How to get a resource path as a string from the project folder - WP7
Tag : chash , By : rhyhann
Date : March 29 2020, 07:55 AM
To fix the issue you can do I have this XML file, which I am trying to deserialize. var rs = Application.GetResourceStream(new Uri("1.xml", UriKind.Relative));
StreamReader sr = new StreamReader(rs.Stream);
string s = sr.ReadToEnd();
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Java getting file as resource when it's in the project folder
Tag : java , By : mtnmuncher
Date : March 29 2020, 07:55 AM
I hope this helps you . Your config file is in your project, somewhere on the file system. However, Eclipse isn't putting it on the classpath. To force it to be on the classpath, right click your folder and add it as a source folder. Eclipse will then add it to the root of the classpath. You can retrieve it with InputStream is = this.getClass().getResourceAsStream("/config");
/config
/qbooksprintfix/FileChecker
/qbooksprintfxi/FilePurgeHandler
/...
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In Java maven project, how does resource folder and target folder map?
Tag : java , By : Nathan Good
Date : March 29 2020, 07:55 AM
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It seems not possible to include in my project, a path to a folder, that has been imported as a linked resource in my pr
Date : March 29 2020, 07:55 AM
will be helpful for those in need The problem you are facing is how the linked folders are being represented in your project (virtual vs. non-virtual) and how the -I argument is being resolved. The Import filesystem wizard does a very different thing when choosing Create virtual folders or not. "/${ProjName}/src/bar" -> /foo/src/bar/inc
"${workspace_loc:/${ProjName}/src/bar/inc}" -> /tmp/so/git/bar/inc
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java.lang.IllegalStateException: Input resource must exist (reader is in 'strict' mode): class path resource
Tag : java , By : algoRhythm99
Date : March 29 2020, 07:55 AM
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