Question

我在Java中使用此代码在> < />之类的字符之前取空格，并且如果行中有多个空格，则会减少为一个。< / p>

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>

Answer 1

使用带有替换功能的"\\s+"会发生什么回合：string.replaceAll("\\s+", " ")会删除除一个空格之外的所有空格。

X+表达式称为量词：指定X出现一次或多次

修改以回复您的评论：

最简单的方法是随后调用replaceAll("\\s+>", " >")然后调用replaceAll("\\s+(/>)", " />");

String a = "asd     > bhsj       <   />   b";
a = a.replaceAll("\\s+>", " >").replaceAll("\\s+(/>)", " />");
System.out.println(a);

Answer 2

你对字符类的尝试搞砸了（你不能像那样嵌套），你不需要逃避这些字符。

这实现了你想要的（我认为）：

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

这里发生的事情是首先将多个空格字符（\s+）替换为单个空格，然后是空格，后跟>，/>或<（使用删除（由空白替换）。

有没有办法改进这个正则表达式代码？

2 个答案: