e-letter wrote: > The file was changed: > > ... $value=pg_fetch_result($query,1,1); > echo 'all files' . var_dump($value); > ... > > The resultant web page produces: > > bool(false) all files > > The php file was changed again: > > ... $value=pg_fetch_result($query); > echo 'all files' . var_dump($value); > ... > > The resultant web page produces: > > NULL all files > > The error log shows: > > ...PHP Warning: pg_fetch_result(): supplied argument is not a valid > PostgreSQL result resource... > > The objective is to learn how to extract data from a database and > print to a web browser, but not much progress made so far..!
There is a good example of how to use pg_fetch_result in the docs at http://php.net/manual/en/function.pg-fetch-result.php. On the basis of the code shown here, it's a bit hard to determine exactly what your problem is; however the odds are that the error "supplied argument is not a valid PostgreSQL result resource" results from a SQL syntax error, or possibly that you have failed to open a connection to pgsql. However, there are some tools to help you; see http://php.net/manual/en/function.pg-result-error.php For future reference, it helps to post all the code that is relevant to your problem, so in this case it would help, for example, to see how you are making the connection to pgsql and how the $query variable is populated. Cheers -- David Robley A seminar on Time Travel will be held two weeks ago. Today is Boomtime, the 49th day of Discord in the YOLD 3177. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php