| From: | Doug Reed <r(dot)douglas(dot)reed(at)gmail(dot)com> |
|---|---|
| To: | Akshay Joshi <akshay(dot)joshi(at)enterprisedb(dot)com> |
| Cc: | Edson Richter <edsonrichter(at)hotmail(dot)com>, "pgadmin-support lists(dot)postgresql(dot)org" <pgadmin-support(at)lists(dot)postgresql(dot)org> |
| Subject: | Re: pgAdmin 4.20 and 4.21 cannot create foreign key |
| Date: | 2020-05-04 13:03:10 |
| Message-ID: | CAD-XmT9b0nnNR1A6Ju8AAq0urMywV8p9xrpWsT13_RNudiU5ww@mail.gmail.com |
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| Thread: | |
| Lists: | pgadmin-support |
I've seen it as well.
On Mon, May 4, 2020 at 4:18 AM Akshay Joshi <akshay(dot)joshi(at)enterprisedb(dot)com>
wrote:
> Hi Edson
>
> I have tried on PostgreSQL 9.6 and 10, not able to reproduce the issue.
> Can you please provide the exact steps and the example table?
>
> On Sun, May 3, 2020 at 12:20 AM Edson Richter <edsonrichter(at)hotmail(dot)com>
> wrote:
>
>> Using PostgreSQL 9.6 on Linux Centos 7 x64 - yum standard installer.
>> pgAdmin 4.20 and 4.21 on Windows Server 2019 x64 - .exe standard
>> installer.
>>
>> I'm trying to create a foreign key using 4.20 and didn't wored: keep
>> always asking to create a covering index.
>> Tryied in 4.20 and also in 4.21 with same results
>>
>> Follow the screenshot of the error:
>>
>>
>>
>
> --
> *Thanks & Regards*
> *Akshay Joshi*
>
> *Sr. Software Architect*
> *EnterpriseDB Software India Private Limited*
> *Mobile: +91 976-788-8246*
>
--
Regards,
Doug
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