There are further problems with the mySQL query:
You have an error in your SQL syntax; check the manual that corresponds
to your MySQL server version for the right syntax to use near ') AND
CURDATE() = DATE(FROM_UNIXTIME(1239508800)), 1, 0), IF(CURDATE() =
DATE'
This is the whole query, what am
Ron Piggott wrote:
The following code gives me this error message:
Parse error: syntax error, unexpected ':'
in /home/thev4173/public_html/test.php on line 8
It is referencing the : that follows EasterDate (and will eventually get
mad at ChristmasDate)
Ron Piggott wrote:
This is the actual code. What should I have?
Same as when you do a query normally.
$query = SELECT ... WHERE ;
$result = mysql_query($query);
.
On Thu, 2009-09-24 at 13:20 +1000, Chris wrote:
Ron Piggott wrote:
The following code gives me this error message:
Ron Piggott wrote:
Let me try this is a different way.
What is the variable which I am able to retrieve? I am expecting the
result to be a 0, 1 or 2.
Your original problem was getting a parse error.
Your query is not php code - step 1 is to fix that.
Once you've done that, post the new
WOW,
Both of your suggestions worked like magic. Thanks a lot Patrick
-Sashi
Patrick Price wrote:
I believe your problem is that on line 33 you forgot your closing double
quote in the echo statement.
Thanks
patrick
Sent from: Decatur Ga United States.
On Fri, Feb 13, 2009 at 10:02 PM,
On Tue, Apr 15, 2008 at 5:56 PM, Javier Viegas [EMAIL PROTECTED] wrote:
Thanks Daniel i ´ve removed the unexpected else secction. Now it works but
it tells me that there might be a syntax error on the query, this is good
news because now i know i´m having connection with the database but also
On Tue, Apr 15, 2008 at 4:49 PM, Javier Viegas [EMAIL PROTECTED] wrote:
Hi, i have this script wich basically connects to a database and delete a
record according to the Id parameter given. The problem is that when i test
it i get this error:
*Parse error*: syntax error, unexpected T_ELSE
Thanks Daniel i ´ve removed the unexpected else secction. Now it works but
it tells me that there might be a syntax error on the query, this is good
news because now i know i´m having connection with the database but also
there must be something wrong with the query.
$del_str = DELETE FROM libros
To: php-db@lists.php.net
Subject: Re: [PHP-DB] Parse error: syntax error, unexpected T_ELSE
Thanks Daniel i ´ve removed the unexpected else secction. Now
it works but
it tells me that there might be a syntax error on the query,
this is good
news because now i know i´m having connection
Check your placement of all the '.' for a start.
raz
On 6/14/05, babu [EMAIL PROTECTED] wrote:
HI,
Whats the error in this code.
$sqlstmt= CREATE USER .$adduser. PROFILE DEFAULT
IDENTIFIED BY .$addpass. DEFAULT TABLESPACE USERS
TEMPORARY TABLESPACE TEMP
Also, correct me if I'm wrong, but this needs to be split into two
seperate query strings and executed seperately...
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
$sqlstmt= CREATE USER .$adduser. PROFILE DEFAULT
IDENTIFIED BY .$addpass. DEFAULT TABLESPACE USERS
TEMPORARY TABLESPACE TEMP
QUOTA UNLIMITED
ON USERS
ACCOUNT UNLOCK;
GRANT CREATE TABLE TO .$adduser.
GRANT CREATE TRIGGER TO .$adduser.
GRANT CONNECT TO .$adduser.;
for the .'s
: php-db@lists.php.net
Subject: Re: [PHP-DB] parse error in create statement.
Date: Tue, 14 Jun 2005 10:56:24 +0100
Check your placement of all the '.' for a start.
raz
On 6/14/05, babu [EMAIL PROTECTED] wrote:
HI,
Whats the error in this code.
$sqlstmt= CREATE USER .$adduser. PROFILE DEFAULT
- Original message from Aravalli Sai : --
but when i run this code on browser it is giving an
error ... Parse error: parse error in
/home/saravall/.HTML/inv.php on line 63
It would be helpful if you said what the parse error is and which line is 63.
Aravalli Sai wrote:
hi
this is php code for inventory management system
which is done using SQLite.this is an user
interface where an user wil giv values to the text
boxes and it should be saved in the backend database
which is sqlite..
but when i run this code on browser it is giving an
Speaking for myself only, I'm probably going to need to see more code than
this to be able to help. These two lines appear to be OK. The error might be
in what feeds line 187.
-Original Message-
From: Dillon, John [mailto:[EMAIL PROTECTED]
Sent: Thursday, November 20, 2003 9:16 AM
To:
November 2003 14:43
To: 'Dillon, John'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Parse error on array and SQL query
Speaking for myself only, I'm probably going to need to see more code than
this to be able to help. These two lines appear to be OK. The error might be
in what feeds line 187
Dillon, John wrote:
On line 188 and similar:
line 187 $this=$altIDs[$i];
line 188 $query .= ID='$this';
I get the error: Parse error: parse error in
/home/jdillon/public_html/provreport.php on line 188
Could this be because some setting has been changed on my shared host?
$this is normally
Dillon, John wrote:
On line 188 and similar:
line 187$this=$altIDs[$i];
line 188$query .= ID='$this';
I get the error: Parse error: parse error in
/home/jdillon/public_html/provreport.php on line 188
Could this be because some setting has
Dillon, John wrote:
Tried that, doesn't work with:
$query=INSERT INTO ReplyTbl VALUES (' . $monster1 . ', ' .
$monster2 .
', ' . $monster3 . ', NULL, NULL);
Don't suppose they use 'monster' as a reserved word...
How does it not work again??
--
---John Holmes...
Amazon Wishlist:
.
-Original Message-
From: John W. Holmes [mailto:[EMAIL PROTECTED]
Sent: Thursday, November 20, 2003 1:09 PM
To: Dillon, John
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Parse error on array and SQL query
Dillon, John wrote:
Tried that, doesn't work with:
$query=INSERT INTO ReplyTbl
Hi Micella
You need to either slash-escape your insert query or use single quotes:
$query=insert into dene (day,start_time) values (\2003.09.25\,\22:30\) ;
or
$query=insert into dene (day,start_time) values ('2003.09.25','22:30') ;
steve
M|cella Erdem Efe wrote:
I am a new programmer of
[EMAIL PROTECTED] wrote:
IBI I am having a problem with a parse error.
IBI error:
IBI Parse error: parse error, unexpected ')' in
IBI /Users/timbest/Sites/cajunmikes/inc/menupage.php on line 75
IBI Fatal error: Cannot instantiate non-existent class: menu in
IBI
If statements use curly braces. Do this:
if ($packages == 1){
if ($airporttransfer == car)
$airporttransfer2 = 12.00);
if ($airporttransfer == bus)
$airporttransfer2 = 10.00);
if ($airporttransfer == none)
$airporttransfer2 = 0.00);
}
On Wed, 2003-02-26 at 16:42, Chris Payne wrote:
Hi
if ($packages == 1){
if ($airporttransfer == car){
$airporttransfer2 = 12.00};
if ($airporttransfer == bus){
$airporttransfer2 = 10.00};
if ($airporttransfer == none){
$airporttransfer2 = 0.00};
};
() round brackets round the condition {} braces around the 'do it' bit :)
HTH
Peter
There aren't any closing quotes on your $message variable or semicolon
to tell the parser that you're done with the assignment statement.
-Micah
On Mon, 2002-12-02 at 15:05, Chase wrote:
When trying to execute the following script to send the contents of a table
via mail() I am getting
There is a single quote and semicolon on the line under /html that I
thought would be defining the end of the assignment.
Should I have used double quotes instead of single?
Chase
Micah Stevens [EMAIL PROTECTED] wrote in message
.
Your close, but it's just not going to work like you think it will.
Jim
---Original Message---
From: Chase
Date: Monday, December 02, 2002 04:34:18 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Parse Error...
There is a single quote and semicolon on the line under /html
ll.
Jim
---Original Message---
From: Chase
Date: Monday, December 02, 2002 04:34:18 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Parse Error...
There is a single quote and semicolon on the line under /html that I
thought would be defining the end of the assignment.
Should I have used doubl
Remember, the print() function sends stuff to the browser. If you wish
to make the code output to the $message variable, you must use
additional assignment statements.
I made some quick changes to the code below, try that:
?
$server = ;
$user = ;
$pass = ;
$db = ;
$table = ;
?
?
$to = CKnott
Thank you!! This appears to have solved it... I didn't even think of using
the $message variable instead of the print()... Sheesh... I still have
LOTS to learn!!
Chase
Micah Stevens [EMAIL PROTECTED] wrote in message
try:
?
$origVar = 100;
echo POriginal value is.$origVar./p;
$origVar += 25;
echo PAdded a value, now it's .$origVar./p;
$origVar -= 12;
echo PSubtracted a value, now it's .$origVar./p;
$origVar .= chickens;
echo PFinal answer: .$origVar./p;
?
Cami
-Original Message-
From: Jennifer
Hi Jenn,
The code looked ok to me , so I copied and pasted the code you posted and it
worked for me.. Where are you getting the parsing error? Is it the whole
code anyway?
Gurhan
-Original Message-
From: Jennifer Downey [mailto:[EMAIL PROTECTED]]
Sent: Thursday, February 21, 2002 9:33
Is $cars an array field? If not, you are trying to compare $cars to an
list/array of values (I am not sure this would work even if $cars was an
array field) 'WHERE $cars
= $car_type\,\$car_model\,\$car_year\,\$car_price\,\$car_vin\,\$dlr
_num\);'.
Normally, you would have to compare each
Here is the code from the file that queries the db and pulls the current
contents of the db and provides a check box form for each record to delete
the items in the db. I dont know if this will help but like I said before
any insight would be great. Thanks in advance.
Jas
?php
require
Not enough information given to figure out what's going on, but looking at
your query i see:
WHERE $cars =
$car_type\,\$car_model\,\$car_year\,\$car_price\,\$car_vin\,\$dlr
_num\);
do you keep 6 different values separated by commas in the field referenced
by $cars??? If this is correct what
, February 21, 2002 12:48 AM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Parse Error
Here is the code from the file that queries the db and pulls the current
contents of the db and provides a check box form for each record to delete
the items in the db. I dont know if this will help but like I
ll don't know what the test table looks like? Does
it
only have one column with all the infos about cars populated in it???
Gurhan
-Original Message-
From: jas [mailto:[EMAIL PROTECTED]]
Sent: Thursday, February 21, 2002 12:48 AM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Par
: [PHP-DB] Parse Error
Here is the dump for the cars table...
test (
id varchar(30) NOT NULL auto_increment,
car_type varchar(30),
car_model varchar(30),
car_year varchar(15),
car_price varchar(15),
car_vin varchar(25),
dlr_num varchar(25),
PRIMARY KEY (id),
KEY id (id
deleted associate each checkbox with the car's id.
Hope this helps..
Gurhan
-Original Message-
From: jas [mailto:[EMAIL PROTECTED]]
Sent: Thursday, February 21, 2002 1:02 AM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Parse Error
Here is the dump for the cars table...
test (
id
According to your code, if the checkbox next to the word 'remove' is checked,
then the value of $cars passed to the form is checkbox. Then in the
done2.php3 page, you are trying to delete where checkbox='array of values
listed'.
Is there a field in your database called checkbox? If so,
the way to start this is to just add a couple of letters to the message and
see if you're still getting an error -
try (on line 357)...
$message .=nobr\nHello world;
keep it all on one line, and then see if you're still getting an error - if
you are then i don't think this line is the one you
On Thu, 2001-12-13 at 23:49, Rob Day wrote:
As far as I can see, you're missing '; in line 360, unless it continues
further down and terminates the $message string there.
Here is the offending code:
353 if ($moderation == no){
354 $message .= checked;
355 }
356
357$message
Rob -
I think I see what you're doing: you're concatenating $message with something
else (the HTML output following line 357?). Problem is, you're just stating
the right hand side of what should be an assignment. The concatenation should
be:
$variable = $variable .= 'HTML output';
and it looks
Dag Jelle,
Als je printf(); gebruikt moet je dit wel correct doen, met print(); zou
het goed moeten gaan.
Dus regel 24 wordt: print ($resultaat[0]);
(Excuse me muttering Dutch between the two of us :)
M.vr.gr.
Roel Mulder
At 21:32 10-11-2001 +0100, you wrote:
Hi all,
I have recently
Hi Jelle,
- Original Message -
I have recently installed an apache server, PhP4 and MySql on my win2000 machine (I
know, I should be running
linux; That will have to come with time). But now I am trying to learn PhP - MySQL.
When running the script
below I get the error message:
Parse
[EMAIL PROTECTED]
To: Jelle Ferwerda [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Saturday, November 10, 2001 10:39 PM
Subject: Re: [PHP-DB] parse-error MySQL - PhP @ win2000
Dag Jelle,
Als je printf(); gebruikt moet je dit wel correct doen, met print(); zou
het goed moeten gaan.
Dus regel 24
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