Reputation: 5314
I have a weird problem :
in src/main/resources i have a "template.xlsx" file.
If i do this :
InputStream is = new ClassPathResource("template.xlsx").getInputStream();
Or this :
InputStream is = ClassLoader.getSystemResourceAsStream("template.xlsx");
Or this :
InputStream is = getClass().getResourceAsStream("/template.xlsx");
When i try to create a workbook :
Workbook wb = new XSSFWorkbook(is);
I get this error :
java.util.zip.ZipException: invalid block type
BUT, when i get my file like this :
InputStream is = new FileInputStream("C:/.../src/main/resources/template.xlsx");
It works !
What is wrong ? I can't hardcode the fullpath to the file.
Can someone help me with this ?
Thanks
Upvotes: 12
Views: 13703
Reputation: 356
I had the same issue, you probably have a problem with maven filtering.
This code load the file from source, unfiltered
InputStream is = new FileInputStream("C:/.../src/main/resources/template.xlsx");
This code load the file from the target directory, after maven has filtered the content
InputStream is = getClass().getResourceAsStream("/template.xlsx");
You should not filter binary files like excel and use two mutually exclusive resource sets as described at the bottom of this page maven resources plugin
Upvotes: 24
Reputation: 4693
haven't you try accessing it like
InputStream is = new FileInputStream("/main/resources/template.xlsx");
?
Upvotes: 0