Select variable in query - PLEASE HELP!

database results - info display

[akelly65]

hi there,

I am quite new to php, was wondering if anyone could give me a pointer to why this code is not working, I am trying to get each row of my database to be displayed on a different page, e.g example.com/query.php?=2 (3,4,5 etc) I assume this is how you go about this, any help or suggestions would be amazing.

<?php
// Connects to your Database
mysql_connect("localhost", "login", "password") or die(mysql_error());
mysql_select_db("news") or die(mysql_error());
// Query
$id = $_GET['id'];
$data = mysql_query("SELECT * FROM `news` WHERE `id`='" .$id. "'");

or die(mysql_error());

\\
Get parse error code on
$data = mysql_query("SELECT * FROM `news` WHERE `id`='" .$id. "'");

[Dilly]

Try this

$data = mysql_query("SELECT * FROM `news` WHERE `id`=".$id);

[akelly65]

thank you

seems to be working thanks, but it is now telling me there is an error with a simple:

or die(mysql_error());

if I delete that

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

the line below that?

[Dilly]

take off the semi colon at the end of this

$data = mysql_query("SELECT * FROM `news` WHERE `id`=".$id);

then put this back in

or die(mysql_error());

so it should be this

$data = mysql_query("SELECT * FROM `news` WHERE `id`=".$id)
or die(mysql_error());

show me the code where you use the mysql_fetch_array()

[edited by: Dilly at 7:21 pm (utc) on Sep. 12, 2007]

[akelly65]

took out the semi colon, returned below.

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 '' at line 1

haha, sorry to be a pain, it just dosn't seem to be going right, feel like I am banging my head against a wall.

[Receptional Andy]

WHERE `id`=".$id

Try

WHERE `id`=$id"

[edited by: Receptional_Andy at 7:40 pm (utc) on Sep. 12, 2007]

[akelly65]

Can you spot any other errors, have tried all the suggestions, still not parsing properly, is there any better way to write this code? or is this correct for what I am trying to achieve?

<?php
// Connects to your Database
mysql_connect("localhost", "login", "password") or die(mysql_error());
mysql_select_db("news") or die(mysql_error());

// Query
$id = $_GET['id'];
$data = mysql_query("SELECT * FROM `news` WHERE `id`=$id");
or die(mysql_error());

// Results
while($info = mysql_fetch_array( $data ))
{
Print "";
Print "<center><b>".$info['title'] . " </b> - headline<br><br>";
Print "<b>".$info['story'] . " <br>";
}
Print "</table>";
?>

[Receptional Andy]

$data = mysql_query("SELECT * FROM `news` WHERE `id`=$id");
or die(mysql_error());

Looks like there is a misplaced semi-colon in that section:

$data = mysql_query("SELECT * FROM `news` WHERE `id`=$id") or die(mysql_error());

[russkern]

Here is how I write a similar script

...

if((isset($_GET['n_id'])) && (is_numeric($_GET['n_id'])) ) {
$id=$_GET['n_id'];

include_once ('./includes/mysql_connect.php');
$query="SELECT headline, text, post_date FROM news WHERE n_id=$id";
$result=mysql_query($query);
$num=mysql_num_rows($result);
$row=(mysql_fetch_array($result, MYSQL_ASSOC));

} else {
echo '<h1> Page Error!</h1> <p> This page was accessed in error.</p>';
include ('./footer.html');
exit ();
}

etc.......

This only selects 1 article... but it may be helpful.

I then report it pretty much as you have

[Receptional Andy]

is there any better way to write this code?

A much more complex question! It depends what the code is for. If it's for learning PHP, then your code is fine. If it's for deploying onto a website, there's a way to go ;)

[akelly65]

thanks all!

got it working using below, thanks russkern for the tip also.

$data = mysql_query("SELECT * FROM `news` WHERE `id`='.$id'") or die(mysql_error());