error: cant explain

zcollvee

I was shocked at the response I got from this community. Its awesome!

<?php
require("tableusers.php");
require("config.php");
echo "CONNECTED TO DATABASE1";

$result = mysql_query("SELECT type, avatar FROM users WHERE username=".$id."");
while($row = mysql_fetch_array($result)){
$avatar = $row['avatar'];
$type = $row['type'];
echo "AVATAR:".$row['avatar']."<br /> TYPE:
".$row['type']."<br />";
} 
echo 'Logged in as:'.$id.'<br />
Type: '.$type.'<br />
avatar:<img src="'.$avatar.'">';
if( $type == "admin"){
mysql_close($db);
$contents = $_POST['contents'];
$title = $_POST['title'];
$date = date(d, m, Y, g, i, s, a);
echo "$date";
$id = $username;
echo "Attempt 1 SAVING:
".$title."<br />
WITH CONTENTS: ".$contents."<br />";
$avatar = $row['avatar'];
require("tableblogposts.php");
require("config.php");
echo "CONNECTED TO DATABASE2";
$query = "INSERT INTO blogposts (title, contents, by, avatar, date)
VALUES('$title', '$contents', '$username', '$avatar', '$date')";
mysql_query($query);
echo "SAVED";
echo '<meta HTTP-EQUIV="REFRESH" content="5; url=index.php?message=BLOG+SAVED!">';
}else{
echo "NOT PERMITTED TO VIEW THIS";
}
?>

This is to save a blog post.
What I want it to do is.
Open up the users table and read the "TYPE" and "AVATAR" feilds and save them on a string. Then I wanted to check if the users "TYPE" is admin. if it is then it goes on...
Then I want to close the table and open up the blog post table. where I want the
title, contents, date, avatar, by to be saved. and then redirect to the index page where it will show the message "BLOG SAVED"

I've made it echo all the data because I wanted to see where it goes wrong.
BTW this is the form:

<form name="newblog" action="saveblog.php" method="post">
<b>Blog Title: </b><input type="text" name="title"><font color="red">*</font><br />
<b>Content: </b><textarea name="contents" wrap="physical" rows="15" cols="20">
</textarea><font color="red">*</font><br />
username: <b><? echo "$id"; ?></b><br />
<input type="submit" value="Save blogpost!"></form>

laserlight

How does it not work?

Also, you should indent and space out your code for readability, e.g.,

<?php
require "tableusers.php";
require "config.php";
echo "CONNECTED TO DATABASE1";

$result = mysql_query("SELECT type, avatar FROM users WHERE username=" . $id);
while ($row = mysql_fetch_array($result)) {
    $avatar = $row['avatar'];
    $type = $row['type'];
    echo "AVATAR:" . $row['avatar']."<br /> TYPE: " . $row['type'] . "<br />";
}

echo 'Logged in as:' . $id . '<br />Type: '.$type.'<br />avatar:<img src="'.$avatar.'">';

if ($type == "admin") {
    mysql_close($db);
    $contents = $_POST['contents'];
    $title = $_POST['title'];
    $date = date(d, m, Y, g, i, s, a);
    echo $date;
    $id = $username;
    echo "Attempt 1 SAVING: " . $title
        . "<br />WITH CONTENTS: " . $contents . "<br />";

$avatar = $row['avatar'];
require "tableblogposts.php";
require "config.php";
echo "CONNECTED TO DATABASE2";
$query = "INSERT INTO blogposts (title, contents, by, avatar, date)
VALUES('$title', '$contents', '$username', '$avatar', '$date')";
mysql_query($query);
echo "SAVED";
echo '<meta HTTP-EQUIV="REFRESH" content="5; url=index.php?message=BLOG+SAVED!">';
} else {
    echo "NOT PERMITTED TO VIEW THIS";
}
?>

At a glance, the problem seems to be that you are not clear on the logical flow of your script. For example, you close the database connection prematurely. You also seem to be retrieving the data for just one user, yet you do so in a loop.

halojoy

zcollvee wrote:
I was shocked at the response I got from this community. Its awesome!

I've made it echo all the data because I wanted to see where it goes wrong.
BTW this is the form:

Great!
If more people would debug using simple
small echoes in parts of their php scripts
we could mainly have time to deal with other more difficult problems.

zcollvee, at latestdot.dk,
You say: to see where it goes wrong

Why dont you give us some info?
How can we help welland quickly,
when we do not know what 'DID goes wrong'😕

What is the secret, man ... or woman 😃

Finally a typical halojoy script, under development and testing.
Well, sort of.
I think Anybody can see My METHOD, I use often 🙂

<?php

$var = 'blue';
$this = 'bluu';
echo 'point1- my script did start!';// just debugging

echo $var; echo ' '; echo $this; // just debugging
if( $var == $this ){

echo 'point2 - they match';// just debugging

}else{

echo 'point3 - no they dont';// just debugging

}

if( $logged == true ){

echo 'point4 - is logged in';// just debugging

}

exit( '<br />EXIT PHP. My script actually !!! did come to THE END' );

?>

regars 🙂
halojoy

zcollvee

Er ok.
Its this page:
http://coolcenter.latestdot.net/test/newblogpost.php
You have to be logged in to view it.
Then I write the blog and then it has to take me to http://coolcenter.latestdot.net/test/saveblog.php
and then it says this:

CONNECTED TO DATABASE1
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /host/z/c/o/zcollvee/test/saveblog.php on line 7

Basically what I want it to do is "READ" the username($id) and then access the user table and search for the username($id). Once found I need it to take the TYPE and AVATAR column data out and store it in a string called $type and $avatar. Then it has to save the title($title), by($id), date($date), avatar($avatar), and contents($contents) but onfy if the type=admin

laserlight

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /host/z/c/o/zcollvee/test/saveblog.php on line 7

This warning is due to one or more of these:

Failed to connect to database server.
Failed to select database.
Failed to execute the associated SQL statement.