What are the reasons for invalid SQL Server object name…

  1. The object name contains illegal characters: SQL Server object names can only contain letters, numbers, underscores, and dollar signs, and cannot start with a number. If the object name contains other special characters or illegal characters, it will result in an invalid object name.
  2. Object name is too long: The length of SQL Server object names is limited, and the specific limit depends on the version and settings of the database. If the object name exceeds the limit, it will result in an invalid object name.
  3. Object name conflicts with reserved keyword: SQL Server contains reserved keywords used to represent specific syntax structures or functions. If an object name is the same as one of these reserved keywords, it will render the object name invalid.
  4. Duplicate object names are not allowed in the same database. If an object with the same name already exists, creating another object with the same name will result in an invalid object name.
  5. Object name casing mismatch: By default, SQL Server does not differentiate between the casing of object names. However, in some cases, the database configuration may cause sensitivity to the casing of object names. If there is a mismatch in the casing of object names, it will result in the object name being invalid.
  6. The pattern to which the object belongs does not exist: Objects in SQL Server can be organized by patterns, which are part of the object’s name. If the pattern to which the object belongs does not exist, it will result in an invalid object name.
  7. The database to which the object belongs does not exist: Objects in SQL Server must belong to a database. If the database to which the object belongs does not exist, it will result in an invalid object name.
  8. The server to which the object belongs does not exist: in a cross-server query or distributed database environment, objects can belong to different servers. If the server to which the object belongs does not exist or cannot be accessed, it will result in an invalid object name.
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