AppGini Community Forums

Posted: **2025-06-13 21:56**

I have two tables, and article and article_status table. The article_status table has four fields in addition to the primary key.

When the article table is inserted, I run this code in the article_after_insert:

Code: Select all

        $sql = "INSERT INTO article_status (article_id, status, user, created, notes) VALUES (".$data['selectedID'].", 'Draft', '".$memberInfo['username']."', NOW(), 'Initial Entry')";
        sql($sql, $eo);

The correct values are entered into the article_status table.

However, the status does not appear in the child table on the article detail view. I can add another status through the web interface and it will appear but the one I inserted with code never does.

Posted: **2025-06-14 03:14**

Did you set_record_owner('TABLENAME', $newPrimaryKey, $memberID) after manually inserting data?

Posted: **2025-06-16 19:53**

No, I did not. Is the new primary key returned from sql or is there another call I need to make to get that?

Posted: **2025-06-16 23:01**

Code: Select all

$pk = db_insert_id(db_link())

Tip: there is a very useful insert() function which takes TABLENAME and a data array

AppGini PHP functions reference.
https://share.google/Oer1HRpsadtyBQpPS

Posted: **2025-06-17 15:02**

Thank you! I ended up using the insert function though it's not documented in the AppGini PHP functions reference.

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Child table not appearing

Child table not appearing

Re: Child table not appearing

Re: Child table not appearing

Re: Child table not appearing

Re: Child table not appearing