On Wednesday 29 May 2002 21:45, [EMAIL PROTECTED] wrote:
> Notepad at the moment as I can't change the settings of my computer due to
> lack of privilages
You have my sympathy :)
--
Jason Wong -> Gremlins Associates -> www.gremlins.com.hk
Open Source Software Systems Integrators
* Web Design & H
Notepad at the moment as I can't change the settings of my computer due to
lack of privilages
- Original Message -
From: "Jason Wong" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, May 29, 2002 11:26 PM
Subject: Re: [PHP] Parse Error(newbie)
&g
CTED]]
> Sent: 29. toukokuuta 2002 16:20
> To: [EMAIL PROTECTED]
> Subject: Re: [PHP] Parse Error(newbie)
>
>
> Line 81 is $query.
>
> I showed it in my first email.
>
> I suspect this line is the one causing the trouble(It is about four
> lines above the start of t
mysql_query("delete from conf_event where time < ( time() );
You're missing the ending " there.
Niklas
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
Sent: 29. toukokuuta 2002 16:20
To: [EMAIL PROTECTED]
Subject: Re: [PHP] Parse Error(ne
---
From: "Jason Wong" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, May 29, 2002 11:11 PM
Subject: Re: [PHP] Parse Error(newbie)
> On Wednesday 29 May 2002 21:13, [EMAIL PROTECTED] wrote:
> > Here is the function:
> >
> > function print_new_st
On Wednesday 29 May 2002 21:13, [EMAIL PROTECTED] wrote:
> Here is the function:
>
> function print_new_story(){
> $query = "SELECT * FROM news ORDER BY id DESC LIMIT 1";
> $result = mysql_query($query);
> $num_results = mysql_num_rows($result);
>
> for ($i=0; $i < $num_results; $i++)
> {
>
in \\johnh\c\co2 busters\includes\functions.php on
line 81
- Original Message -
From: "Nick Wilson" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, May 29, 2002 11:08 PM
Subject: Re: [PHP] Parse Error(newbie)
> -BEGIN PGP SIGNED MESSAGE-
>
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
* and then [EMAIL PROTECTED] declared
> I get a parse error on this line
>
> $query = "SELECT * FROM news ORDER BY id DESC LIMIT 1";
>
> I believe that it is an error in the SQL statement
Nothing wrong with that. Let's have the whole error msg
[EMAIL PROTECTED] wrote:
> $query = "SELECT * FROM news ORDER BY id DESC LIMIT 1";
>
> I believe that it is an error in the SQL statement
Nope, looks fine. Show us the previous few lines, the error is probably in
those somewhere.
--
Stuart
--
PHP General Mailing List (http://www.php.net/)
To
Look for errors in the line above this one. PHP doesn't know what SQL is...it's just a
string to PHP.
---John Holmes...
- Original Message -
From: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Wednesday, May 29, 2002 9:03 AM
Subject: [PHP] Parse Error(newbie)
I get a parse error on this line
$query = "SELECT * FROM news ORDER BY id DESC LIMIT
1";
I believe that it is an error in the SQL
statement
Thanks in advance
JJ Harrison[EMAIL PROTECTED]www.tececo.com
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www
Another thing you may want to try is the following line:
$query = "select * from news WHERE id = '" . $_get['id'] . "'";
Matt
On Mon, 27 May 2002 [EMAIL PROTECTED] wrote:
> I know it is probably something obvious but the following gives me a parse error and
>as a newbie I am having trouble l
try this:
$query = "select * from news WHERE id = '$_get['id']'";
the double quote right before the $ is ending the string, from that point
on everything is an error.
Matt
On Mon, 27 May 2002 [EMAIL PROTECTED] wrote:
> I know it is probably something obvious but the following gives me a par
PROTECTED]]
> Sent: Monday, May 27, 2002 2:18 AM
> To: [EMAIL PROTECTED]
> Cc: [EMAIL PROTECTED]
> Subject: Re: [PHP] Parse Error(Newbie)
>
> On Mon, 27 May 2002 [EMAIL PROTECTED] wrote:
> > I know it is probably something obvious but the following gives me a
> >
On Mon, 27 May 2002 [EMAIL PROTECTED] wrote:
> I know it is probably something obvious but the following gives me a
> parse error and as a newbie I am having trouble locating it.
>
> $query = "select * from news WHERE id = "$_get['id']"";
A lot of people have answered this already, but just for
-Original Message-
I know it is probably something obvious but the following gives me a
parse error and as a newbie I am having trouble locating it.
$query = "select * from news WHERE id = "$_get['id']"";
-Original Message-
Any time you end up with two " characters together is a
d this
any way, that it's the line above the line the message says it is..
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
Sent: Monday, 27 May 2002 3:47 PM
To: [EMAIL PROTECTED]
Subject: [PHP] Parse Error(Newbie)
I know it is probably something obv
om news WHERE id = \"$_get['id']\"";
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
Sent: Monday, May 27, 2002 3:47 PM
To: [EMAIL PROTECTED]
Subject: [PHP] Parse Error(Newbie)
I know it is probably something obvious but the following gives
I know it is probably something obvious but the
following gives me a parse error and as a newbie I am having trouble locating
it.
$query = "select * from news WHERE id =
"$_get['id']"";
JJ Harrison[EMAIL PROTECTED]www.tececo.com
--
PHP General Mailing List (http://www.php.net/)
To unsub
On jeudi, mai 23, 2002, at 01:27 , Nick Wilson wrote:
>> I get a parse error on line 114 can anyone see the problem?
>>
>> Sorry if it is something obvious :}
>> for ($i=0; $i <$num_results; $i++)
>>
>> 110. {
>> 111. $row = mysql_fetch_array($result);
>> 112. echo '';
>
> Have you tried escaping
why now just write:
for ($i=0; $i <$num_results; $i++) {
$row = mysql_fetch_array($result);
echo "link text";
> -Original Message-
> From: Nick Wilson [mailto:[EMAIL PROTECTED]]
> Sent: 23 May 2002 12:28 PM
> To: [EMAIL PROTECTED]
> Subject:
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
* and then [EMAIL PROTECTED] declared
> I get a parse error on line 114 can anyone see the problem?
>
> Sorry if it is something obvious :}
> for ($i=0; $i <$num_results; $i++)
>
> 110. {
>
> 111. $row = mysql_fetch_array($result);
>
> 112. e
I get a parse error on line 114 can anyone see the
problem?
Sorry if it is something obvious :}
for ($i=0; $i <$num_results; $i++)
110. {
111. $row = mysql_fetch_array($result);
112. echo '';
115. echo $row['description'];
116. echo '';
}
JJ Harrison[EMAIL PROTECTED]www.tececo.com
--
PHP G
23 matches
Mail list logo
