Thanks Everyone...
After I sent that...I got thinking about doing both queries in one statement.
So thats what I did.
Its working fine...
Here is the updated code:
'$tstamp' and
egw_cal.cal_id=egw_cal_dates.cal_id", $db);
if ($event = mysql_fetch_array($events)) {
echo "\n";
echo "\n";
This is a join - Read up on them, they're very useful and don't require
the overhead of a sub-query.
SELECT egw_cal.* FROM egw_cal_dates
LEFT JOIN egw_cal using (cal_id)
where egw_cal_dates.cal_start > $tstamp
AND egw_cal.cal_category = '501'
-Micah
On 02/12/2007
Try this as your SQL. It should give you all the results, then you can use PHP
to sort it all out.
SELECT * FROM egw_cal WHERE cal_category='501' and cal_id in (SELECT cal_id
FROM egw_cal_dates where cal_start > $tstamp)
-TG
= = = Original message = = =
Hello Everyone
Got a simple / st
Matthew Ferry wrote:
Hello Everyone
Got a simple / stupid question.
Worked on this all night. I'm over looking something very basic here.
The query "event_time" brings back the calendar id for each event that is
pending in the future.
ie 12, 13, 14, 26 (There could be 100 of them out
Chris Carter wrote:
What wrong with this syntax, its not giving any error on runtime but I am
facing a blank page while paging.
$query=" SELECT * FROM gurgaonmalls WHERE mallname = '$mallname' limit $eu,
$limit ";
Have you tried...
echo " $query ";
...to unsure the variables have the values
OK, this makes my day clear!!
I have versión 3.23.49-3 of MySQL
Thanks Dwight!
-Original Message-
From: Dwight Altman [mailto:[EMAIL PROTECTED]
Sent: Jueves, 28 de Septiembre de 2006 11:32 a.m.
To: php-db@lists.php.net
Subject: RE: [PHP-DB] SQL query
Check your version. Subselects
Check your version. Subselects were only added in MySQL Version 4.1.
Regards,
Dwight
> -Original Message-
> From: Edwin Cruz [mailto:[EMAIL PROTECTED]
> Sent: Thursday, September 28, 2006 10:53 AM
> To: 'Miguel Guirao'; php-db@lists.php.net
> Subject: RE: [PH
Make sure that your second query is returning only one row, if it dont
help, try this:
$query="select email from usuarios where userName in (select username
from fussv where folio = 'FUSS-130-2006')"
MySQL think that you second query returns more than 1 row, that's why
mysql dont accept your quer
From: "Muhammed Mamedov" <[EMAIL PROTECTED]>
> Try this
>
> "SELECT DISTINCT(file_name), Count(file_name) AS cnt FROM $table_name
WHERE
> date
> > BETWEEN '2003-10-01' AND '2003-12-31' group by file_name order by cnt;
> > desc"
If you're GROUPing by "file_name" then you don't need DISTINCT(file
Try this
"SELECT DISTINCT(file_name), Count(file_name) AS cnt FROM $table_name WHERE
date
> BETWEEN '2003-10-01' AND '2003-12-31' group by file_name order by cnt;
> desc"
Regards,
Muhammed Mamedov
tmchat.com
- Original Message -
From: <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent:
use alias for 'Count(file_name)' to use in order by clause
F.x.
"SELECT DISTINCT(file_name), Count(file_name) as file_count FROM $table_name
WHERE date BETWEEN '2003-10-01' AND '2003-12-31' group by file_name order by
file_count desc"
Hope that solves it
Nitin
- Original Message -
From:
Ah...
you can name the count()..
live and learn, cheers dude...
"brett king" <[EMAIL PROTECTED]>
15/01/2004 11:17
Please respond to
<[EMAIL PROTECTED]>
To
<[EMAIL PROTECTED]>, <[EMAIL PROTECTED]>
cc
Subject
RE: [PHP-DB] SQL query...
"SELECT DIS
"SELECT DISTINCT(file_name), Count(file_name) FROM $table_name WHERE date
BETWEEN '2003-10-01' AND '2003-12-31' group by file_name order by ???
desc"
In the above sql statement, I'm trying to achieve:
1. select all file names, between two dates.
2. list them, and order by the highest number o
Create a form for editing the record
Then on the display funtion just put a link on each record to that form
and pass the id of that record like edit
On the edit form just grab the data of the $id passed on the url and put
those values on the input fields like
Than just save the form result into t
> i'm starting out in this whole thing, and i'm trying to program a zine
> that
> uses a sql database for content and calls the content into a template.
> what
> i am unclear on how to do is set up the links so that I can get the
info
> from the database. anyone willing to help me out?
www.youdom
id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
Adam
On Sat, 7 Sep 2002, Bryan McLemore wrote:
> Hi Guys I have written this SQL Query :
>
> CREATE TABLE tasks (id INT AUTO_INCREMENT, name VARCHAR(50), desc TEXT, address
>VARCHAR(50), startDate DATE, lastWork DATE, progress
It would unless you told it not to. Set a flag
$lastContactType='';
Then on the first page you display something where then display it and then
if ($lastContactType != $row['contactType']){
echo $row['contactType'];
$lastContactType=$row['contactType'];
} // if ($lastContactType !
Hello Chip,
Could you please send your program to dig more.And also
send your database schema.
Do you want to
select title,name from your_table_name where title='title1' and
name='blue2';
Is the title and name field are unique.If it so,then the above query will
helps you.
Regard
The problem is they are parts of different rows. In a relational database rows only
know about other rows through relationships. You have either:
1. Designed your database incorrectly.
or
2. Are a genius with a unique concept that I do not grasp.
<>< Ryan
-Original Message-
From: [EM
Thanks I will try that.
"Beau Lebens" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> try replacing
>
> WHERE display=$custcatergory");
>
> with
>
> WHERE display='$custcatergory'");
>
> and also tru echoing the value of that SQL statement (assign it to a var
>
try replacing
WHERE display=$custcatergory");
with
WHERE display='$custcatergory'");
and also tru echoing the value of that SQL statement (assign it to a var
first) to make sure you are getting the right thing.
HTH
/b
// -Original Message-
// From: CrossWalkCentral [mailto:[EMAIL PR
l.com/doc/A/N/ANSI_diff_Sub-selects.html
-Original Message-
From: Pierre
Sent: Mon 11/12/2001 12:20 AM
To: Gonzalez, Lorenzo; [EMAIL PROTECTED]
Cc:
Subject: Re: [PHP-DB] SQL query
Thanks but I th
Thanks but I think subselect is not possible with MYSQL.
Pierre
- Original Message -
From: Gonzalez, Lorenzo <[EMAIL PROTECTED]>
To: Pierre <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Monday, November 12, 2001 1:08 PM
Subject: RE: [PHP-DB] SQL query
> in other
in other RDBMs this is easily done with a subselect - don't know if it's
doable in MySQL or not, someone else can confirm, or you can try it
yourself...
select * from table1 where table1.id not in (select table2.id);
-Lorenzo
-Original Message-
From: Pierre
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