爪哇 - 除去XML属性的双引号 [英] Java - Removing the double quotes in XML attributes

问题描述

我有一个XML字符串，我通过REST调用获得。然而，一些属性已损坏的值。例如：

I have an xml string which I get via a REST call. However, some of the attributes have corrupted values. For example:

<property name="foo" value="Some corrupted String because of "something" like that"/>

我怎么能代替双引号不是没有$ P $通过的值pceded = 或 /> 用单引号不follown并得到一个有效的XML串出的，在Java 6的损坏呢？

How can I replace double-quotes either not preceded by value= or not follown by /> with a single quote and get a valid XML string out of that corrupted one in Java 6?

的编辑：的

我试图修改此预测先行/后向正则表达式是用于VisualBasic中。但由于转义字符我猜的不合，我无法创建它的Java版本。在这里，它是：

I have tried to modify this lookahead/lookbehind regex that was used for VisualBasic. But because of the incompatibility of escape characters I guess, I could not create the Java version of it. Here it is:

（小于？= ^ [^] *，（大于[^] *[^] *）* [^] *）（？！\\ S + \\ W + = | \\ S * [/？]？＆GT;）|（小于？！？（= ^？\\ W + =）] *（大于[^] *[^] *）* [^] * $）

推荐答案

您可以使用下面的正则表达式：

You can use the following regex:

\s+[\w:.-]+="([^"]*(?:"(?!\s+[\w:.-]+="|\s*(?:\/?|\?)>)[^"]*)*)"

请参阅正则表达式演示。这将匹配任何属性的名称/值对捕捉到后者的1组，我们可以回调内改变。

See regex demo. It will match any attribute name/value pair capturing the latter into Group 1 that we can change inside a callback.

下面是一个的Java code演示：

String s =  "<?xml version=\"1.0\" encoding=\"UTF-8\"?> <resources> <resource> <properties> <property name=\"name\" value=\"retrieveFoo\"/>\n<property name=\"foo\" value=\"Some corrupted String because of \"something\" like that\"/>";
StringBuffer result = new StringBuffer();
Matcher m = Pattern.compile("(\\s+[\\w:.-]+=\")([^\"]*(?:\"(?!\\s+[\\w:.-]+=\"|\\s*(?:/?|\\?)>)[^\"]*)*)\"").matcher(s);
while (m.find()) {
    m.appendReplacement(result, m.group(1) + m.group(2).replace("\"", "&quot;") + "\"");
}
m.appendTail(result);
System.out.println(result.toString());

输出：

＆LT;？XML版本=1.0编码=UTF-8＆GT？; ＆LT;资源＆GT; ＆LT;资源＆GT; ＆LT;性状＆gt; ＆LT;属性名=名字值=retrieveFoo/＆GT;
＆LT;属性名=foo的值=因为与放一些损坏的字符串; QUOT;东西＆安培; QUOT;像/＆GT;

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