[Libreoffice-bugs] [Bug 126639] New: Auto_increment issue editing Base

[bugzilla-daemon at bugs.documentfoundation.org]

https://bugs.documentfoundation.org/show_bug.cgi?id=126639

            Bug ID: 126639
           Summary: Auto_increment issue editing Base
           Product: LibreOffice
           Version: 6.1.2.1 release
          Hardware: x86-64 (AMD64)
                OS: Windows (All)
            Status: UNCONFIRMED
          Severity: normal
          Priority: medium
         Component: Base
          Assignee: libreoffice-bugs at lists.freedesktop.org
          Reporter: happy_hour at web.de

Description:
I tried to create a new table in Base (actually 6.2.5, but happend also
before). I defined a primary Key with auto increment. If I don't put any other
column in the table, everything is ok. If I try to add a column, the auto
increment option is set to no - and I can't edit any setting of any column
after saving. In former versions, the auto increment worked. When I create a
table with the assistant and fill in to the auto increment option
AUTO_INCREMENT according to the help file, there occurs an error note that
AUTO_INCREMENT is unknown. If I don't write the text into the option, the
column is created with no autoincrement, despite I marked the option in the
dialogue for the primary key. If You need some screenshots, feel free to
contact me.

Steps to Reproduce:
see Description

Actual Results:
primary key is created without autoincrement, columns are not to be edited any
more

Expected Results:
autoincrement setting should be preserved even after editing columns. Columns
should be to edit after saving


Reproducible: Always


User Profile Reset: No



Additional Info:
Auto-Increment-Ausdruck

>From HELP (German Version):
Geben Sie den SQL-Befehlsparameter ein, der die Datenquelle anweist, ein
bestimmtes Integer-Datenfeld als Auto-Increment-Datenfeld zu behandeln. Der
folgende MySQL-Befehl verwendet beispielsweise den AUTO_INCREMENT-Befehl zum
Hochzählen des "id"-Feldes, wenn durch den Befehl ein Datenfeld erstellt wird:

CREATE TABLE "table1" ("id" INTEGER AUTO_INCREMENT)

Für dieses Beispiel müssen Sie AUTO_INCREMENT in das Feld Auto-Increment-Befehl
eingeben.

This throwed me an error

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