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doc: Make UPDATE FROM examples consistent 

The original first half of the example used an employees table and an
accounts.sales_person foreign key column, while the second half (added
in commit 8f889b1083) used a salesmen table and accounts.sales_id
for the foreign key.  This makes everything use the original names.

Author: Dagfinn Ilmari Mannsåker <ilmari@ilmari.org>
Discussion: https://postgr.es/m/87o81vqjw0.fsf@wibble.ilmari.org
This commit is contained in:
Daniel Gustafsson 2022-03-29 14:53:20 +02:00
parent ebc8b7d441
commit 3785d8e98b
1 changed files with 6 additions and 6 deletions
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12
doc/src/sgml/ref/update.sgml
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@ -387,23 +387,23 @@ UPDATE employees SET sales_count = sales_count + 1 WHERE id =
<para>
Update contact names in an accounts table to match the currently assigned
salesmen:
salespeople:
<programlisting>
UPDATE accounts SET (contact_first_name, contact_last_name) =
(SELECT first_name, last_name FROM salesmen
WHERE salesmen.id = accounts.sales_id);
(SELECT first_name, last_name FROM employees
WHERE employees.id = accounts.sales_person);
</programlisting>
A similar result could be accomplished with a join:
<programlisting>
UPDATE accounts SET contact_first_name = first_name,
contact_last_name = last_name
FROM salesmen WHERE salesmen.id = accounts.sales_id;
FROM employees WHERE employees.id = accounts.sales_person;
</programlisting>
However, the second query may give unexpected results
if <structname>salesmen</structname>.<structfield>id</structfield> is not a unique key, whereas
if <structname>employees</structname>.<structfield>id</structfield> is not a unique key, whereas
the first query is guaranteed to raise an error if there are multiple
<structfield>id</structfield> matches. Also, if there is no match for a particular
<structname>accounts</structname>.<structfield>sales_id</structfield> entry, the first query
<structname>accounts</structname>.<structfield>sales_person</structfield> entry, the first query
will set the corresponding name fields to NULL, whereas the second query
will not update that row at all.
</para>