(Document type: Appointment, Operation: create) ERROR - odoo-12

If i create new record with diffrerent user(limited access) this prompt error will show. Why?.
I already declare user_1 in security (csv).
access_emergency_nurse,Emergency Nurse,model_hms_appointment,acs_hms.user_1,1,1,1,0

This is error occured due to record rule i think you define record rule for this model and that is causing problem.Thanks

Related

Can't find data record in database table course_categories in moodle

[I'm getting this error in my moodle][1]
And I can't go in any activity like security and any other activity so I can solve
[1]: https://i.stack.imgur.com/83ofA.png

Parse Server Upgrade to v4.2: Unable to ensure uniqueness for user email addresses

I have recently upgraded parse-server to v4.2. I have also updated Mongo to v4.2 (It's a Replica Set configuration). Now when parse-server is raised, the following error appears "Unable to ensure uniqueness for user email addresses: Tried to ensure field uniqueness for a class that already has duplicates."
The detailed error is this:
{
"message": "Unable to ensure uniqueness for user email addresses: Tried to ensure field uniqueness for a class that already has duplicates.",
"code": 137,
"level": "warn",
"stack": "Error: Tried to ensure field uniqueness for a class that already has duplicates.\n at _adaptiveCollection.then.catch.error (/parse/node_modules/parse-server/lib/Adapters/Storage/Mongo/MongoStorageAdapter.js:569:15)\n at <anonymous>\n at process._tickDomainCallback (internal/process/next_tick.js:229:7)"
The error is caused by the _User class that has the username and email fields with the same data.
Any ideas on how to fix the problem or make the parse-server not do this check.
The error message means that a unique index could not be created on the email field.
The unique index ensures that in the email field a value can only occur once, in other words, that it occurs uniquely. MongoDB cannot create such an index if there are already duplicate values in the email field that would violate the rule of uniqueness, in other words, an identical email address already appears multiple times in the collection.
Keep in mind that Parse Server 4.0 introduced a case-insensitive index for the the field email and username. If you already have a user with email email#example.com and another one with EMAIL#example.com, the index creation will also fail.
The solution is to remove the duplicate email entries and restart Parse Server, so that index creation will be attempted again.

Issue with Vendor and Employee Business Rules

According to the documentation, there is a Business Rule for Vendor and Employee that says
The name, first name, or last name field should not be blank.
Considering that NAME is required for create, the question is ¿Does this mean that ALL three properties have to have a value on Create?
We understood that, so we ran some tests. Creating a record with the three properties populated has no problem. The issue comes when we try so insert a null value for GivenName and FamilyName (first name and last name).
Quickbooks seems to take the NULL values as valid, and when inserting the record is not returning any error. The problem is that the record that was inserted never shows in the QBD UI, but is seems like it persists somewhere, because if we try to insert it again we get a "duplicate name" error.
Not sure if this is an issue or an expected behaviour... Any hints??
Thanks
QuickBooks for windows has some legacy behavior to be aware of. The First, Last and Name combines for a unique identifier accross Customers, Vendors and Employees. So you should fill in ALL the information you have. Any attempts to insert a record with the same F or L name, will fail if there is a Customer, Vendor or Employee with the same F & L.
Second, you mentioned that the record doesn't show in QuickBooks for Windows?
Did it sync successfully? Did you check the error state of the record you inserted?
Did you look at the sync logs to see if it failed to sync?
Lastly, you need to check all three objects for a user with the same F & L name if you are getting a duplicate error. Normally you would check to see if the customer/vendor/employee exists firsts so you can update the existing or determine if it is in fact a new record.
hope that helps
Jarred

Magento: Item (Mage_Sales_Model_Order) with the same id "1" already exist

After I migrate my Magento shop from one hosting provider to another one. Before I start to migrate a delete all my test orders and put counter of order number to 0.
I have big problem. When I want to go in back office into sales->order I get an error "There has been an error processing your request
Exception printing is disabled by default for security reasons.
Error log record number: 213759415"
If I go into var/report and open file "213759415" I see 36 lines the first one are:
a:5:{i:0;s:64:"Item (Mage_Sales_Model_Order) with the same id "1" already exist";i:1;s:5345:"#0 /usr/home/xxx/domains/xxx.com/public_html/lib/Varien/Data/Collection/Db.php(576): Varien_Data_Collection->addItem(Object(Mage_Sales_Model_Order))
I have Magento 1.7.0.2
Any ideas?
Many thanks!
I found in db, tabel called "onestepcheckout" where I forgot to delete test orders.
Make sure you properly deleted all test orders.
You can use the following module to do this:
http://www.magentocommerce.com/magento-connect/asperience-deleteorders.html

VersionConflictException get the conflicting entry id

If a VersionConflictException is thrown the stacktrace contains the conflicting ID in the string:
com.google.gdata.util.VersionConflictException: Conflict Duplicate
insert - page with name "EF9BDFEFBCF2AD70A68B57EFC4621D6D"
already exists - existing page id is 8279809882450151638
I would like to get a handle to the conflicting entry. My current approach is to fetch the id from the string and fetch the entry through the SitesService.getEntry().
Is there a better way to do this? Either getting a handle to the entry from the exception or a method to fetch the ID of the conflicting entry? If Google decides to change the error message my method will break.
client.getEntry(new URL(getSiteFeedUrl() + location), SiteEntry.class);
Works, location is the last part of the site URL