How to join with an empty string?

[bravoecho]

I would like to use the join function with an empty string as the separator, but I cannot find a way. If I leave the first argument empty, as in join(,field1,field2), the behaviour becomes unpredictable depending on the type of the other arguments (e.g. literal string or field names), but more often than not it will drop field1.

As a very contrived example, if I run:

 echo '{ "filename": "a.txt", "size": "1MB" }
{ "filename": "b.txt", "size": "0.5MB" }
{ "filename": "c.txt", "size": "1.5MB" }
{ "filename": "d.txt", "size": "0.3MB" }' \
> | dasel -r json -w csv 'all().mapOf(file_size,size,file_path,this().join(,string(/some/dir/),filename)).merge()'

...then I get...

file_path,file_size
a.txt,1MB
b.txt,0.5MB
c.txt,1.5MB
d.txt,0.3MB

...whereas I would expect...

file_path,file_size
/some/dir/a.txt,1MB
/some/dir/b.txt,0.5MB
/some/dir/c.txt,1.5MB
/some/dir/d.txt,0.3MB

I have tried a combination of escaping and so on, but just can't get it to work.

Thanks if you can help! 🙂

[TomWright]

You should be able to use string() with no args to represent an empy string, however I wouldn't join with an empty string in this case, rather join with /.

E.g. This worked for me:

$ echo '{ "filename": "a.txt", "size": "1MB" }
{ "filename": "b.txt", "size": "0.5MB" }
{ "filename": "c.txt", "size": "1.5MB" }
{ "filename": "d.txt", "size": "0.3MB" }' | dasel -r json -w csv 'all().mapOf(file_path,join(/,string(/some/dir),filename),file_size,size).merge()'
file_path,file_size
/some/dir/a.txt,1MB
/some/dir/b.txt,0.5MB
/some/dir/c.txt,1.5MB
/some/dir/d.txt,0.3MB

[TomWright]

As a side note, I just discovered the separator for join isn't parsed as a selector, so it should infact be join(,x,y) to join x and y without any spaces.

I can do further testing on this in the future but the answer above should solve your problem.