![]() ![]() Read it, and how it uses various parts to generate the uuid. When stored as a string, UUIDs require no. Some reading material about the uniqueness: Perhaps the easiest way to store a UUID to a database is to create a char(36) column and store the UUID as a string. ![]() Their uniqueness does not depend on a central registration. Maybe you mean something else? For instance, if you use UUID() to generate somethingthat should be unique (like a primary key, or a Unique field etc), and you've previously added the same number (like for instance you called UUID() once, but inserted something twice), then you'll just get the default error you get when adding non-unique content to a place that should be unique. When generated according to the standard methods, UUIDs are, for practical purposes, unique. Values, even if these calls are performed on two separate computers Two calls to UUID() are expected to generate two different datasource db is convention - however, you can give your data source any name - for example, datasource mysql or datasource data. You can only have one datasource block in a schema. There would be no "regenerate" code available: the function is designed to create unique keys even across computers, so how could it even know its result was not unique?Ī UUID is designed as a number that is globally unique in space and UUID values are unique across tables, databases, and even servers that allow you to merge rows from different databases or distribute databases across servers. The following providers are available: sqlite postgresql mysql sqlserver mongodb cockroachdb Remarks. UUID is by definition, 'collision-less', that is the. The problem of duplicate entries in the database is not that uuid is not unique, it is that a new uuid is not being generated for a different node revision id, and that should not go in. It's supposed to be unique and it should be always, as far as I know. The problem of uniqueness is about generating different uuids, and that part is working perfectly afaik. ![]() Well, if you call UUID() twice and get the same results, the most problematic thing would be that "stuff is broken" (tm). ![]()
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