I'm calling the generateTopKFrequentPattern() method of FP growth mining in the Apache Mahout library. In that call, when I use Resources.getResource("FILENAME"), I get the above error saying the file is not found.
My directory structure is Application/src/FILENAME. In the Eclipse window, from run configurations, the default classpath is "Application". I tried using an absolute classpath, "Application/src/FILENAME" and "src/FILENAME". I Still have the same error.
Can anyone please help me with this?
#Thilo, /FILENAME didn't work.
The file should be in the same folder as the classpath (location of the .class file). Either moving the data file to the location of classpath or changing the location of classpath to that of the data file solved the error.
Earlier I had set the classpath location same as the data file location. Somehow, the configuration was reset in Eclipse.
In my case the classpath is "Application/bin" and file is in "Application/src/FILENAME". Specifying file path relative to classpath say "../src/FILENAME" or "/../src/FILENAME" also didn't work for me.
The issue is that the resource, in this case the file, is not in the build path.
If you are using Eclipse, right click the file, and select Add to Build Path.
Related
I have a question regarding file handling.
I automate a page using selenium, and I need to upload a file in this page.
I want to put the file in resource folder and read it's path in the test (since many OS and path will be different to any computer WIN/MAC).
I put the file manually in the resource folder, and it put it in:
X:\Project_11_01_2021\src\test\resources
when I used the ClassLoader and try to find the file it not found it,
I saw that if I manually put it in this path it find it, found.
X:\Project_11_01_2021\out\test\resources
the problem is that I am using git and if I add to the resources it upload to git and every one will get the change, and when I put in out\test\resources it is not displayed in the source tree to commit to git.
is their a way that classLoader will search in the first location? and not in the second?
[][path that worked]
[][when here not worked]
/******* test *******/
public void entertax() throws Exception {
WebDriver deiver2 = getWebDriver();
Thread.sleep(1000);
ClassLoader classLoader = getClass().getClassLoader();
String path = classLoader.getResource("TAX12.pdf").getPath();
System.out.println("\n\n path is " + path);
deiver2.switchTo()
.activeElement();
deiver2.findElement(By.xpath("//input[#type='file']"))
.sendKeys(
"X:\\Project_11_01_2021\\out\\test\\resources\\fw8TAX12.pdf");
System.out.println("END");
}
This issue looks more like an IDE configuration problem.
Your code classLoader.getResource("TAX12.pdf") looks correct (1. using classLoader.getResource() method && 2. specifying the correct relative path within the resources folder).
I assume why the code fails to find the file is due to the fact that the resources folder is not part of your applications classpath (I'm assuming your trying to run the code from your IDE which apparently seems to be IntelliJ).
I'm not regularly using IntelliJ, but you can specify the classpath settings in the Module settings.
In the Module settings specify the resources folder as a resource:
When successfully added the resources folder to the classpath it should display an icon like this in the project explorer:
After adding the resources folder to your classpath everything should work.
EDIT:
In case you're using Maven also make sure to specify the resources folder as such in the pom.xml file correspondingly.
Also make sure to spell the name of the file you're looking for "TAX12.pdf" correctly, since it seems to differ in your code and in your screenshots (not sure if it differs only because posting it here or also in your real code base).
I'm trying to read a json file in a Resources directory and I use the following:
jsonObject = this.readJson(this.getClass().getClassLoader().getResource("jsonFileName").getPath());
In the IDE it runs correctly but when I build thw jar and try to run it by java -jar jarName I get a "File not found" Error and when I checked the path, it looks like this:
...projectName/target/projectName-1.0-SNAPSHOT.jar!/kb/is/identity.json
When running on the IDE the paths looks like this:
...projectName/target/classes/kb/is/identity.json
It's important to use getResourceAsStream, not getResource.
Have a look at How getClassLoader().getResourceAsStream() works in java
getResource("jsonFileName") - in this case root directory is project name using it under Idea, but when you run it under jar - I think that root path is User Home.
If I remember correctly, you can fix it with using / in resource path.
E.g. when using Maven, you have a resource directory. You have identity.json in the resource root. Using getClass().getResourceAsStream("/identity.json") receive this file (in Idea and in jar), because when you build a jar, all resources are copied to the root of the jar file.
I have a maven project with these standard directory structures:
src/main/java
src/main/java/pdf/Pdf.java
src/test/resources
src/test/resources/files/x.pdf
In my Pdf.java,
File file = new File("../../../test/resources/files/x.pdf");
Why does it report "No such file or dirctory"? The relative path should work. Right?
Relative paths work relative to the current working directory. Maven does not set it, so it is inherited from whatever value it had in the Java process your code is executing in.
The only reliable way is to figure it out in your own code. Depending on how you do things, there are several ways to do so. See How to get the real path of Java application at runtime? for suggestions. You are most likely looking at this.getClass().getProtectionDomain().getCodeSource().getLocation() and then you know where the class file is and can navigate relative to that.
Why does it report "No such file or dirctory"? The relative path should work. Right?
wrong.
Your classes are compiled to $PROJECT_ROOT/target/classes
and your resources are copied to the same folder keeping their relative paths below src/main/resources.
The file will be located relative to the classpath of which the root is $PROJECT_ROOT/target/classes. Therefore you have to write in your Pdf.java:
File file = new File("/files/x.pdf");
Your relative path will be evaluated from the projects current working directory which is $PROJECT_ROOT (AFAIR).
But it does not matter because you want that to work in your final application and not only in your build environment. Therefore you should access the file with getClass().getResource("/path/to/file/within/classpath") which searches the file in the class path of which the root is $PROJECT_ROOT/target/classes.
No the way you are referencing the files is according to your file system. Java knows about the classpath not the file system if you want to reference something like that you have to use the fully qualified name of the file.
Also I do not know if File constructor works with the classpath since it's an abstraction to manage the file system it will depend where the application is run from. Say it is run from the target directory at the same level as source in that case you have to go one directory up and then on src then test the resources the files and finally in x.pdf.
Since you are using a resources folder I think you want the file to be on the classpath and then you can load a resource with:
InputStream in = this.getClass().getClassLoader()
.getResourceAsStream("<path in classpath>");
Then you can create a FileInputStream or something to wrap around the file. Otherwise use the fully qualiefied name and put it somewere like /home/{user}/files/x.pdf.
I am using IDEA 14 to follow a simple Java tutorial (JDBC). As part of this, I am storing some configuration properties in a file called jdbcTutorial.properties. When I put this in the root directory of the project, I can read it with the following:
Properties props = new Properties();
props.load(new FileInputStream("jdbcTutorial.properties"));
However, as soon as I move it to any other directory in the project, I get the error "No such file or directory". This happens regardless of whether I specify a relative or absolute path:
Maybe there are more standard ways to use config files, but I would really like to understand the behavior I am observing. Thanks for helping!
By default root directory would be added to your project's build path. Since the directory in which you are putting the file is not added in your project's build path jvm is unable to find the file. You have two options:
Add the folder where you are putting your prop to build path.
Access the file with full path i.e. /home/user/workspace/....
When you build a project, IDEA takes the files in the resources directory and puts them in the executable jar. So to get an input stream from that file, you need to get it directly from inside the jar. Instead of FileInputStream, use
getClass().getResourceAsStream("jdbcTutorial.properties")
When no path is supplied for a file, java assumes that this file is in the project's root folder. So the relative path "." always points to this folder. When the file is somewhere else, put the appropriate relative path before the path name, like "./files/configuration/jdbcTutorial.properties".
It work also when you put an absolute file path before the path name, like "C:/Users/Me/Documents/Java/workspace/thisProject/files/configuration/jdbcTutorial.properties".
So I see there has already been a post very similar to this issue, however I am in a situation where I have no power to specify the location of this file within my jar and so am hoping someone is aware of a solution to get around this.
So I currently use the following snippet to obtain a file as an input stream, the file 'plugin.xml' is located at the root of the jar and I cannot change this location as another piece of software (dynatrace) creates this file and determines its location.
the standard snippet:
InputStream is = JmxPlugin.class.getResourceAsStream("/plugin.xml");
Now I am aware that the issue is that the ClassLoader is picking up the first file which matches the name 'Plugin.xml' along the classpath (one which isn't in my jar, yay).
Can anyone think of a way to ensure I pick up the correct file without having to move it? The relative path of my class in the jar is com/something/jmx/JmxPlugin.class.
(Id rather not have to unpack the jar in memory).
Many thanks for any contributions,
I'm not absolutely sure, but seems like Thread.currentThread().getContextClassLoader().getResourceAsStream("/plugin.xml") may solve your issue. If not, you'll have to create your own ClassLoader and resolve the issue there.
The simplies way is to move your jar in classpath to be the first containing Plugin.xml,
Another approach is to use getResource() to locate your jar file:
URL myJar=JmxPlugin.class.getResource("/"+JmxPlugin.class.getName().replace(".","/")+".class");
then use this URL to open jar file and extract Plugin.xml.