我必须拆分这个字符串:
00016282000079116050
它有预定义的块。应该就是这样:
00016 282 00 0079 116 050
我制作了这段代码:
String unformatted = String.valueOf("00016282000079116050");
String str1 = unformatted.substring(0,5);
String str2 = unformatted.substring(5,8);
String str3 = unformatted.substring(8,10);
String str4 = unformatted.substring(10,14);
String str5 = unformatted.substring(14,17);
String str6 = unformatted.substring(17,20);
System.out.println(String.format("%s %s %s %s %s %s", str1, str2, str3, str4, str5, str6));
它有效,但我需要让这段代码更具代表性/更漂亮。
使用Java 8流或正则表达式的东西应该是好的。有什么建议吗?
答案 0 :(得分:9)
您可以使用正则表达式来编译具有6个组的Pattern
,类似于
String unformatted = "00016282000079116050"; // <-- no need for String.valueOf
Pattern p = Pattern.compile("(\\d{5})(\\d{3})(\\d{2})(\\d{4})(\\d{3})(\\d{3})");
Matcher m = p.matcher(unformatted);
if (m.matches()) {
System.out.printf("%s %s %s %s %s %s", m.group(1), m.group(2), m.group(3),
m.group(4), m.group(5), m.group(6));
}
输出(根据要求)
00016 282 00 0079 116 050
或,正如评论中指出的那样,您可以使用Matcher.replaceAll(String)
之类的
if (m.matches()) {
System.out.println(m.replaceAll("$1 $2 $3 $4 $5 $6"));
}
答案 1 :(得分:4)
更面向java的方法是使用循环和预定长度的数组。
int[] lims = {5,3,2,4,3,3};
int total = 0;
String original = "00016282000079116050";
for (int i = 0 ; i < lims.length ; i++) {
System.out.print(original.substring(total, total+=lims[i]) + " ");
}
00016 282 00 0079 116 050
答案 2 :(得分:0)
String unformatted = "00016282000079116050";
int[] splittingPoints = new int[]{5,3,2,4,3,3};
int charactersProcessed = 0;
int howManyBeforeSplit = 0;
int chunksDone = 0;
for (char c : unformatted.toCharArray()) {
if ((charactersProcessed++ - chunksDone) == splittingPoints[howManyBeforeSplit]) {
System.out.print(" ");
chunksDone += splittingPoints[howManyBeforeSplit++];
}
System.out.print(c);
}