java - xml字符串java之间的子字符串提取
问题描述
我有一个大字符串,它是 XML 的表示形式。我正在尝试按如下方式提取节点数据:
String textToExtract = "<FnAnno>\r\n" +
" <PropDesc F_ANNOTATEDID=\"{60431964-0000-C411-9979-E6A21CEE873F}\" F_BACKCOLOR=\"0\" F_BORDER_BACKMODE=\"2\" F_BORDER_COLOR=\"0\" F_BORDER_STYLE=\"0\" F_BORDER_WIDTH=\"1\" F_CLASSID=\"{5CF11941-018F-11D0-A87A-00A0246922A5}\" F_CLASSNAME=\"Text\" F_CREATOR=\"req92333\" F_ENTRYDATE=\"2018-06-19T13:15:43.0000000-05:00\" F_FONT_BOLD=\"true\" F_FONT_ITALIC=\"false\" F_FONT_NAME=\"arial\" F_FONT_SIZE=\"12\" F_FONT_STRIKETHROUGH=\"false\" F_FONT_UNDERLINE=\"false\" F_FORECOLOR=\"0\" F_HASBORDER=\"true\" F_HEIGHT=\"0\" F_ID=\"{60431964-0000-C411-9979-E6A21CEE873F}\" F_LEFT=\"3.430379746835443\" F_MODIFYDATE=\"2018-06-19T13:15:49.0000000-05:00\" F_MULTIPAGETIFFPAGENUMBER=\"1\" F_NAME=\"-1-1\" F_PAGENUMBER=\"1\" F_TEXT_BACKMODE=\"2\" F_TOOLTIP=\"0043007200650061007400650064002000420079003A002000720065007100390032003300330033002C0020002000430072006500610074006500640020004F006E003A002000320030003100380020004A0075006E0065002000310039002C002000310033003A00310035003A00340033002C0020005500540043002D0035\" F_TOOLTIPTRANSFERENCODING=\"hex\" F_TOP=\"1.3291139240506329\" F_WIDTH=\"0\">\r\n" +
" <F_CUSTOM_BYTES/>\r\n" +
" <F_POINTS/>\r\n" +
" <F_TEXT Encoding=\"unicode\">005400680069007300200069007300200061002000740065007300740020000A00280041006200680069006c0061007300680020004d007500740068007500720061006a00200036002f00310039002f00320030003100380029</F_TEXT>\r\n" +
" </PropDesc>\r\n" +
"</FnAnno>";
String extractedString =textToExtract.substring(textToExtract.indexOf("=\"unicode\">"),textToExtract.indexOf("</F_TEXT>")).replaceFirst("=\"unicode\">", "");
Result is 005400680069007300200069007300200061002000740065007300740020000A00280041006200680069006c0061007300680020004d007500740068007500720061006a00200036002f00310039002f00320030003100380029
为了提高效率,我想使用 Pattern 和 matcher 来提取子字符串。下面是我正在努力的代码:
Pattern pattern = Pattern.compile("\\bEncoding=.*?\\.*F_TEXT\\b");
Matcher matcher = pattern.matcher(textToExtract);
while (matcher.find()){
extractedString = (matcher.group());
}
上面的结果是 Encoding="unicode">005400680069007 我需要再次截断。
如何仅获取 之间的数据<F_TEXT Encoding=\"unicode\"> and </F_TEXT>
?我在学校的正则表达式中遇到了问题,甚至现在在工作中也遇到了问题:(我想我需要大量练习。
谢谢。
解决方案
如果您总是要在相同的 XML 标记之间检索数据,那么您不必担心将其解析为数据结构。你有正确的想法。如果速度是你所追求的,只需抓住你知道会在那里的标记之间的绳子。
但是,您的方式正在浪费一些周期。
textToExtract.substring(textToExtract.indexOf("=\"unicode\">"),textToExtract.indexOf("</F_TEXT>")).replaceFirst("=\"unicode\">", "");
让我们分解一下:
// loops through the array until "=\"unicode\">" is found
int startIndex = textToExtract.indexOf("=\"unicode\">");
// loops through the array again, until "</F_TEXT>" is found
int endIndex = textToExtract.indexOf("</F_TEXT>");
//loop through the array, copying the bytes to a new array to form a new String
String substr = textToExtract.substring(startIndex,endIndex);
//loop through the array to find and replace "=\"unicode\">" with nothing
String data = substr.replaceFirst("=\"unicode\">", "");
你在同一个数组中循环了很多。
一旦你知道起点在哪里,就不需要再从头开始搜索了。相反,从那个起点开始寻找。然后,一旦你有了子字符串的起点和终点,你就可以简单地得到它。
// we know what precedes the substring we want
String anchor = "<F_TEXT Encoding=\"unicode\">";
// so we use it to get the start point, looping once, up to that point
int start = textToExtract.indexOf(anchor)+anchor.length();
// we know the end point won't be before the start point, so start where it left off
int end = start;
// count each character from that point until the next XML tag starts
while (textToExtract.charAt(end) != '<') { end++; }
// now we have what we need to simply get the substring
String data = textToExtract.substring(start,end);
这将产生大约 60% 的性能提升。
编辑:为了完成,让我们解决正则表达式
正则表达式很棒,而且在脚本中很有趣,但是对于这样的事情效率很低。如果您可以避免使用正则表达式,请这样做。我倾向于只使用它来“快速而肮脏” - 在编码时间而不是执行时间方面快速。阅读正则表达式引擎的工作原理。这真的很有趣,但你会明白为什么它是最后的手段。
/* this pattern will look for the XML tag.
** then, it will match [^>]+
** [...] will match a single character that matches SOMETHING inside the "character class."
** [^...] will match a single character that is NOT something inside the character class.
** [^>]+ will match as many characters as it can that do not match '>'
** putting this expression inside brackets tells the engine we want to capture it to be referenced later.
** '<' at the end just ensures we capture up until that point.
*/
// create the pattern
Pattern pattern = Pattern.compile("<F_TEXT Encoding=\"unicode\">([^>]+)<");
// get a matcher for it
Matcher matcher = pattern.matcher(textToExtract);
// if we find a match
if (matcher.find()) {
// we can use group(1) to refer to our first capture group
// group(0) will always return the full string matched, but we don't want the tags.
String data= matcher.group(1);
}
推荐阅读
- c# - 如何在 ASP NET Core 中处理 GET 请求
- azure - 如何使用 powershell 在另一个 yaml 中使用一个 yaml 的变量?
- linux - 选择性解压,忽略现有的绝对文件夹结构,tar到一个目录
- makefile - 为子目录编写通用 Makefile 规则
- javascript - 如何让 Webpack 从我的 node_modules 加载文件夹?
- python - Scrapy 从一个文件运行多个蜘蛛
- ios - 在手机上本地存储数据是否安全?
- .net - 转换为 Lamar - 注册 - 使用 StructureMap 的这段代码的等价物是什么?
- http - 如何检查在 TLS/SSL 握手期间是否进行主机名验证?
- bazel - 如何将规则限制为 cpp 工具链的子集?