-
parse() function is not throwing an exception for this string `cly63t164000245zw008pggon';select 1;`
```js
const idSchema = z.array(z.string().cuid(), {
message: "Each ID must be a valid CUID…
-
I encountered a problem with the isCuid function in version 2.2.2. Specifically, the function returns true for non-CUID strings.
```javascript
const cuid = require('cuid');
console.log(cuid.isC…
-
I notice that the documentation is slightly ambiguous about this point, and I'm curious if there is a way around it. As far as I can tell, the `path` column follows a zero-padded representation of the…
-
### Bug description
imagine this:
```
model A {
id String @default(cuid())
tenantId String
bId String?
b B? @relation(fields: [tenantId, bId], references: [tenantId, id…
-
**Describe the bug**
I am unsure if this implementation is the same algorithm as the [original](https://github.com/paralleldrive/cuid2/tree/main).
**To Reproduce**
The entropy is computed dif…
-
**Description and expected behavior**
Schema
```
model Asset {
@@map('assets')
id String @id @default(cuid())
url String
downloadUrl String
pathname …
-
Due to security concerns (which also exist in database auto-increment ids, uuid/guid, and most other id standards), the Cuid standard is now deprecated in favor of [Cuid2](https://github.com/paralleld…
-
**Description and expected behavior**
If you specify `@@auth` on a model but also have a `User` model defined, Zenstack uses the `User` model for the `AuthUser` type.
```
model AuthUser {
@@…
-
This is probably a very similar issue to https://github.com/joe-re/sql-language-server/issues/129 in that it simply requires extension of the parser.
The code I'm hacking on:
```sql
CREATE OR R…
-
Due to security concerns (which also exist in database auto-increment ids, uuid/guid, and most other id standrads), the Cuid standard is now deprecated in favor of [Cuid2](https://github.com/paralleld…