使用Google Guava:
Maven:
<dependency>
<artifactId>guava</artifactId>
<groupId>com.google.guava</groupId>
<version>14.0.1</version>
</dependency>
示例代码:
Strings.padStart("129018", 10, '0') returns "0000129018"
基于@Haroldo Macêdo的答案,我在我的自定义Utils
类中创建了一个方法,如下所示:
/**
* Left padding a string with the given character
*
* @param str The string to be padded
* @param length The total fix length of the string
* @param padChar The pad character
* @return The padded string
*/
public static String padLeft(String str, int length, String padChar) {
String pad = "";
for (int i = 0; i < length; i++) {
pad += padChar;
}
return pad.substring(str.length()) + str;
}
Utils.padLeft(str, 10, "0");
。int maxDigits = 10;
String str = "129018";
String formatString = "%"+n+"s";
String str2 = String.format(formatString, str).replace(' ', '0');
System.out.println(str2);
这在大多数情况下都可以工作
String s = Integer.toBinaryString(5); //Convert decimal to binary
int p = 8; //preferred length
for(int g=0,j=s.length();g<p-j;g++, s= "0" + s);
System.out.println(s);
int pad = 4;
char[] temp = (new String(new char[pad]) + "129018").toCharArray()
Arrays.fill(temp, 0, pad, '0');
System.out.println(temp)
固定长度为10的右侧填充: String.format("%1$-10s", "abc") 固定长度为10的左侧填充: String.format("%1$10s", "abc")
这是一个基于String.format的解决方案,适用于字符串并且适合变长。
public static String PadLeft(String stringToPad, int padToLength){
String retValue = null;
if(stringToPad.length() < padToLength) {
retValue = String.format("%0" + String.valueOf(padToLength - stringToPad.length()) + "d%s",0,stringToPad);
}
else{
retValue = stringToPad;
}
return retValue;
}
public static void main(String[] args) {
System.out.println("'" + PadLeft("test", 10) + "'");
System.out.println("'" + PadLeft("test", 3) + "'");
System.out.println("'" + PadLeft("test", 4) + "'");
System.out.println("'" + PadLeft("test", 5) + "'");
}
int number = -1;
int holdingDigits = 7;
System.out.println(String.format("%0"+ holdingDigits +"d", number));
刚在面试中被问到这个问题........
我下面的回答,但是上面提到的方法更好->
String.format("%05d", num);
我的回答是:
static String leadingZeros(int num, int digitSize) {
//test for capacity being too small.
if (digitSize < String.valueOf(num).length()) {
return "Error : you number " + num + " is higher than the decimal system specified capacity of " + digitSize + " zeros.";
//test for capacity will exactly hold the number.
} else if (digitSize == String.valueOf(num).length()) {
return String.valueOf(num);
//else do something here to calculate if the digitSize will over flow the StringBuilder buffer java.lang.OutOfMemoryError
//else calculate and return string
} else {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < digitSize; i++) {
sb.append("0");
}
sb.append(String.valueOf(num));
return sb.substring(sb.length() - digitSize, sb.length());
}
}
请检查我的代码,它适用于整数和字符串。
假设我们的第一个数字是129018。我们想要在它前面添加零,使得最终字符串的长度为10。你可以使用以下代码:
int number=129018;
int requiredLengthAfterPadding=10;
String resultString=Integer.toString(number);
int inputStringLengh=resultString.length();
int diff=requiredLengthAfterPadding-inputStringLengh;
if(inputStringLengh<requiredLengthAfterPadding)
{
resultString=new String(new char[diff]).replace("\0", "0")+number;
}
System.out.println(resultString);
我已经使用了这个:
DecimalFormat numFormat = new DecimalFormat("00000");
System.out.println("Code format: "+numFormat.format(123));
结果:00123
希望你觉得它有用!