[SOLVED] Display data from database

[wrathican]

hi

thanks for looking at my post

what i need help with is displaying data from a database field.

the data to be displayed is determined by the value of a variable.

i am using a switch statement to define the variable $content.

 

the query is something like this:

SELECT * FROM table WHERE field=$content

 

is this right?

 

the table has multiple fields, ID, Title, Page and content.

and i will only want to echo the Title and Content fields.

 

what would the best way to do this be?

 

this thing is i will be closing the php tag after the query has been exceuted and then reopening it when i need to use the Title and content. is this possible or does php think it doesnt need it after i close the php tag?

[Wuhtzu]

What you need are the very basics of interacting with MySQL from PHP. I will suggest you read some tutorials on the subject:

 

Get data from database: (what you are asking about in your thread)

http://www.php-mysql-tutorial.com/php-mysql-select.php

 

 

General PHP and MySQL:

http://www.php-mysql-tutorial.com/mysql-tutorial/index.php

 

Best regards

Wuhtzu

[aniesh82]

hello

 

Close or open a php tag is not a problem.

 

 

Regards

Aniesh Joseph

anieshjoseph@gmail.com

[wrathican]

hmm i have tried the site you suggested but i keep getting this error:

 

 

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/fhlinux179/l/ls12style.co.uk/user/htdocs/cycleyorks/cycleyorks/index.php on line 12

 

i am using this code:

<?php 
include 'switch.inc';
include 'misc.inc';
include 'opendb.inc'; 

//sets the query
$query  = "SELECT * FROM cy_content WHERE cont_page=".$content;

//sets the result as the query
$result = mysql_query($query);

while($row = mysql_fetch_assoc($result))
{
$title = $row['cont_title'];
$maincontent = $row['cont_content'];
}

?>

and then echoing the maincontent and title variables later down my page but nothing shows in return

 

any idea whats wrong?

[wrathican]

argh! ive tried so much to correct the problem but its not working

my query doesnt return any result as far as i can see

 

the query is this :

$query  = "SELECT * FROM cy_content WHERE cont_page=".$content;

 

the $content variable is declared in a switch statement before the query and the default is set to 'index'.

 

i then have the result variable containing the mysql_query and $query:

$result = mysql_query($query);

 

and then i have the while statement:

while($row = mysql_fetch_array($result, MYSQL_ASSOC))

{

$title = $row['cont_title'];

$maincontent = $row['cont_content'];

}

 

but even when i just echo $result i do not see anything.

 

this is a screen shot of my tables list

 

this is a screen shot of the data in my cy_content table

 

this is my database connection info file:

<?php 
$host = "**************";
$user = "*********";
$pass = "******";
$dbname = "ls12style";
?>

 

this is my open DB file:

<?php
$con = mysql_connect($host, $user, $pass) or die                      ('Error connecting to mysql');
mysql_select_db($dbname);
?>

 

both of the above files are .inc files and are included at the start of the php code.

 

this is the switch statement i am using:

the 'a' is from the url which is set when i click a link, the switch statement is fine there are no erros with it. its for referencing purposes.

switch($_GET['a'])

{

case "1":
$homepic = "homeoff.jpg";
$aboutpic = "abouton.jpg";
$coursepic = "coursesoff.jpg";
$tourpic = "toursoff.jpg";
$hirepic = "hireoff.jpg";
$accompic = "accomoff.jpg";
$datespic = "datesoff.jpg";
$bookingpic = "bookingoff.jpg";
$bringpic = "bringoff.jpg";
$contactpic = "contactoff.jpg";
$commentpic = "commentsoff.jpg";
$gallerypic = "galleryoff.jpg";
$challengespic = "challengesoff.jpg";
$linkspic = "linksoff.jpg";

//this is the variable that decides on what content to show
$content = "about";
break;

 

the reason i have posted all this is so that you can look at everything and see if all the variables are the same, and there are no errors

[Illusion]

$content="about';

there are no matching records in table cy_content with cont_title as 'about'. Obviously no results would be returned.

[wrathican]

$content="about';

there are no matching records in table cy_content with cont_title as 'about'. Obviously no results would be returned.

my query isnt asking about cont_title equalling about

 

its asking cont_page to equal it

 

well i have gotten rid of the error i had because i realised i had missed out some quotes.

 

now i have something different...

when i echo my query i get this:

SELECT * FROM cy_content WHERE cont_page='index'Resource id #6

the 'Resource is #6' is a new thing... i havent ever seen it happen before.

any idea what that means

and i still return no results

[Illusion]

Sorry for my mistake and

Resource is #6 is the value stored in $result. You might be echoed $result also.

add this code after $result = mysql_query($query)

if (!$result) {
    echo 'Could not run query: ' . mysql_error();
    exit;
}
if(mysql_num_rows($result)==0)
{
echo 'No results are found';
}

then post what you are getting.

[wrathican]

i still get this.....

 

SELECT * FROM cy_content WHERE cont_page='index'

Resource id #6

 

and i dont get anything else...

[Illusion]

Post php code you are using .

[wrathican]

<?php echo $query; echo "<br>"; 
	if (!$result) {
    echo 'Could not run query: ' . mysql_error();
    exit;
}
if(mysql_num_rows($result)==0)
{
echo 'No results are found';
}?>

 

and sorry i dont get the resource is #6 anymore

 

but i do not get anything from this, but by the looks of it im not meant to....

[Illusion]

I don't know what you have understand form my posts what I mean is, use the following code

<?php 
if ((include 'switch.inc') == 'OK') {
    echo 'OK';}
if ((include 'misc.inc') == 'OK') {
    echo 'OK';}
if ((include 'opendb.inc') == 'OK') {
    echo 'OK';}


//sets the query
$query  = "SELECT * FROM cy_content WHERE cont_page=".$content;

//sets the result as the query
$result = mysql_query($query);
if (!$result) {
    echo 'Could not run query: ' . mysql_error();
    exit;
}
if(mysql_num_rows($result)==0)
{
echo 'No results are found';
}

while($row = mysql_fetch_assoc($result))
{
$title = $row['cont_title'];
$maincontent = $row['cont_content'];
}
?>

If you are getting three OKs then your files are included and

if you got the error "Could not run query:  . mysql_error()" there went something wrong with the query

tell me what you are getting........

[wrathican]

ah i am sorry for my mistake

altough i really am not seeing anything

i do not even see the OK's

i really dont know what is happening here

ive been working on this all day

[wrathican]

ahh thanks

 

i dont know how you fixed it but its working now

 

thanks alot