recipetool: create: use ${BP} for subdir for binary packages
If we use ${BP} for the subdirectory, the default value of S will work rather than having to have an ugly value derived from the package file name in both places. This does mean that we have to assume the default though (we can't just let the normal logic work because the value of BP is the default until later on, so the replacement doesn't work). (From OE-Core rev: 13bc2438d61c345a8f229b9d83bf36a14d08916f) Signed-off-by: Paul Eggleton <paul.eggleton@linux.intel.com> Signed-off-by: Richard Purdie <richard.purdie@linuxfoundation.org>
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@ -359,7 +359,7 @@ def create_recipe(args):
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if args.binary:
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# Assume the archive contains the directory structure verbatim
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# so we need to extract to a subdirectory
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fetchuri += ';subdir=%s' % os.path.splitext(os.path.basename(urlparse.urlsplit(fetchuri).path))[0]
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fetchuri += ';subdir=${BP}'
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srcuri = fetchuri
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rev_re = re.compile(';rev=([^;]+)')
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res = rev_re.search(srcuri)
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@ -566,7 +566,9 @@ def create_recipe(args):
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lines_before.append('SRCREV = "%s"' % srcrev)
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lines_before.append('')
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if srcsubdir:
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if srcsubdir and not args.binary:
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# (for binary packages we explicitly specify subdir= when fetching to
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# match the default value of S, so we don't need to set it in that case)
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lines_before.append('S = "${WORKDIR}/%s"' % srcsubdir)
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lines_before.append('')
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