Re: Trouble with SQL statement using variable
От | jeff fitzmyers |
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Тема | Re: Trouble with SQL statement using variable |
Дата | |
Msg-id | 0F1A0EF2-14F3-11D6-9D8D-00306569F51E@earthlink.net обсуждение исходный текст |
Ответ на | Trouble with SQL statement using variable (Jeff Self <jself@nngov.com>) |
Список | pgsql-php |
Have you tried: echo "CONN: $conn<br>"; echo "SQL: $sql<br>"; to see what is being passed to postgres? Jeff On Tuesday, January 29, 2002, at 10:34 AM, Jeff Self wrote: > I'm having trouble getting an SQL statement to use a variable. Here's > the statement: > > $sql = "SELECT emp_fname FROM employee WHERE username = '$username'"; > $result = pg_exec($conn,$sql); > if (!$result) { > exit; > } > I get the following warning from this: > > Warning: Supplied argument is not a valid PostgreSQL link resource in > /var/www/personnel/include/functions.inc on line 25 > > Is there another way to use a variable in an SQL statement. This format > works with MySQL. > > -- > Jeff Self > Information Technology Analyst > Department of Personnel > City of Newport News > 2400 Washington Ave. > Newport News, VA 23607 > 757-926-6930 > > > ---------------------------(end of broadcast)--------------------------- > TIP 5: Have you checked our extensive FAQ? > > http://www.postgresql.org/users-lounge/docs/faq.html >
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