[FIX] Properly redirect in case of invalid url slugs

Use werkzeug's api instead of doing hackish url replacement

bzr revid: fme@openerp.com-20140310130530-e86cdwhdca9cs9g3
This commit is contained in:
Fabien Meghazi 2014-03-10 14:05:30 +01:00
parent 35d87dd19b
commit 5e036bbf7c
1 changed files with 12 additions and 15 deletions

View File

@ -78,21 +78,18 @@ class ir_http(orm.AbstractModel):
return self._dispatch()
def _postprocess_args(self, arguments):
url = request.httprequest.url
for arg in arguments.itervalues():
if isinstance(arg, orm.browse_record) and isinstance(arg._uid, RequestUID):
placeholder = arg._uid
arg._uid = request.uid
try:
good_slug = slug(arg)
if str(arg.id) != placeholder.value and placeholder.value != good_slug:
# TODO: properly recompose the url instead of using replace()
url = url.replace(placeholder.value, good_slug)
except KeyError:
return self._handle_exception(werkzeug.exceptions.NotFound())
if url != request.httprequest.url:
werkzeug.exceptions.abort(werkzeug.utils.redirect(url))
def _postprocess_args(self, arguments, rule):
for arg, val in arguments.items():
# Replace uid placeholder by the current request.uid
if isinstance(val, orm.browse_record) and isinstance(val._uid, RequestUID):
val._uid = request.uid
try:
_, path = rule.build(arguments)
assert path is not None
except Exception:
return self._handle_exception(werkzeug.exceptions.NotFound())
if path != request.httprequest.path:
return werkzeug.utils.redirect(path)
def _handle_exception(self, exception=None, code=500):
if isinstance(exception, werkzeug.exceptions.HTTPException) and hasattr(exception, 'response') and exception.response: