Обсуждение: php insert problem
<snip>
$database = pg_connect ("host=localhost dbname=rbak user=nobody") or die("database connection failed");
$result = pg_exec ($database, "INSERT into employees (emp_name, emp_email) VALUES ('$emp_name','$emp_email')";
if (!$result) {
echo "query died\n";
exit;
}
</snip>
the above code produces a parse error. "parse error, unexpected ';' in blah.php on line X"
line X is the $result = pg_exec ($database, "INSERT into employees (emp_name, emp_email) VALUES
('$emp_name','$emp_email')";line. I've started at this long enough, and it's time for other eyes to look. any ideas?
similarSELECT queries work just fine.
dan
Hi Dan,
You're missing a close parentheses ")" in the pg_exec line... :)
Roj Niyogi
pgHoster.com
PostgreSQL web hosting
dan radom wrote:
><snip>
>
> $database = pg_connect ("host=localhost dbname=rbak user=nobody") or die("database connection failed");
> $result = pg_exec ($database, "INSERT into employees (emp_name, emp_email) VALUES ('$emp_name','$emp_email')";
>
> if (!$result) {
> echo "query died\n";
> exit;
> }
>
></snip>
>
>the above code produces a parse error. "parse error, unexpected ';' in blah.php on line X"
>
>line X is the $result = pg_exec ($database, "INSERT into employees (emp_name, emp_email) VALUES
('$emp_name','$emp_email')";line. I've started at this long enough, and it's time for other eyes to look. any ideas?
similarSELECT queries work just fine.
>
>dan
>
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>
You need to close your parenthesis after the last double quote.
On Wed, 22 May 2002, dan radom wrote:
> <snip>
>
> $database = pg_connect ("host=localhost dbname=rbak user=nobody") or die("database connection failed");
> $result = pg_exec ($database, "INSERT into employees (emp_name, emp_email) VALUES ('$emp_name','$emp_email')";
>
> if (!$result) {
> echo "query died\n";
> exit;
> }
>
> </snip>
>
> the above code produces a parse error. "parse error, unexpected ';' in blah.php on line X"
>
> line X is the $result = pg_exec ($database, "INSERT into employees (emp_name, emp_email) VALUES
('$emp_name','$emp_email')";line. I've started at this long enough, and it's time for other eyes to look. any ideas?
similarSELECT queries work just fine.
>
> dan
>
> ---------------------------(end of broadcast)---------------------------
> TIP 6: Have you searched our list archives?
>
> http://archives.postgresql.org
>
-Chadwick
You're missing a closing )
$result = pg_exec ($database, "INSERT into employees (emp_name, emp_email)
VALUES ('$emp_name','$emp_email')");
Peter
----- Original Message -----
From: "dan radom" <dan@radom.org>
To: <pgsql-php@postgresql.org>
Sent: Wednesday, May 22, 2002 4:06 PM
Subject: [PHP] php insert problem
> <snip>
>
> $database = pg_connect ("host=localhost dbname=rbak user=nobody") or
die("database connection failed");
> $result = pg_exec ($database, "INSERT into employees (emp_name,
emp_email) VALUES ('$emp_name','$emp_email')";
>
> if (!$result) {
> echo "query died\n";
> exit;
> }
>
> </snip>
>
> the above code produces a parse error. "parse error, unexpected ';' in
blah.php on line X"
>
> line X is the $result = pg_exec ($database, "INSERT into employees
(emp_name, emp_email) VALUES ('$emp_name','$emp_email')"; line. I've
started at this long enough, and it's time for other eyes to look. any
ideas? similar SELECT queries work just fine.
>
> dan
>
> ---------------------------(end of broadcast)---------------------------
> TIP 6: Have you searched our list archives?
>
> http://archives.postgresql.org
>
>
>
Hi,
It seems you forgot closing the php function arguments with ')':
$result = pg_exec ($database, "INSERT into employees
(emp_name, emp_email) VALUES ('$emp_name','$emp_email')");
This should work
Regards
--
Matteo Beccati
http://www.phpadsnew.com
http://phppgads.sourceforge.net
Should it not be
$result = pg_exec ($database, "INSERT into employees (emp_name, emp_email) VALUES ('$emp_name','$emp_email');");
Cheers
On Wed, 22 May 2002, dan radom wrote:
> <snip>
>
> $database = pg_connect ("host=localhost dbname=rbak user=nobody") or die("database connection failed");
> $result = pg_exec ($database, "INSERT into employees (emp_name, emp_email) VALUES ('$emp_name','$emp_email')";
>
> if (!$result) {
> echo "query died\n";
> exit;
> }
>
> </snip>
>
> the above code produces a parse error. "parse error, unexpected ';' in blah.php on line X"
>
> line X is the $result = pg_exec ($database, "INSERT into employees (emp_name, emp_email) VALUES
('$emp_name','$emp_email')";line. I've started at this long enough, and it's time for other eyes to look. any ideas?
similarSELECT queries work just fine.
>
> dan
>
> ---------------------------(end of broadcast)---------------------------
> TIP 6: Have you searched our list archives?
>
> http://archives.postgresql.org
> --
> Virus scanned by edNET.
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Hey, dan... are you getting these messages? It seems we have a consensus.
Do you still get an error?
On Wed, 22 May 2002, Chris Thompson wrote:
> Should it not be
>
> $result = pg_exec ($database, "INSERT into employees (emp_name, emp_email) VALUES ('$emp_name','$emp_email');");
>
> Cheers
>
> On Wed, 22 May 2002, dan radom wrote:
>
> > <snip>
> >
> > $database = pg_connect ("host=localhost dbname=rbak user=nobody") or die("database connection failed");
> > $result = pg_exec ($database, "INSERT into employees (emp_name, emp_email) VALUES ('$emp_name','$emp_email')";
> >
> > if (!$result) {
> > echo "query died\n";
> > exit;
> > }
> >
> > </snip>
> >
> > the above code produces a parse error. "parse error, unexpected ';' in blah.php on line X"
> >
> > line X is the $result = pg_exec ($database, "INSERT into employees (emp_name, emp_email) VALUES
('$emp_name','$emp_email')";line. I've started at this long enough, and it's time for other eyes to look. any ideas?
similarSELECT queries work just fine.
> >
> > dan
> >
> > ---------------------------(end of broadcast)---------------------------
> > TIP 6: Have you searched our list archives?
> >
> > http://archives.postgresql.org
> > --
> > Virus scanned by edNET.
> >
>
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>
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-Chadwick