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PHP - Database Tables Relationship
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 Similar TutorialsI have a table for logins and a table for addresses. I'm using 1 form to create the username/password (which is being inserted into the "logins" database), and the persons name, address, etc. which is being inserted into the "addresses" database. What do I need to do to make it so that when a user log's in to the site and make's it to the checkout page, that their address is tied to their username? Ok, something I haven't done yet, I've programmed the majority of my site so far and have yet needed to utilize my relationships. Hopefully I set the tables up correct with foreign keys, primary keys, constraints, and references. This is what I need to do, should be relatively easy for most people here. I have one table... Friends with columns... friend_id - id of the friend users_id - id of the user who added the friend I have another table Statuses with columns... users_id - id of the user who posted the status statuses - the status posted whens - the time the status posted ------ I have statuses properly posting to a person's profile.php page but, I want the homepage of the site (same concept as Facebook) to post the status of your friends. I mean, what's the purpose of a status if your friends can't see them. ------ So, I need to somehow extract all of a users friends from the friends table, and use those friends to insert the statuses of friends on someone's homepage. Much help is greatly appreciated. If you wish to see how my tables are set up and if they are correct, that would be appreciated, just ask, and I'll go get my sql query of those tables for you to look at. THANKS! By the way, my site is at... http://inyoursocialnetwork.netne.net - the domain name and the site name are just temporary fill ins. If you have any ideas or suggestions for a good name, please, please, please let me know. It's driving me nuts I can't come up with a good name. I would prefer a short one. Shorter is easier to type and generally easier to remember. Hi,
I want each new incident created into the t_incidents table by the user to be associated with the AgencyID, a foreign key in the t_users table.
The problem is I do not know how this relationship will work and whether I will need a junction table.
The two tables a
t_Users +----------+----------+------------+ |UserID | AgencyID |User name | +----------+----------+------------+ |1 | 1 |john | +----------+----------+------------+ |2 | 1 |andrew | +----------+----------+------------+ t_Agencies +----------+------------+ |AgencyID |agency name | +----------+------------+ |1 |police | +----------+------------+ |2 |immigration | +----------+------------+I will appreciate your advice. Joseph Edited by josephbupe, 13 October 2014 - 03:04 AM. Hi all I have 3 tables Table_1, Table_2 and Table_3 Table_1 is a list of countries, with name and country_id Table_2 is a table that has 3 fields, id, name and description Table 3 is a table that has name, Table_2_id So what I need to do: Display the name and description field of Table_2 in a form Loop through the countries table and display each as an input box and display on the same form When I fill out the form details, the name/description must be inserted into Table_2, creating an id The input boxes data then also needs inserting into Table_3, with the foreign key of Table_2_id So a small example would be: Name: testing Description: this is a test Country of Australia: Hello Country of Zimbabwe: Welcome This means that in Table_2, I will have the following: ============================= | id | name | description | 1 | testing | this is a test ============================= Table_3 ============================= | Table_2_id | name | country_id | 1 | Hello | 20 | 1 | Welcome | 17 ============================= 20 is the country_id of Australia 17 is the country_id of Zimbabwe Code: Generating the input fields dynamically: $site_id = $this->settings['site_id']; $options = ''; $country_code = ''; $query = $DB->query("SELECT country_code, country_id, IF(country_code = '".$country_code."', '', '') AS sel FROM Table_1 WHERE site_id='".$this->settings['site_id']."' ORDER BY country_name ASC"); foreach ($query->result as $row) { $options .= '<label>' . 'Test for ' . $this->settings['countries'][$row['country_code']] . '</label>' . '<br />'; //$row['country_id'] is the country_id from Table_1 $options .= '<input style="width: 100%; height: 5%;" id="country_data" type="text" name="' . $row['country_id'] . '" value="GET_VALUE_FROM_DB" />' . '<br /><br />'; } echo $options; This outputs: Code: [Select] Textareas go here...... <label>Test for Australia</label> <input type="text" value="" name="20" id="country_data" style="width: 100%; height: 5%;"> <label>Test for Zimbabwe</label> <input type="text" value="" name="17" id="country_data" style="width: 100%; height: 5%;"> Now, I need to insert the value of the input field and it's country_id (20 or 17) into Table_3 and also Table_2_id. This then means I could get the value from Table_3 to populate 'GET_VALUE_FROM_DB' But I'm at a loss on how I'd do this. Could someone help me with this? Thanks The following metadata... <?xml version="1.0" encoding="utf-8"?> <doctrine-mapping xmlns="http://doctrine-project.org/schemas/orm/doctrine-mapping" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://doctrine-project.org/schemas/orm/doctrine-mapping https://www.doctrine-project.org/schemas/orm/doctrine-mapping.xsd"> <entity name="NotionCommotion\App\Account\PublicAccountIdIncrementor" table="public_account_id_incrementor"> <id name="account" association-key="true"/> <field name="chart_id" type="integer"> <options> <option name="default">0</option> </options> </field> <field name="source_id" type="integer"> <options> <option name="default">0</option> </options> </field> <one-to-one field="account" target-entity="NotionCommotion\App\Account\Account" fetch="LAZY" inversed-by="idPublicIncrementor"> <join-columns> <join-column name="id" referenced-column-name="id" nullable="false"/> </join-columns> </one-to-one> </entity> </doctrine-mapping> <?xml version="1.0" encoding="utf-8"?> <doctrine-mapping xmlns="http://doctrine-project.org/schemas/orm/doctrine-mapping" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://doctrine-project.org/schemas/orm/doctrine-mapping https://www.doctrine-project.org/schemas/orm/doctrine-mapping.xsd"> <entity name="NotionCommotion\App\Account\Account" table="account"> <id name="id" type="integer"/> <one-to-one field="idPublicIncrementor" target-entity="NotionCommotion\App\Account\PublicAccountIdIncrementor" mapped-by="account"/> </entity> </doctrine-mapping> creates the following methods...
<?php
class Account
{
/**
* Set idPublicIncrementor.
*
* @param \NotionCommotion\App\Account\PublicAccountIdIncrementor|null $idPublicIncrementor
*
* @return Account
*/
public function setIdPublicIncrementor(\NotionCommotion\App\Account\PublicAccountIdIncrementor $idPublicIncrementor = null)
{
$this->idPublicIncrementor = $idPublicIncrementor;
return $this;
}
/**
* Get idPublicIncrementor.
*
* @return \NotionCommotion\App\Account\PublicAccountIdIncrementor|null
*/
public function getIdPublicIncrementor()
{
return $this->idPublicIncrementor;
}
}
As seen, Account::idPublicIncrementor allows NULL value. Is this by design since PublicAccountIdIncrementor's primary key is Account's primary key? If not, how do I instruct Doctrine to prevent NULL? Hi guys, Is it possible to insert into two tables in a database? Thanks 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; ?> How we can make the connection to ODBC in PHP If any shares the coding part then it would be great. (Table to database) I have two tables. Let's call the first one items and the second one item categories. Now the items table would look something like this: id(auto_increment id) name description categoryid The item categories table would look something like this: categoryid name description An entry in the items table would look something like this: categoryid = 1,2 name = Thing description = something id = 1 Say I want to retrieve records that contain "1" in the categoryid column. How would I do that? 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 Hi everyone. I would like to do this: In page1.php, I need to make a few checkboxes, which when the user clicks "next" button, transfers the data (which checkboxes were clicked) to a database table. Then, when the user clicks a button "next", he is moved to another page, page2.php, from which he is automatically transferred to page3.php. At page3.php those values are gathered from the database, and the user has to click "next" button again, then those values are written into another table, in the same or another database. How can I do this? I've not much knowledge with neither SQL, neither PHP. I hope you will help me make this work, I'm making a little system for my HTML website. Thanks, if you do! 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($line['grouping']==0) echo '<tr><th colspan="7">Names to Know</th></tr>'; } This topic has been escorted to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=354341.0 Hey All, Been banging my head into the wall on this one. I have 2 tables one for users 'myMembers' and one for products 'products'. Each table has a auto increment id. The myMembers id is the user id and the products table id is for the product id. I have a row in the products table for agent_id. I would like the agent_id to be filled with the id from the myMembers table. I took a look at the manual; still do not understand how to take the id from the myMembers table then place that id into the agent_id; so the products(id) can be listed under the specific members id(agent_id) in the products table. So far my script for the products table inserts items correctly, but does not file under the specific agent_id. Here is the script for entering items to the products table. Thanks for the guidance! <?php // Script Error Reporting error_reporting(E_ALL); ini_set('display_errors', '1'); ?> <?php // Delete Item Question to Admin, and Delete Product if they choose if (isset($_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 $date_added</em> <a href='inventory_edit.php?pid=$id'>edit</a> • <a href='inventory_list.php?deleteid=$id'>delete</a><br />"; } } else { $product_list = "You have no products yet"; } ?> This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=353621.0 This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=305902.0 I have a relational table call sales with prodcut_id and custmer_id tables. I also have a product and customer tables. products table with product_id as an auto increment and customer with custumer_id I want to join through the sales tables all t he fields of row 1 from tables customer and products and pull it at once. This is a member login script and I want to display the products by members. So far I have this query to display the products once the member is login in. $userid is the id of the customer coming from the $userid= $_SESSION['customer_id']; $mysqlSales="SELECT products.* FROM products JOIN sales ON (products.product_id = procuct_id ) WHERE sales.customer_id = '$userid'"; so far that statement is not working where should I have some type of incoherance with the english statement above expressing what I want the query to do. Hi, I have a problem. i want to show the current date and time in my tables of the database. I want to show it because i want to show it on my report generated in php. By the way i do not have a "date" field in my form. Can anyone help me out? Thanks, Heshan. Hello,
I want to show checkbox is checked when there is entry of that id in a table in my database.
I have 2 tables page and access_level. I am getting data from page table and displays it in <ul><li> tag with checkbox to select all or only few. After selecting the checkbox, i will store only selected checkbox value in access_level table along with table id. Page link and page name details will be stored in page table.
Now if i want to edit , i should display all the pages which is there in page table and i should also mark checked to those which are already stored in access_level table.
I am using LEFT OUT JOIN, It displays all the pages. But it is not displaying the check mark to those which are already selected.
Here is my code
<?php $s1 = mysql_query("SELECT pages.page_id, pages.code, pages.page, pages.href, access_level.aid, access_level.page_id as pgid, access_level.department, access_level.position, access_level.active FROM pages LEFT OUTER JOIN access_level ON pages.page_id=access_level.page_id WHERE access_level.department=".$department." AND access_level.position=".$position." AND pages.code='sn'") or die(mysql_error());
while($s2 = mysql_fetch_array($s1)) { ?>
<tr><td><li><?php echo $s2['page']; ?> </td><td><input type="checkbox" name="sn[]" value="<?php echo $s2['page_id']; ?>"
<?php if($row['pgid'] === $s2['page_id']) echo 'checked="checked"';?> />
here is my pages table
pages.JPG 26.55KB
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access_level
access_level.JPG 19.09KB
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In access_level table i do not have page ids 8 and 9. But i want to display that also from pages table and for 1 to 7 and 10 i should display check mark.
How i can achieve this? Please Help
This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=308347.0 |
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