我喜欢用零填充字符串的解决方案

String.format("%1$" + length + "s", inputString).replace(' ', '0');

with length = "8" and inputString = "Apple"

使用方法String::repeat的解决方案(Java 11)

String str = "Apple";
String formatted = "0".repeat(8 - str.length()) + str;

如果需要，将8更改为其他数字或将其参数化

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。)

String input = "Apple";
StringBuffer buf = new StringBuffer(input);

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

String output = buf.toString();

我喜欢用零填充字符串的解决方案

String.format("%1$" + length + "s", inputString).replace(' ', '0');

with length = "8" and inputString = "Apple"

public static String lpad(String str, int requiredLength, char padChar) {
    if (str.length() > requiredLength) {
        return str;
    } else {
        return new String(new char[requiredLength - str.length()]).replace('\0', padChar) + str;
    }
}