0

我在 Java 中编写了这段代码，以便在某些字符之前使用空格> < />，如果行中有多个空格，空格将减少为一个。

所以我的代码是。

return source.replaceAll("\\s{2,}", " ").replaceAll("[[\\>][\\s]][[\\s][\\>]]", ">").replaceAll("[[\\<][\\s]][[\\s][\\<]]", "<").replaceAll("[\\s][/>]", "/>");

那么我可以改进我的正则表达式代码吗？

我的测试代码是..

<?xml version="1.0" encoding="UTF-8"?>
<nfeProc xmlns="http://www.portalfiscal.inf.br/nfe"  versao="2.00" > 
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<protNFe xmlns="http://www.portalfiscal.inf.br/nfe" versao="2.00"><infProt><tpAmb>1</tpAmb><verAplic>13_0_94</verAplic><chNFe>31130821483359001109550010002217051649842350</chNFe><dhRecbto>2013-08-07T01:52:47</dhRecbto><nProt>131131174998806</nProt><digVal>JOjHIJzegFkL7ZfGkfAgSwP0+s4=</digVal><cStat>100</cStat><xMotivo>Autorizado o uso da NF-e</xMotivo></infProt></protNFe>
</nfeProc>
4

2 回答 2

1

"\\s+"使用替换功能怎么样：string.replaceAll("\\s+", " ")这将删除除一个之外的所有空白。

X+表达式称为量词：指定X发生一次或多次

编辑：回应您的评论：

最简单的方法是进行两次后续调用replaceAll("\\s+>", " >")then replaceAll("\\s+(/>)", " />");

String a = "asd     > bhsj       <   />   b";
a = a.replaceAll("\\s+>", " >").replaceAll("\\s+(/>)", " />");
System.out.println(a);
于 2013-11-10T13:10:03.947 回答
0

您对字符类的尝试搞砸了（您不能像那样嵌套它们），并且您不需要转义这些字符。

这实现了（我认为）你想要的：

return source.replaceAll("\\s+", " ").replaceAll(" (?=/?>|<)", "");

这里发生的事情是首先将多个空白字符 ( \s+) 替换为单个空格，然后删除>,/>或<(使用前瞻断言) 后跟的空格 (替换为空白)。

于 2013-11-10T13:29:05.493 回答