Query not working - INSERT INTO - php

Question

So i recently starting learning PHP, and now im trying to get code into my database.
The data is input from the user through a form.
here is my code:
if(isset($_POST['submit'])) {
$blogtitle = $_POST['blogTitle'];
$blogcategory = $_POST['blogCategory'];
$blogcontent = $_POST['blogContent'];
// aanmaak date van de blog
$blogdate = date("d/m.Y");
// Checkt of alle velden zijn ingevuld
if (!empty($blogtitle) && !empty($blogcategory) && !empty($blogcontent)) {
//echo "je zit nu bij de query";
$addBlogQuery = mysql_query("INSERT INTO blog (blog_ID, blog_title, blog_category, blog_content, blog_date)
VALUES (NULL, $blogtitle, $blogcategory, $blogcontent, blogdate )");
if ($addBlogQuery) {
echo "blog added successfully";
}
else {
echo "something went wrong";
}
}
else {
$this->notFilledErrorAction();
}
}
For some reason it's not adding any data my database. My connection to my database is working properly, and i dont see a mistake in my query.
Does someone see an error in this code? or could help me figure out the problem?

Answer 1

You have a syntax error in your INSERT statement:
$addBlogQuery = mysql_query("INSERT INTO blog (blog_ID, blog_title, blog_category, blog_content, blog_date)
VALUES (NULL, $blogtitle, $blogcategory, $blogcontent, blogdate )");
You are missing a $ here------------------------^
You also need to wrap your variables in ' single quotes:
$addBlogQuery = mysql_query("INSERT INTO blog (blog_ID, blog_title, blog_category, blog_content, blog_date)
VALUES (NULL, '$blogtitle', '$blogcategory', '$blogcontent', '$blogdate' )");
Furthermore, the mysql_* API is now deprecated. Please read the big red box here. You should start using MySQLi or PDO now whilst it is still relatively easy to change.

Answer 2

You need to enclose strings ,date and DATETIME values with single quotes (').
And you have not enclosed in your SQL.
Please modify your SQL as:
$addBlogQuery = mysql_query("INSERT INTO blog (blog_ID, blog_title, blog_category, blog_content, blog_date)
VALUES (NULL, '$blogtitle', '$blogcategory', '$blogcontent', 'blogdate' )");

Answer 3

Mysql support only following date format:
YYYY-mm-dd
but your code has different format
$blogdate = date("d/m.Y");
Try the following:
$blogdate = date("Y-m-d");
And you have passed ID null, i think you have selected ID as primary key. primary key can not be null. if your ID field support auto-increment you don't need to pass anything.
$addBlogQuery = mysql_query("INSERT INTO blog (blog_title, blog_category, blog_content, blog_date)
VALUES ($blogtitle, $blogcategory, $blogcontent, blogdate )");
Hopefully it will work.

Answer 4

See the changes below and try again:
$blogdate = date("Y-m-d");
$addBlogQuery = mysql_query("INSERT INTO `blog` (`blog_title`, `blog_category`, `blog_content`, `blog_date`)
VALUES ('$blogtitle', '$blogcategory', '$blogcontent', '$blogdate' )");

Answer 5

Use more secure way , Use PDO - Stop using MYSQL_* it's deprecated
PDO escapes itself, you doesn't need to use mysql_real_escape_string
<?php
$user="root";
$pass="";
$db = new PDO('mysql:host=hostname;dbname=databasename', $user, $pass); //establish new connection
$sql ="INSERT INTO blog (blog_ID, blog_title, blog_category, blog_content, blog_date)
VALUES (NULL, ?, ?, ?, ?)";
try{
$stmt = $db->prepare($sql);
$stmt->execute(array($a, $b, $c, $d));
if($stmt->rowCount()>0){
//done
}
}catch(PDOException $e){
echo $e->getMessage();
?>
Before inserting into database you should sanitize you data to prevent SQL injection and XSS. Use this function:
function sanitize($data){
$data= htmlentities(strip_tags(trim($data)));
return $data;
}

Answer 6

Try following query there is no need to use php date function mysql has native support to date and time functions
INSERT INTO blog
(`title`, `category`, `content`, `date`, `id`)
VALUES
('Title here', 'category here','blog content here', NOW(), 1);
here is SQL test SQL Fiddle

Answer 7

$addBlogQuery = mysql_query("INSERT INTO blog (blog_title, blog_category, blog_content, blog_date) VALUES ($blogtitle, $blogcategory, $blogcontent, blogdate )");
It seems your mistake is insert the value of blog_id by NULL. blog_id column is the primary key. If you insert blog_id by NULL, then the data can't push to your database. Since blog_id is primary key, you don't need to insert blog_id manually. It will automatically inserted.