escaping newline in multi-line strings

[Lorenzo Bettini]

Hi

If I have a multi line string

'''
foo.
bar
'''

is there a way of escaping the newline char so that the result is actually

foo.bar

?

I'm asking since in my xtend2 based generator I'd like to split the
multi line string just for formatting purposes, but I don't want the
newline appear in the result

thanks in advance
Lorenzo

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[Sven Efftinge]

Hi Lorenzo,

'''
foo.«
»bar
'''

would do the trick.

Sven

[Lorenzo Bettini]

Uh! Cool! :)
is it documented somewhere?

thanks Sven!
cheers
Lorenzo

[huan...@gmail.com]

Hi Sven and Lorenzo,

I have the following code, xtend always generates a new line after c.value.toString(), do you know how can I avoid that new line? Thanks!

'''

«IF !consts.empty»

            «FOR c : consts»

                «IF c.name == "foo"»

                "«c.value.toString()»"

                «ENDIF»

            «ENDFOR»

        «ENDIF»

'''

[Sven Efftinge]

c.value seems to have the new line. You can remove any surrounding whitespace with trim().

   c.value.toString.trim

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[Sebastian Zarnekow]

Hi,

your template has a newline after «c.value.toString()» thus it will always emit the newline. If you want to drop that, you'd need to make it a one-liner, e.g. «c.consts.filter[name == 'foo'].join[c.value.toString]». If you see more than one newline char for each c, that it's the value itself, that contains it.

Best,

Sebastian

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[Sven Efftinge]

You can tell which solution solves your problem by the position of the newline.

If it is as you said behind c.value.toString the newline is in the value.

If you want to get rid of the newline after the “ then Sebastian’s tip will help.

[Sebastian Zarnekow]

My bad, I didn't recognize the quotes around your «c.value.toString» thus the join won't work as described but you'd have to provide before, after, and separator, too.

[minna hu]

Thank you very much for your tips! I troubleshooted, the value itself does not contain newline. We need to convert the code into one liner to avoid the new line.

I have tried to translate my code into one-liner, however, it generated compilation error. Do you know how to fix it? Thanks! I am not sure how to use "before, after, and separator", please let me know. Thanks!

I need to get the constant with its name equals to "foo", and then get value of this constant.

Old code:

'''

«IF !consts.empty»

            «FOR c : consts»

                «IF c.name == "foo"»

                "«c.value.toString()»"

                «ENDIF»

            «ENDFOR»

        «ENDIF»

'''

to one-liner code:

"«consts.filter[name == "foo"].into(new ArrayList)[0].getValue().toString()»"

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Best Regards,

Minna Hu

[minna hu]

Oh, it complained about ".into(new ArrayList)[0]". Thanks.

[Stefan Oehme]

There is no "[0]" syntax in Xtend. Please use ".head" or ".get(0)"

[minna hu]

.into(new ArrayList) didn't work for me. I removed .into(new ArrayList), changed to get(0) and used the iterator returned by the filter method directly, it worked, thank you, guys!

Here is the working version:

"«consts.filter[name == "foo"].get(0).getValue().toString()»"

[Sven Efftinge]

Just in case you are interested in yet another improvement :-)

  «const.findFirst[name==“foo”].value.toString»

… will do it as well.

[minna hu]

Yes, findFirst also worked for me, and it makes the code clearer, thank you, Sven.