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refactor: Check parent validity using parent meta

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This commit is contained in:
Suraj Shetty 2021-10-28 13:41:12 +05:30
parent 95258bf9f2
commit cb76118268
1 changed files with 5 additions and 8 deletions
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13
frappe/permissions.py
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@ -97,7 +97,7 @@ def has_permission(doctype, ptype="read", doc=None, verbose=False, user=None, ra
if not perm:
perm = false_if_not_shared()
return perm
return bool(perm)
def get_doc_permissions(doc, user=None, ptype=None):
"""Returns a dict of evaluated permissions for given `doc` like `{"read":1, "write":1}`"""
@ -592,10 +592,7 @@ def has_child_table_permission(child_doctype, ptype="read", child_doc=None,
def is_parent_valid(child_doctype, parent_doctype):
filters = {
"options": child_doctype,
"fieldtype": "Table",
"parent": parent_doctype
}
return not frappe.is_table(parent_doctype) and (frappe.db.exists("DocField", filters, cache=True)
or frappe.db.exists("Custom Field", filters, cache=True))
from frappe.core.utils import find
parent_meta = frappe.get_meta(parent_doctype)
child_table_field_exist = find(parent_meta.get_table_fields(), lambda d: d.options == child_doctype)
return not frappe.is_table(parent_doctype) and child_table_field_exist