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while checking links prior to deleting, if link is found, display parent if it exists, else show child record's name

version-14
Anand Doshi преди 13 години
родител
ревизия
fdc1c2b1e7
променени са 1 файла, в които са добавени 2 реда и са изтрити 2 реда
  1. +2
    -2
      py/webnotes/model/__init__.py

+ 2
- 2
py/webnotes/model/__init__.py Целия файл

@@ -50,12 +50,12 @@ def check_if_doc_is_linked(dt, dn):
else: else:
item = None item = None
try: try:
item = sql("select name from `tab%s` where `%s`='%s' and docstatus!=2 limit 1" % (link_dt, link_field, dn))
item = sql("select name, parent from `tab%s` where `%s`='%s' and docstatus!=2 limit 1" % (link_dt, link_field, dn))
except Exception, e: except Exception, e:
if e.args[0]==1146: pass if e.args[0]==1146: pass
else: raise e else: raise e
if item: if item:
webnotes.msgprint("Cannot delete %s <b>%s</b> because it is linked in %s <b>%s</b>" % (dt, dn, link_dt, item[0][0]), raise_exception=1)
webnotes.msgprint("Cannot delete %s <b>%s</b> because it is linked in %s <b>%s</b>" % (dt, dn, link_dt, item[0][1] or item[0][0]), raise_exception=1)


@webnotes.whitelist() @webnotes.whitelist()
def delete_doc(doctype=None, name=None, doclist = None, force=0): def delete_doc(doctype=None, name=None, doclist = None, force=0):


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