Aug 1, 2012

Php Mysql blog tutorial, creating a simple and secured blog

php mysql blog

On today's tutorial we will talk about creating a blog on php and mysql

Table of contest:
1. Connecting to mysql database
2. Creating mysql tables
3. Coding the blog

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Connecting to mysql database is simple, in just few lines of code

Before we start i suggest you read Php Mysql posting to database for better understanding of mysql.

<?php
// connecting to host
mysql_connect("host", "user", "password"); // replace this with your data
mysql_select_db("your_database"); // database name here
?>


Creating database, tables and columns
Now let's build the mysql table.
First build a database if you haven't, and dont forget to edit the connection details at the top.
Click on databases
php mysql blog
Set a name to it and click create.
Now we must build the tables:
mysql create tables
Click 'go', a new table will popup, complete those fields like in the pocture below:
Scroll down and click save.
Last one, click on posts and insert tab at top, write something to title, and id live empty.

Now lets continue coding the blog

<?php
// connecting to host
mysql_connect("host", "user", "password"); // replace this with your data
mysql_select_db("your_database"); // database name here

// lets build a blog
if(isset($_GET['post'])){
$post = $_GET['post'];

// first of all we will check string length
if(strlen($post) > 11){ // if post id is bigger than 11 charachters
die('Blog post nof found.');
}
// now we will make sure that the post id is numeric and this is a nice security method
if(is_numeric($post)){ // is numeric allows numbers only
$post = (int)$post; // and the int function, which replace every
// string to its correspoing number

// for the tutorial im gonna add mysql_real_escape_string
// but is not really needed in this case
$post = mysql_real_escape_string($post); // final sqli defense

// final part
$query = mysql_query("SELECT title FROM posts WHERE id=$post LIMIT 1");
while($row = mysql_fetch_array($query)){
echo $row['title'];
}

}
else{ // if post is not numeric then
die('Blog post nof found.'); // post does not exist
}

}
else{
// if post is not submitted display them all
$query = mysql_query("SELECT title, id FROM posts");
while($row = mysql_fetch_array($query)){
echo $row['title'];
echo $row['id'];
echo '<a href="?post='.$id.'">'.$title.'</a><br>';
}
}
?>
Few last tips!

I used LIMIT 1 on the first mysql query.
That's done for 2 main pruposes:
1. we are only displaying 1 result
2. it's a nice speed optinization for mysql
To display blog posts by newest ones replace


SELECT title, id FROM posts
with
SELECT title, id FROM posts ORDER BY id DESC
It's a simple blog who does the work.
If you want to learn more about sql injection we suggest Sql Injection prevention tutorial

79 comments:

  1. Excellent blog very nice and unique information related to Php Mysql blog tutorial. Thanks for sharing this information.

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  2. Was having some trouble understand how to pull the information from the DB and show its result, but after reading your tutorial i have a complete understanding of it now, great commenting. Thanks!

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  3. Excelente el blog tanto la info como el diseño y la estetica del mismo, lo estoy poniendo entre mis favoritos, saludos cordiales.-

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  4. Hi there. PHP mysql is the growing trend now. It is a quite nice post PHP professionals. Thanks for sharing it.

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  5. Awesome post for Freshers and PHP professionals those who are in search of job or someone working in Non IT sector and desire to extend their skill sets to be a PHP Web Developer.
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  6. wow now i understood it very well. thank you

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  7. This comment has been removed by a blog administrator.

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  8. I am really glad I've found this information. Nowadays bloggers publish only about gossips and net and this is really irritating. A good blog with interesting content, this is what I need. Thank you

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  9. HI, for some reason i get the following error. Can you help please|
    ( ! ) Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\wampXXXXXXXXXXXXXXXXXXXX (deleted sorry!)
    Call Stack
    # Time Memory Function Location
    1 0.0015 371600 {main}( ) ..\newblog08052013.php:0
    2 0.0364 388776 mysql_fetch_array ( ) ..\newblog08052013.php:37

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  10. that was great tnx very much :)))))))))

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  11. Thanks a lot.Through this post I learned lot of new things to create my new blog.

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  15. pls provide...platforms for others to blog also

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  16. Nice post. I usually use PDO to avoid SQL injection.

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  19. Nice blog post! But if you are newbie to mysql then you should learn Basics of MySQL and PHP communication in detail. Thank you.

    Kind Regards,
    Sahil Sharma
    www.webdevelopersahil.com

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  20. hi i have the actual problem, when i click to a post , give me this error Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/CCCCC/public_html/index.php on line 84 Line 84 is // final part
    $query = mysql_query("SELECT title, FROM posts WHERE id=$post LIMIT 1");
    while($row = mysql_fetch_array($query)){
    echo $row['title'];
    }
    please help!

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    with regards,
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