I want to discover all xml files that my ClassLoader is aware of using a wildcard pattern. Is there any way to do this?
我想发现我的ClassLoader使用通配符模式知道的所有xml文件。有没有办法做到这一点?
5 个解决方案
#1
5
It requires a little trickery, but here's an relevant blog entry. You first figure out the URLs of the jars, then open the jar and scan its contents. I think you would discover the URLs of all jars by looking for `/META-INF/MANIFEST.MF'. Directories would be another matter.
它需要一些小技巧,但这是一个相关的博客条目。首先找出罐子的URL,然后打开罐子并扫描其内容。我想你会通过查找`/META-INF/MANIFEST.MF'来发现所有罐子的URL。目录将是另一回事。
#2
7
A Spring ApplicationContext
can do this trivially:
Spring ApplicationContext可以做到这一点:
ApplicationContext context = new ClassPathXmlApplicationContext("applicationConext.xml");
Resource[] xmlResources = context.getResources("classpath:/**/*.xml");
See ResourcePatternResolver#getResources, or ApplicationContext.
请参阅ResourcePatternResolver#getResources或ApplicationContext。
#3
6
List<URL> resources = CPScanner.scanResources(new PackageNameFilter("net.sf.corn.cps.sample"), new ResourceNameFilter("A*.xml"));
put the snippet in your pom.xml
把代码段放在你的pom.xml中
<dependency>
<groupId>net.sf.corn</groupId>
<artifactId>corn-cps</artifactId>
<version>1.0.1</version>
</dependency>
#4
1
A JAR-file is just another ZIP-file, right?
JAR文件只是另一个ZIP文件,对吧?
So I suppose you could iterate the jar-files using http://java.sun.com/javase/6/docs/api/java/util/zip/ZipInputStream.html
所以我想你可以使用http://java.sun.com/javase/6/docs/api/java/util/zip/ZipInputStream.html来迭代jar文件
I'm thinking something like:
我想的是:
ZipSearcher searcher = new ZipSearcher(new ZipInputStream(new FileInputStream("my.jar"))); List xmlFilenames = searcher.search(new RegexFilenameFilter(".xml$"));
ZipSearcher searcher = new ZipSearcher(new ZipInputStream(new FileInputStream(“my.jar”)));列出xmlFilenames = searcher.search(new RegexFilenameFilter(“。xml $”));
Cheers. Keith.
#5
-4
Well, it is not from within Java, but
好吧,它不是来自Java内部,而是
jar -tvf jarname | grep xml$
jar -tvf jarname | grep xml $
will show you all the XMLs in the jar.
将显示jar中的所有XML。
#1
5
It requires a little trickery, but here's an relevant blog entry. You first figure out the URLs of the jars, then open the jar and scan its contents. I think you would discover the URLs of all jars by looking for `/META-INF/MANIFEST.MF'. Directories would be another matter.
它需要一些小技巧,但这是一个相关的博客条目。首先找出罐子的URL,然后打开罐子并扫描其内容。我想你会通过查找`/META-INF/MANIFEST.MF'来发现所有罐子的URL。目录将是另一回事。
#2
7
A Spring ApplicationContext
can do this trivially:
Spring ApplicationContext可以做到这一点:
ApplicationContext context = new ClassPathXmlApplicationContext("applicationConext.xml");
Resource[] xmlResources = context.getResources("classpath:/**/*.xml");
See ResourcePatternResolver#getResources, or ApplicationContext.
请参阅ResourcePatternResolver#getResources或ApplicationContext。
#3
6
List<URL> resources = CPScanner.scanResources(new PackageNameFilter("net.sf.corn.cps.sample"), new ResourceNameFilter("A*.xml"));
put the snippet in your pom.xml
把代码段放在你的pom.xml中
<dependency>
<groupId>net.sf.corn</groupId>
<artifactId>corn-cps</artifactId>
<version>1.0.1</version>
</dependency>
#4
1
A JAR-file is just another ZIP-file, right?
JAR文件只是另一个ZIP文件,对吧?
So I suppose you could iterate the jar-files using http://java.sun.com/javase/6/docs/api/java/util/zip/ZipInputStream.html
所以我想你可以使用http://java.sun.com/javase/6/docs/api/java/util/zip/ZipInputStream.html来迭代jar文件
I'm thinking something like:
我想的是:
ZipSearcher searcher = new ZipSearcher(new ZipInputStream(new FileInputStream("my.jar"))); List xmlFilenames = searcher.search(new RegexFilenameFilter(".xml$"));
ZipSearcher searcher = new ZipSearcher(new ZipInputStream(new FileInputStream(“my.jar”)));列出xmlFilenames = searcher.search(new RegexFilenameFilter(“。xml $”));
Cheers. Keith.
#5
-4
Well, it is not from within Java, but
好吧,它不是来自Java内部,而是
jar -tvf jarname | grep xml$
jar -tvf jarname | grep xml $
will show you all the XMLs in the jar.
将显示jar中的所有XML。