oe.license: add is_included convenience function

Given a license string and whitelist and blacklist, determine if the
license string matches the whitelist and does not match the blacklist.

When encountering an OR, it prefers the side with the highest weight (more
included licenses). It then checks the inclusion of the flattened list of
licenses from there.

Returns a tuple holding the boolean state and a list of the applicable
licenses which were excluded (or None, if the state is True)

Examples:

    is_included, excluded = oe.license.is_included(licensestr, ['GPL*', 'LGPL*'])
    is_included, excluded = oe.license.is_included(licensestr, blacklist=['Proprietary', 'CLOSED'])

(From OE-Core rev: 7903433898b4683a1c09cc9a6a379421bc9bbd58)

Signed-off-by: Christopher Larson <chris_larson@mentor.com>
Signed-off-by: Richard Purdie <richard.purdie@linuxfoundation.org>

1 files changed, 36 insertions, 0 deletions

diff --git a/meta/lib/oe/license.py b/meta/lib/oe/license.py
index 7ab66e762f..3543cfe1f6 100644
--- a/meta/lib/oe/license.py
+++ b/meta/lib/oe/license.py
@@ -3,6 +3,7 @@
 
 import ast
 import re
+from fnmatch import fnmatchcase as fnmatch
 
 class InvalidLicense(StandardError):
     def __init__(self, license):
@@ -60,3 +61,38 @@ def flattened_licenses(licensestr, choose_licenses):
     flatten = FlattenVisitor(choose_licenses)
     flatten.visit_string(licensestr)
     return flatten.licenses
+
+def is_included(licensestr, whitelist=None, blacklist=None):
+    """Given a license string and whitelist and blacklist, determine if the
+    license string matches the whitelist and does not match the blacklist.
+
+    Returns a tuple holding the boolean state and a list of the applicable
+    licenses which were excluded (or None, if the state is True)
+    """
+
+    def include_license(license):
+        return (any(fnmatch(license, pattern) for pattern in whitelist) and not
+                any(fnmatch(license, pattern) for pattern in blacklist))
+
+    def choose_licenses(alpha, beta):
+        """Select the option in an OR which is the 'best' (has the most
+        included licenses)."""
+        alpha_weight = len(filter(include_license, alpha))
+        beta_weight = len(filter(include_license, beta))
+        if alpha_weight > beta_weight:
+            return alpha
+        else:
+            return beta
+
+    if not whitelist:
+        whitelist = ['*']
+
+    if not blacklist:
+        blacklist = []
+
+    licenses = flattened_licenses(licensestr, choose_licenses)
+    excluded = filter(lambda lic: not include_license(lic), licenses)
+    if excluded:
+        return False, excluded
+    else:
+        return True, None