Postgres Error Updating Column Data

Postgres error updating column data

There is no need to wrap table name in double quote "employee", and use single quotes for column values

UPDATE employee   
   SET first_name='ok', last_name='pk', email='ooo', phone='000'
 WHERE employee_id = 1;

See Working Example

Postgres column doesn't exist error on update

Change the double quotes you have around test.pdf to single quotes.

Error: Column does not exist in postgresql for update

In Postgres, double quote stand for identifiers (such as table or column names). Here, you actually want a string literal, so you need single quotes:

UPDATE public.meditech_ar_test4
SET filename = 'text'
WHERE filename is null;

Some databases (namely, MySQL), tolerate double quotes for string literals, while using other characters for identifiers (in MySQL: backticks). However in that regard Postgres follows the rules of standard SQL, which defines double quotes for identifiers. You should just take the habit of always using single quotes for string literals (most databases do support that).

PostgreSQL update data error in excluded.column

The column names from the "excluded" record refer to the columns of the target table. And the target column in the SET expression must not be prefixed with the table name (because you can't update a different table anyway)

So you need to use:

SET id_vertex_nearest = excluded.id_vertex_nearest

Unable to update values in postgresql

As documented in the manual the assignment expressions in an UPDATE are separated by commas:

update test 
  set three = 'aa',  --<< comma here, no AND
      four = 'aa', 
      five = 'aa' 
where one = 1 
  and two = 'a';

Request not updating rows but returns successful

You are running out of space on your storage device. Make room on disk (or whatever you use as storage) before starting the big UPDATE. Delete dispensable files (unrelated to the database). Or shrink your database somehow.

A plain VACUUM might do the job. Or VACUUM FULL (blocks concurrent access) to aggressively shrink physical storage. If you cannot afford to block, consider one of the non-blocking community tools. See:

• Optimize Postgres query on timestamp range

VACUUM FULL preferably not on sirene_eta (the target table) which will reuse dead tuples in the UPDATE anyway (after a plain VACUUM). And make sure VACUUM is not blocked by a long running transaction. See:

• What are the consequences of not ending a database transaction?
• Why does this PostgreSQL transaction give "WARNING: there is no transaction in progress"

Whatever else you do, if you don't expect that all targeted rows actually change, add a WHERE condition to filter empty updates (at full cost!)

UPDATE sirene_eta eta
SET    longitude = geo.x
     , latitude  = geo.y
FROM   sirene_geo geo
WHERE  eta.siret = geo.siret
AND   (eta.longitude IS DISTINCT FROM geo.x  -- !
    OR eta.latitude  IS DISTINCT FROM geo.y)

Might even fix your problem by reducing the work to be done (dramatically). (Turns out, it doesn't in your case.

See:

• How do I (or can I) SELECT DISTINCT on multiple columns?

Error updating field conditionally in PostgreSQL-11

Use the table alias

WHEN s.offline_access_token ...

A table name is not visible when an alias was used. Not related, you should probably use IS NOT DISTINCT FROM instead of ==.

Postgresql ERROR: column ... does not exist In A Simple Query

The problem was, I was writing string with ", not with '.
When I changed the query like:

select *
from "userGroupUserOrganizations"  
where "id" = '9fce8e9b-597a-4100-bb3c-efb86aaa83ae';

it worked

Updating column in table using left join from another table in SQL

If you want to update only the rows in animals that have a matching location in location_costs then use this syntax:

UPDATE animals
SET location_cost = location_costs.costs
FROM location_costs
WHERE location_costs.location = animals.location;

If you want to update all the rows of animals (even the rows without a a matching location in location_costs will be updated to null), then use a correlated subquery:

UPDATE animals
SET location_cost = (
  SELECT location_costs.costs 
  FROM location_costs 
  WHERE location_costs.location = animals.location
);

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