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Scanner cannot find (correct) file path

I'm programming in Java with IntelliJ and have been trying to use the Scanner class to read the file. Even with the correct path, I still get a "No such file or directory" error. Does anyone have any suggestions?
My working directory is /Users/kevinliu/Desktop/test
Here is a picture of how the project is set-up.

Are you trying to create a swing/console application using maven?
If yes, maven is not able to find the source. You have to add it on the pom file. See here on how to add it on pom file.
if no, do you have rights to access the address of the image file? Some times, folder are protected by the OS.
You can also use YourClassName.class.getResource("input/input1.txt") to locate file/s under the directory that your class was in.

Even with the correct path, I still get a "No such file or directory" error.
The path is NOT correct. That path says look for a directory called "src" in the root directory of your computer. That is almost certainly not where the input file lives.
If you are going to use an absolute pathname for a file within the working directory that you stated, it should look like this:
/Users/kevinliu/Desktop/test/src/input/input1.txt
(You can check what it will actually be using a file browser ... outside of Intellij.)
If you want to use a relative pathname, try this
src/input/input1.txt
Notes:
There is no leading "/" on a relative pathname. A leading "/" means it is an absolute pathname. Absolute pathnames start at the root directory.
A relative path is resolved relative to the >>current<< working directory. That will depend on where and how you run the application ...
For a production application, you would not want to refer to a file in the source tree. The end user typically won't have the source tree.
Consider making the path a command line argument or configuration setting for your application.
Consider making the file a "resource" that is part of the application's JAR file. (You would open it a different way ...)
If you ever get a "No such file or directory" message, that means that the path is not correct in some sense. You might be in the wrong place, you might not have permission on a parent directory, the file may have been removed or renamed, there may be a you, or something else. Either way, that error comes from the operating system and the OS doesn't make mistakes about these things. The mistake will be yours (or the user's).

Using relative path in a maven project

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.

Reading file form classpath

I have a maven project and want to read file in it form its class path. The code that i am using is
InputStream is = getClass().getResourceAsStream ("filename.json");
But every time i am getting null inputstreams. I am not sure why ?
The file is places under /src/main/resources. The same folder which contains log4j.xml and it is being picked up decently.
Please note, I am trying to run this file from Eclipse i.e., run or debug mode. No vm arguments or whatsoever.

The Class.getResourceAsStream(String) method looks for the given resource within the same namespace (i.e. package) that the given class is in unless you give it an absolute path (see the API documentation); If it can't find the resource on the classpath in this namespace, it returns null. Since your class is likely inside e.g. com.myproject.resourcemanagement, your resource file has to analogously be under src/main/resources/com/myproject/resourcemanagement, similary to how your class source files are organised (under src/main/java/com/myproject/resourcemanagement).

How to use ClassLoader.getResources() in jar file

Problem statement:
I have a jar file with a set of configuration files in a package mycompany/configuration/file/.
I don't know the file names.
My intention is to load the file names at runtime from the jar file in which they are packaged and then use it in my application.
As far as I understood:
When I use the ClassLoader.getResources("mycompany/configuration/file/") I should be getting all the configuration files as URLs.
However, this is what is happening:
I get one URL object with URL like jar:file:/C:/myJarName.jar!mycompany/configuration/file/
Could you please let me know what I am doing wrong ?

For what you are trying to do I don't think it is possible.
getResource and getResources are about finding named resources within the classloader's classpath, not listing values beneath a directory or folder.
So for example ClassLoader.getResources("mycompany/configuration/file/Config.cfg") would return all the files named Config.cfg that existed in the mycompany/configuration/file path within the class loader's class path (I find this especially useful for loading version information personally).
In your case I think you might almost have half a solution. The URL you are getting back contains the source jar file (jar:file:/C:/myJarName.jar). You could use this information to crack open the jar file a read a listing of the entries, filtering those entries whose name starts with "mycompany/configuration/file/".
From there, you could then fall back on the getResource method to load a reference to each one (now that you have the name and path)

ClassLoader loading the wrong instance of a file

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.