PHP - 3 Pages Linking With A Two Database Tables.

<?php 

$con = database_connect();
$sql = "SELECT *
FROM anime1, episode1
WHERE animeid='$animeid'";
$result = mysql_query($sql);
while ($row = mysql_fetch_assoc($result)) {
$title = $row['title'];
$ep = $row['ep'];
}


 ?>

keep giving me back error

Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\wamp\www\studying\take 2\addin12.php on line 45

i have two tables Loan Application table and Loans table. There is status option and date loan was applied section. After status has been chosen as Pending from the dropdown field and date chosen and loan application saved, it goes under the loans section.

Now, admin has to go and approve the loan and pick the date the loan was approved under the loans section but date can be chosen for the approved date but the status cant be changed to approve because it shows pending. Please what can i do so when i select the date for approval , the status should change to approved automatically

Okay so I have 2 tables in my database. One called user and one called messages. A user logs in to the message board and leaves a message (eg nice website). They write in the author name and the message then after the message is posted it says

"Nice website"
Posted by (author) on (date).

All is good so far. It works. However if you look at my code you will see I have a session started. This session is storing the username of the logged in user. From the column username in the users table. (This table has has an id for each user). Ive played around with the code trying to make it so the user doesnt have to fill in the author box. I want rid of that box So the logged in user just leaves a message then it says "posted by (username) on (date).

Im missing something from my code.
Can anyone tell me what? Please?

<?php
session_start();
mysql_connect("*************", "*****************", "***************");
mysql_select_db("***********************");
$time = time();



//this checks to see if the $_SESSION variable has been not set 
//or if the $_SESSION variable has been not set to true
//and if one or the other is not set then the user gets
//sent to the login page
if (!isset($_SESSION['username'])) {
    header('Location: http://***************.com/login.php');
}

$query = "INSERT INTO messages VALUES( NULL, '". mysql_real_escape_string($_POST['message']) ."', '". mysql_real_escape_string($_POST['username']) ."', '$time' )";if( $result = mysql_query($query) ) {
        if(mysql_affected_rows() > 0 ) {
                echo "Message Posted.<br><a href='messageboard.php'>Return</a>";
        } else {
                echo 'There was an error posting your message. Please try again later.';
        }
} else {
        echo "There was a database error.";
        // comment out next line for live site.
        echo "<br>Query string: $query<br>Returned error: " . mysql_error() . '<br>';
}


;

Here's the scenario:

I have an output (list of fields from a database in a table format). It is a list of items in my game, but I want to make the name be hyperlinked to the information from the database (name, attributes, price, etc.).

How would I create a hyperlink so that when I do click on the item's name, it will open up the item's information to make it editable?

(I have a form that creates the item already, but I want to make it have that information pre-populated for that item so if it needs edited, it could be at that point.)

Thanks!

Hopefully I explained that ok...

Hi,

Currently on my website I have a section where you can select an article held within a database, I still want this to happen but I wouldn't a different word to be linked to the article.
Currently it reads 'Click here to view this entry' underneath the title of the article, I want he title of all the different articles to link to the right article is there anyway to do this?

My code currently reads:

Code: [Select]
<?php

$blog_postnumber = 5;

if(!isset($_GET['page'])) {
$page = 1;
}
else {
$page = (int)$_GET['page'];
}
$from = (($page * $blog_postnumber) - $blog_postnumber);

$sql = "SELECT * FROM cms_article ORDER BY timestamp DESC LIMIT $from, $blog_postnumber";

$result = mysql_query($sql) or print ("Can't select entries from table cms_article.<br />" . $sql . "<br />" . mysql_error());

while($row = mysql_fetch_array($result)) {

    $date = date("l F d Y", $row['timestamp']);

    $title = stripslashes($row['title']);
    $entry = stripslashes($row['entry']);
$id = $row['id'];

if (strlen($entry) > 0) {
    $entry = substr($entry, 0, 0);
    $entry = "$entry<a href=\"journal.php?id=" . $id . "\">Click here to view this entry.</a>";
}

    ?>

Right i want to have friends displayed on peoples profiles but i can only ave there name and not the avater displayed here is the code ive got which doesnt work

        <? 
$picture = mysql_query("SELECT * FROM users WHERE username = '$dip->person'");
$pc = mysql_fetch_object($picture);
$query_friends=mysql_query("SELECT * FROM friends WHERE username='$viewuser' AND type='Friend'");

$rows=mysql_num_rows($query_friends);
if ($rows == "0"){ echo "<center>No friends</center>"; }
$friend = 0;
while($dip=mysql_fetch_object($query_friends)){

echo "&nbsp;&nbsp;<img src='$pc->image' width='50' height='50' border='1'><br><a href='profile.php?viewuser=$dip->person'>$dip->person</a>,"; 
$friend++; 
echo ($friend % 3 == 0)? "<br>" : ""; }
?>


The friends are from the friends database and the avater is from the users database how can i link them so the name and the avater show ive tried this but only the names are displayed and the avaters dont show

Sorry for my bad english but i hope you would understand my query.

How can i set 404 not found for database driven pages..?


I am using header("HTTP/1.0 404 Not Found"); in my 404 error page and ErrorDocument 404 /404.html, whereas 404.html is my error page.

i have an URL http://www.example.com/folder/my-test-for-url-95/ which is opening correctly because it has existence on database. but when i am opening http://www.example.com/folder/my-test-for-url-951/ which has existence on database is also opening without any information but i want here a 404 not found page. How can i do so.


Thanks

Hi guys,

Is it possible to insert into two tables in a database?

Thanks

How we can make the connection to ODBC in PHP If any shares the coding part then it would be great. (Table to database)

 

Hi can anybody help I need to export and download tables from a database into a excel sheet. I have this code and it works what I need is to export specific fields and not just the whole table can anyone help modifying the code to export certain fields within the table please? Here is the code...
Code: [Select]
<?php
$host = 'localhost';
$user = 'user';
$pass = 'password';
$db = 'qdbname';
$table = 'tablename';
$file = 'export';

$link = mysql_connect($host, $user, $pass) or die("Can not connect." . mysql_error());
mysql_select_db($db) or die("Can not connect.");

$result = mysql_query("SHOW COLUMNS FROM ".$table."");
$i = 0;
if (mysql_num_rows($result) > 0) {
while ($row = mysql_fetch_assoc($result)) {
$csv_output .= $row['Field']."; ";
$i++;
}
}
$csv_output .= "\n";

$values = mysql_query("SELECT * FROM ".$table."");
while ($rowr = mysql_fetch_row($values)) {
for ($j=0;$j<$i;$j++) {
$csv_output .= $rowr[$j]."; ";
}
$csv_output .= "\n";
}

$filename = $file."_".date("Y-m-d_H-i",time());
header("Content-type: application/vnd.ms-excel");
header("Content-disposition: csv" . date("Y-m-d") . ".csv");
header( "Content-disposition: filename=".$filename.".csv");
print $csv_output;
exit;
?>

Hi,

I am re-visiting my database table relationship for criminal incidents.

Six tables are involved plus a junction table to make it seven tables in total.
I need someone to review this relationship and advise whether or not it is correctly set up.

The relationships are as follows:

1. A person can belong to more than one incident:

t_persons (one-to-many) t)_incidents_persons


2. A person can only have one nationality by birth

t_persons (one-to-one) t_countries

3. An agency can have more than one incident

t_agencies (one-to-many) t_incidents


4. A status (e.g. closed incident) can belong to more than one incident

t_status (one-to-many_ t_incidents

5. A keyword (i.e. offence type eg theft)

t_offencekeywords (one-to-many) t_incidents

Hey All,

Been banging my head into the wall on this one. I have 2 tables one for users 'myMembers' and one for products 'products'. Each table has a auto increment id. The myMembers id is the user id and the products table id is for the product id. I have a row in the products table for agent_id. I would like the agent_id to be filled with the id from the myMembers table.  I took a look at the manual; still do not understand how to take the id from the myMembers table then place that id into the agent_id; so the products(id) can be listed under the specific members id(agent_id) in the products table.

So far my script for the products table inserts items correctly, but does not file under the specific agent_id. Here is the script for entering items to the products table.

Thanks for the guidance!


<?php
// Script Error Reporting
error_reporting(E_ALL);
ini_set('display_errors', '1');
?>
<?php
// Delete Item Question to Admin, and Delete Product if they choose
if (isset($_GET['deleteid'])) {
   echo 'Do you really want to delete product with ID of ' . $_GET['deleteid'] . '? <a href="inventory_list.php?yesdelete=' . $_GET['deleteid'] . '">Yes</a> | <a href="inventory_list.php">No</a>';
   exit();
}
if (isset($_GET['yesdelete'])) {
   // remove item from system and delete its picture
   // delete from database
   $id_to_delete = $_GET['yesdelete'];
   $sql = mysql_query("DELETE FROM products WHERE id='$id_to_delete' LIMIT 1") or die (mysql_error());
   // unlink the image from server
   // Remove The Pic -------------------------------------------
    $pictodelete = ("../inventory_images/$id_to_delete.jpg");
    if (file_exists($pictodelete)) {
                 unlink($pictodelete);
    }
   header("location: inventory_list.php");
    exit();
}
?>
<?php
// Parse the form data and add inventory item to the system
if (isset($_POST['product_name'])) {
   
    $product_name = mysql_real_escape_string($_POST['product_name']);
   $price = mysql_real_escape_string($_POST['price']);
   $category = mysql_real_escape_string($_POST['category']);
   $subcategory = mysql_real_escape_string($_POST['subcategory']);
   $details = mysql_real_escape_string($_POST['details']);
   // See if that product name is an identical match to another product in the system
   $sql = mysql_query("SELECT id FROM products WHERE product_name='$product_name' LIMIT 1");
   $productMatch = mysql_num_rows($sql); // count the output amount
    if ($productMatch > 0) {
      echo 'Sorry you tried to place a duplicate "Product Name" into the system, <a href="inventory_list.php">click here</a>';
      exit();
   }
   // Add this product into the database now
   $sql = mysql_query("INSERT INTO products (product_name, agent_id price, details, category, subcategory, date_added)
        VALUES('$product_name','$price','$details','$category','$subcategory',now())") or die (mysql_error());
     $pid = mysql_insert_id();
   // Place image in the folder
   $newname = "$pid.jpg";
   move_uploaded_file( $_FILES['fileField']['tmp_name'], "../inventory_images/$newname");
   header("location: inventory_list.php");
    exit();
}
?>
<?php
// This block grabs the whole list for viewing
$product_list = "";
$sql = mysql_query("SELECT * FROM products ORDER BY date_added DESC");
$productCount = mysql_num_rows($sql); // count the output amount
if ($productCount > 0) {
   while($row = mysql_fetch_array($sql)){
             $id = $row["id"];
          $product_name = $row["product_name"];
          $price = $row["price"];
          $date_added = strftime("%b %d, %Y", strtotime($row["date_added"]));
          $product_list .= "Product ID: $id - <strong>$product_name</strong> - $$price - <em>Added $date_added</em> &nbsp; &nbsp; &nbsp; <a href='inventory_edit.php?pid=$id'>edit</a> &bull; <a href='inventory_list.php?deleteid=$id'>delete</a><br />";
    }
} else {
   $product_list = "You have no products yet";
}
?>

Hi there i have a big issues which is starting to cause me lots of stress. I have two tables one which is products and contains the following below and the other which is additionalCategories which the contains is also below.

products

    id                    
    name                                
    image1                         
    image2                            
    image3                            
    image4                         
    image5                            
    description                         
    department                            
    category                            
    subcategory                         
    Price

additonalCategories

         id                     
    departmentID                         
    categoryID                         
    subcategoryID                           
    productID

i have made a products upload script to my site where i can add products too it saves the category and subcategory id's in the fields above then if i wish to add the product to additional categories i do so and again this adds all the data to the addiotnalCategories table.

however i can seems to be able to display this on the view products page i tried the below code but get no luck

Code: [Select]
<?php
session_start();
$username=$_SESSION['username'];
include_once"../includes/db_connect.php";
$cat=$_GET['cat'];
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
<?php 
$query("SELECT products.name AS name, products.price AS price, addtionalCategories.productID AS productID FROM products, addtionalCategories WHERE category='$cat' AND products.id = addtionalCategories.productID");
while($fetch=mysql_fetch_object($query)){
echo"$fetch->name";
}
?>
</body>
</html>

i need a way in which on the page it check the category id then finds all the products that are saved in the additionalCategories table then to find the data saved in the products table such as the name and price etc

please any help will be appreciated thank you for reading
 

This topic has been escorted to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=354341.0

Here is the sample page:
http://hoosierhoopsreport.com/pg-test/

Trying to create a table that has three sections (groupings).  That part I have figured out.  The first grouping is sorted numerically.  The second group is sorted alphabetically based on what part of the state they live in.  It's breaking down this second part that I can't seem to get.  (The third group is just what's left over, sorted alphabetically.)

So it needs to look like this:  (Each group and region have their own header row.  Only Group 2 is broken down into Regions.)

Group 1
Group 2
-Region 1
-Region 2
-Region 3
-Region 4
-Region 5
Group 3

The sample page above, in Group 2, it just shows Region 1 header. 

Code: [Select]
$query = 'SELECT * FROM a_playerRank WHERE year="2015" and position="1" ORDER BY grouping DESC,rankPos,region,nameLast';
$results = mysql_query($query) or trigger_error('MySQL error: ' . mysql_error());

$currentGrouping = false;
$currentRegion = false;

while($line = mysql_fetch_assoc($results)) {

    if($currentGrouping != $line['grouping'])
        {
            //Status has changed, display status header

            $currentGrouping = $line['grouping'];

if($line['grouping']==2)  {

echo '<thead>
<tr>
<th class="num">#</th>
<th class="top">Top 10</th>
<th class="height">HT</th>';

if (current_user_can("access_s2member_level4")){
echo '<th class="level">Level</th>';
}
echo'
<th class="school">City (School)</th>';

if (current_user_can("access_s2member_level4")){
echo '<th class="summer">Summer Team</th>';
}
echo'
<th class="college">College</th>
</tr>
</thead>';
}

if($line['grouping']==1) {echo '<tr><th colspan="7">Best of the Rest</th></tr>';


if($currentRegion != $line['region'])
{
$currentRegion = $line['region'];

echo '<tr><th colspan="7">Region' .$line['region'] . '</th></tr>';}


}

if($line['grouping']==0) echo '<tr><th colspan="7">Names to Know</th></tr>';

}