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PHP / MySQL select data and split on pagesThis tutorial is going to show you how to SELECT data from a MySQL database, split it on multiple pages and display it using page numbers. We have MySQL table called “students” holding 90 records with the following fields: $sql = "SELECT * FROM students ORDER BY name ASC LIMIT 0, 20"; returns 20 records sorted by name starting from the first record. This next query $sql = "SELECT * FROM students ORDER BY name ASC LIMIT 50, 20"; shows 20 records sorted again by name but this time it will start from the 50th record. Next thing to do is to make a PHP file called pagination.php which will show the first 20 records from our table. The code below selects and then prints the data in a table. <?php if (isset($_GET["page"])) { $page = $_GET["page"]; } else { $page=1; }; $start_from = ($page-1) * 20; $sql = "SELECT * FROM students ORDER BY name ASC LIMIT $start_from, 20"; $rs_result = mysql_query ($sql, $connection); ?> <table> <tr><td>Name</td><td>Phone</td></tr> <?php while ($row = mysql_fetch_assoc($rs_result)) { ?> <tr> <td><? echo $row["Name"]; ?></td> <td><? echo $row["PhoneNumber"]; ?></td> </tr> <?php }; ?> </table> Now, when you open pagination.php in your web browser you will see table showing the first 20 records from your ‘students’ table. The first 2 lines of the above code if (isset($_GET["page"])) { $page = $_GET["page"]; } else { $page=1; }; $start_from = ($page-1) * 20; are used to create a $start_from variable depending on the page that we want to view. Later you will see that we will pass a “page” value using the URL (e.g. pagination.php?page=2) to go to different pages. Next we need to find out the total amount of records in our table and the number of pages that we will need. To do this we run another query using COUNT() function. $sql = "SELECT COUNT(Name) FROM students"; $rs_result = mysql_query($sql,$connection); $row = mysql_fetch_row($rs_result); $total_records = $row[0]; The $total_records is now equal to the number of records that we have in our database, in our case 90. We have 20 records per page so the total number of pages that will be needed is 5 (4 pages with 20 records and last page will have 10 records). Calculating the amount of pages needed using PHP can be done using ceil() function. $total_pages = ceil($total_records / 20);We divide the total number of records by records per page and then the ceil() function will round up the result. Now we have 2 new variables - $total_records equal to 90 and $total_pages equal to 5. To print page numbers and associate URLs to each number we will use for() cycle. <?php for ($i=1; $i<=$total_pages; $i++) { echo "<a href='pagination.php?page=".$i."'>".$i."</a> "; }; ?> Above code will print numbers from 1 to 5 and for each number will create different link. At the end you should have a file like this (remember to add the MySQL connection string): <?php if (isset($_GET["page"])) { $page = $_GET["page"]; } else { $page=1; }; $start_from = ($page-1) * 20; $sql = "SELECT * FROM students ORDER BY name ASC LIMIT $start_from, 20"; $rs_result = mysql_query ($sql,$connection); ?> <table> <tr><td>Name</td><td>Phone</td></tr> <?php while ($row = mysql_fetch_assoc($rs_result)) { ?> <tr> <td><? echo $row["Name"]; ?></td> <td><? echo $row["PhoneNumber"]; ?></td> </tr> <?php }; ?> </table> <?php $sql = "SELECT COUNT(Name) FROM students"; $rs_result = mysql_query($sql,$connection); $row = mysql_fetch_row($rs_result); $total_records = $row[0]; $total_pages = ceil($total_records / 20); for ($i=1; $i<=$total_pages; $i++) { echo "<a href='pagination.php?page=".$i."'>".$i."</a> "; }; ?> This pagination.php file will print a table with maximum 20 records per page and at the bottom 5 page numbers each pointing to a page showing different 20 records. Do not forget that for a small fee I can add pagination to all your PHP files. Let me know if you need help with this and I will give you a quote. COMMENTS
Moses I need to create a database linked to a PHP Calendar which allows maximum of 2 bookings between 9 am and 3pm and will check if date is available before accepting booking. Many thanks ioppo good article, thanks Jason Great tutorial, I found it better to do it this way. Tutorials ------------- $sql = "SELECT COUNT(Name) FROM students"; $rs_result = mysql_query($sql,$connection); $row = mysql_fetch_row($rs_result); $total_records = $row[0]; $total_pages = ceil($total_records / 20); ------------- Mine ------------- $table = mysql_query("SELECT * FROM students"); $total_records = mysql_num_rows($table); $total_pages = ceil($total_records / 20); ------------- ANSWER FOR ONLY 1 PAGE WORKING Those who had problems with 2nd and other pages not working, but showing the first page only: I had the same issue and for me it was about this line: $sql = "SELECT * FROM students ORDER BY name ASC LIMIT $start_from, 20"; i had it this way $sql = "SELECT * FROM students ORDER BY name ASC LIMIT 0, 20"; So "$start_from" was missing completely. It was because this tutorial started with different code and it changed later on. Hope this help. Compare codes if you have some issues. This is working completely now. All pages. ;) Sambath Hi all clever..! I am a beginner in PHP. I want help that: I have 100 data in Mysql database. when I displayed all and I want to make a link to views each individual record in detail page. How do I code? 1. code to link from a row. 2. cod to get each record from that link. Thank for help... please help me.. Ricardo Thank you!!! I've been searching for this code for a long time! But i do have one question... How do i select the latest records in the tabel? maryjeykle nice! very useful. thanks for sharing! Flynn Excellent post, very useful and easy to follow! custom php mysql programmer Guys, thanks a lot... your code helps lot for my code... shak Warning: mysql_query() expects parameter 2 to be resource, null given in C:\xampp\htdocs\db\1\showraw1234.php on line 18 Warning: mysql_fetch_assoc() expects parameter 1 to be resource, null given in C:\xampp\htdocs\db\1\showraw1234.php on line 23 Name Phone Warning: mysql_query() expects parameter 2 to be resource, null given in C:\xampp\htdocs\db\1\showraw1234.php on line 35 Warning: mysql_fetch_row() expects parameter 1 to be resource, null given in C:\xampp\htdocs\db\1\showraw1234.php on line 36 POST A COMMENT
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