while (buf.length() < 8) {
  buf.insert(0, '0');
}

String output = buf.toString();

以防你不得不在没有库的帮助下做这件事:

("00000000" + "Apple").substring("Apple".length())

(工作,只要你的字符串不超过8个字符。)

public class PaddingLeft {
    public static void main(String[] args) {
        String input = "Apple";
        String result = "00000000" + input;
        int length = result.length();
        result = result.substring(length - 8, length);
        System.out.println(result);
    }
}

这并不漂亮,但很有效。如果你有apache commons,我建议你使用它

if (val.length() < 8) {
  for (int i = 0; i < val - 8; i++) {
    val = "0" + val;
  }
}