[PHP-DB] mysql_num_rows, resource id #
In my code below, I am trying to verify that the query is selecting data from
both rows of my answers table. I have run the query on my MySQL CLI and
getting answers from both rows, but running this script I get $rows = 0. I
can't figure out why its not returning 2 for the number of rows. It is not
giving me any error msgs such as cannot connect.
Also can tell me where I can find documentation on resource id #, such as
resource(5)?
?PHP
DEFINE (host,localhost);
DEFINE (user,root);
DEFINE (password,pasword);
DEFINE (database,questions);
$connection=mysql_connect(host,user,password) or die ('Could not connect' .
mysql_error() );
$dbConnect=mysql_select_db('questions',$connection);
if (!$dbConnect) {die ('Could not connect to database' . mysql_error() );}
$result = mysql_query(Select answer from answers where
studentID='A123456789') or die(mysql_error());
$rows = mysql_num_rows($result);
echo $rows;
?
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Re: [PHP-DB] mysql_num_rows, resource id #
Richard Dunne wrote:
In my code below, I am trying to verify that the query is selecting data from both rows of my answers table. I have run the query on my MySQL CLI and getting answers from both rows, but running this script I get $rows = 0. I can't figure out why its not returning 2 for the number of rows. It is not giving me any error msgs such as cannot connect.
Also can tell me where I can find documentation on resource id #, such as
resource(5)?
?PHP
DEFINE (host,localhost);
DEFINE (user,root);
DEFINE (password,pasword);
DEFINE (database,questions);
$connection=mysql_connect(host,user,password) or die ('Could not connect' .
mysql_error() );
$dbConnect=mysql_select_db('questions',$connection);
if (!$dbConnect) {die ('Could not connect to database' . mysql_error() );}
$result = mysql_query(Select answer from answers where
studentID='A123456789') or die(mysql_error());
$rows = mysql_num_rows($result);
echo $rows;
?
Still no luck then. Try selecting the DB without the connection parameter:
$dbConnect=mysql_select_db('questions');
And don't forget to set error_reporting(E_ALL) above your code.
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Fwd: Re: [PHP-DB] mysql_num_rows, resource id #
This much is working:
?PHP
ini_set('error_reporting',E_All);
DEFINE (host,localhost);
DEFINE (user,root);
DEFINE (password,password);
DEFINE (database,questions);
$connection=mysql_connect(host,user,password) or die ('Could not connect' .
mysql_error() );
$dbConnect=mysql_select_db('questions');
if (!$dbConnect) {die ('Could not connect to database' . mysql_error() );}
$result = mysql_query(Select answer from answers where
studentID='A123456789') or die(mysql_error());
$rows = mysql_num_rows($result);
echo $rows . \n;
$answer = mysql_fetch_row($result);
foreach($answer as $a)
{
echo $a;
}
?
The output is
1
1
which is correct.
One small step at a time, hopefully forward.
Richard.
---BeginMessage---
Richard Dunne wrote:
In my code below, I am trying to verify that the query is selecting data from both rows of my answers table. I have run the query on my MySQL CLI and getting answers from both rows, but running this script I get $rows = 0. I can't figure out why its not returning 2 for the number of rows. It is not giving me any error msgs such as cannot connect.
Also can tell me where I can find documentation on resource id #, such as
resource(5)?
?PHP
DEFINE (host,localhost);
DEFINE (user,root);
DEFINE (password,pasword);
DEFINE (database,questions);
$connection=mysql_connect(host,user,password) or die ('Could not connect' .
mysql_error() );
$dbConnect=mysql_select_db('questions',$connection);
if (!$dbConnect) {die ('Could not connect to database' . mysql_error() );}
$result = mysql_query(Select answer from answers where
studentID='A123456789') or die(mysql_error());
$rows = mysql_num_rows($result);
echo $rows;
?
Still no luck then. Try selecting the DB without the connection parameter:
$dbConnect=mysql_select_db('questions');
And don't forget to set error_reporting(E_ALL) above your code.
---End Message---
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Re: [PHP-DB] Not updating certain fields in same row
Hi Everyone
Thanks for all the info and help with this, I have decided to write a
separate function for changing the password. That way I can compare
the original password with the one inputted in the database, and then
change it after both have been the old and new password have been
verified :)
So Thanks! :)
On Mar 25, 2008, at 12:59 PM, Jason Pruim wrote:
Hi everyone,
I am attempting to update a record for a login system while leaving
certain fields untouched if they arn't changed, and am running into
issues.
Basically what I want to do, is say I have these fields:
Field1
Field2
Field3
Field4
I update Field1 and Field3 but not Field2 and Field4. What I want to
do is change the values in Field1 and Field3 without touching the
values in Field2 and Field4.
I have tried this code:
$tab = \t;
if (!isset($_POST['txtLoginName']) ||
empty($_POST['txtLoginName'])) {
$loginName = mysqli_real_escape_string($chpwpostlink,
$_POST['txtLoginName']);
}
else
{
$loginName = $tab;
}
which works the fields that I've changed, but if I don't submit a
value in the form it sets the field to be blank in MySQL. Which is
what I am trying to avoid. Any ideas?
--
Jason Pruim
Raoset Inc.
Technology Manager
MQC Specialist
3251 132nd ave
Holland, MI, 49424-9337
www.raoset.com
[EMAIL PROTECTED]
--
Jason Pruim
Raoset Inc.
Technology Manager
MQC Specialist
3251 132nd ave
Holland, MI, 49424-9337
www.raoset.com
[EMAIL PROTECTED]
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[PHP-DB] Resource id #5
Can someone explain how I can translate Resource id #5 which is what I am
getting from the code below?
$result = mysql_query(Select answer from answers) or die(mysql_error());
$resultArray = explode(',',$result);
for ($i=0;$isizeof($resultArray);$i++)
{
echo $resultArray[$i];
}
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RE: [PHP-DB] Resource id #5
Been a positive integer, it jeans that the SQL query was succesfully
executed, read the function description.
__
Miguel Guirao Aguilera, Linux+, ITIL
Sistemas de Información
Informática R8
Ext. 7540
-- -Original Message-
-- From: Richard Dunne [mailto:[EMAIL PROTECTED]
-- Sent: Thursday, March 27, 2008 11:51 AM
-- To: php-db@lists.php.net
-- Subject: [PHP-DB] Resource id #5
--
-- Can someone explain how I can translate Resource id #5 which is what I
-- am getting from the code below?
--
-- $result = mysql_query(Select answer from answers) or
-- die(mysql_error());
-- $resultArray = explode(',',$result);
-- for ($i=0;$isizeof($resultArray);$i++)
-- {
-- echo $resultArray[$i];
-- }
--
--
--
-- --
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Re: [PHP-DB] Resource id #5
On 27 Mar 2008, at 17:51, Richard Dunne wrote:
Can someone explain how I can translate Resource id #5 which is what
I am getting from the code below?
$result = mysql_query(Select answer from answers) or
die(mysql_error());
$resultArray = explode(',',$result);
for ($i=0;$isizeof($resultArray);$i++)
{
echo $resultArray[$i];
}
For the love of $DEITY, please read the frickin' manual: http://php.net/mysql
. You've been asking similar questions around this topic for the past
few days and you clearly haven't moved forward in your understanding.
The mysql_query function returns a resource. If you print a resource
you get the text Resource id #n where n is replaced with its ID. To
make use of this resource you need to use functions like
mysql_fetch_array, mysql_fetch_assoc or one of the many others
detailed in, you guessed it, the manual. Try it, you might like it.
-Stut
--
http://stut.net/
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Re: [PHP-DB] Table optimization ideas needed
On Thu, Mar 27, 2008 at 1:55 PM, Chris [EMAIL PROTECTED] wrote: Good idea. But I wonder whether calling the trigger each insert will loose any performance. It's going to affect things slightly but whether it'll be noticable only you can answer by testing. Another option I sometimes see is set up a replicated slave and run your reports off that instead of the live system, then: 1) it won't bog the live db down 2) it doesn't really matter how many queries you run 3) it doesn't really matter how long they take to run Hm... This makes sense. :) I indeed have several slaves running. Great. Thank you. Thank you for all your help. -- Postgresql php tutorials http://www.designmagick.com/ -- Regards, Shelley
