Java 正则表达式去除被视为字符串的前导零

regex to strip leading zeros treated as string

提问by Horse Voice

I have numbers like this that need leading zero's removed.

我有这样的数字需要删除前导零。

Here is what I need:

这是我需要的：

00000004334300343-> 4334300343

00000004334300343-> 4334300343

0003030435243-> 3030435243

0003030435243-> 3030435243

I can't figure this out as I'm new to regular expressions. This does not work:

我无法弄清楚这一点，因为我是正则表达式的新手。这不起作用：

(^0)

采纳答案by Rohit Jain

You're almost there. You just need quantifier:

您快到了。你只需要量词：

str = str.replaceAll("^0+", "");

It replaces 1 or more occurrences of 0 (that is what +quantifier is for. Similarly, we have *quantifier, which means 0 or more), at the beginning of the string (that's given by caret - ^), with empty string.

它用空字符串替换 1 次或多次出现的 0（这就是+量词的用途。类似地，我们有*量词，表示 0 或更多），在字符串的开头（由插入符号 - 给出^）。

回答by anubhava

If you know input strings are all containing digits then you can do:

如果您知道输入字符串都包含数字，那么您可以执行以下操作：

String s = "00000004334300343";
System.out.println(Long.valueOf(s));
// 4334300343

Code Demo

代码演示

By converting to Longit will automatically strip off all leading zeroes.

通过转换为Long它会自动去除所有前导零。

回答by SubOptimal

Another solution (might be more intuitive to read)

另一种解决方案（可能更直观阅读）

str = str.replaceFirst("^0+", "");

^- match the beginning of a line
0+ - match the zero digit character one or more times

^- 匹配行首
0+ - 匹配零位字符一次或多次

A exhausting list of pattern you can find here Pattern.

您可以在此处找到详尽的模式列表Pattern。

回答by MaxZoom

The correct regex to strip leading zeros is

去除前导零的正确正则表达式是

str = str.replaceAll("^0+", "");

This regex will match 0character in quantity of one and moreat the string beginning. There is not reason to worry about replaceAllmethod, as regex has ^(begin input) special character that assure the replacement will be invoked only once.

此正则表达式将匹配字符串开头的0字符数量one and more。没有理由担心replaceAll方法，因为正则表达式具有^（开始输入）特殊字符，可确保仅调用一次替换。

Ultimatelyyou can use Java build-in feature to do the same:

最终，您可以使用 Java 内置功能来执行相同的操作：

String str = "00000004334300343";
long number = Long.parseLong(str);
// outputs 4334300343

The leading zeros will be stripped for you automatically.

将自动为您去除前导零。

回答by Luis Colorado

\b0+\Bwill do the work. See demo\banchors your match to a word boundary, it matches a sequence of one or more zeros 0+, and finishes not in a word boundary (to not eliminate the last 0in case you have only 00...000)

\b0+\B会做的工作。请参阅演示\b将您的匹配锚定到单词边界，它匹配一个或多个零的序列0+，并且不在单词边界中结束（0如果您只有，则不消除最后一个00...000）

回答by Artur Dumchev

Accepted solution will fail if you need to get "0" from "00". This is right one:

如果您需要从“00”获取“0”，则已接受的解决方案将失败。这是对的：

str = str.replaceAll("0+(?!$)", "");

"^0+(?!$)" means match one or more zeros if it is not followed by end of string.

"^0+(?!$)" 表示匹配一个或多个零，如果后面没有跟字符串结尾。