Php, sqli problem to insert data

Question

0.00/5 (No votes)

See more:

Hello guys,
I am new to PHP and sqli, i write this code to insert data into my database.

Connection establish successful but record are not able to insert.
Kindly help to find error in the code.
thanks

What I have tried:

PHP

<?php
include("connect.php");
if (isset($_POST['submit'])){
	$title=$_POST['p_title'];
	$date=date('d-m-y');
	$name=$_POST['p_name'];
	$country=$_POST['p_country'];
	$contactno=$_POST['p_contactno'];
	$email=$_POST['p_email'];
	$image=$_FILES['p_image']['name'];
	$image_type=$_FILES['p_image']['type'];
	$image_size=$_FILES['p_image']['size'];
	$image_tmp=$_FILES['p_image']['tmp_name'];
	$content=$_POST['p_content'];
	
	if($title == '' or $name==''){
		echo " alert('All Field Required')";
		exit();
	}
	if($image_type == "image/jpeg" or $image_type == "image/png"
	or $image_type == "image/gif"){
		if($image_size<=500000){
			move_uploaded_file($image_tmp,"images/$image_name");
		}else{
			echo " alert ('Image size is larger')";
		}
}else {
	echo "alert ('Image type is invalid')";
}
	$query = "INSERT INTO testimonial (post_title, post_date, post_author, post_image, post_content, post_country, post_phno, post_email) 
	valuse ('$title','$date','$name','$image','$content','$country','$contactno','$email')";
	
	$run = mysqli_query($con, $query);
	if($run){
		echo " alert ('Record Added') ";
		header ('refresh:3; url: blog.php');
	}else{
		echo " alert ('Data Not Sent') ";
		header ('refresh:3; location: localhost/insert_post.php');
	}
}
?>

Posted 28-Jun-18 1:22am

Member 13890780

Updated 2-Jul-18 5:21am

Richard MacCutchan

v2

Add a Solution

Comments

Richard MacCutchan 28-Jun-18 8:03am

What happens when you run your code?

2 solutions

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Patrice T · Answer 1 · 2018-06-28T01:36:00

PHP

$query = "INSERT INTO testimonial (post_title, post_date, post_author, post_image, post_content, post_country, post_phno, post_email)
valuse ('$title','$date','$name','$image','$content','$country','$contactno','$email')";

Not a solution to your question, but another problem you have.
Never build an SQL query by concatenating strings. Sooner or later, you will do it with user inputs, and this opens door to a vulnerability named "SQL injection", it is dangerous for your database and error prone.
A single quote in a name and your program crash. If a user input a name like "Brian O'Conner" can crash your app, it is an SQL injection vulnerability, and the crash is the least of the problems, a malicious user input and it is promoted to SQL commands with all credentials.
SQL injection - Wikipedia[^]
SQL Injection[^]
SQL Injection Attacks by Example[^]
PHP: SQL Injection - Manual[^]
SQL Injection Prevention Cheat Sheet - OWASP[^]
How can I explain SQL injection without technical jargon? - Information Security Stack Exchange[^]

muhammad sufiyan · Answer 2 · 2018-07-02T05:21:00

TRY THIS CODE

<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$sql = "INSERT INTO MyGuests (firstname, lastname, email)
VALUES ('aaa', 'bbb', 'abc@example.com')";

if ($conn->query($sql) === TRUE) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();
?>

Php, sqli problem to insert data

2 solutions

Solution 1

Solution 2

Add your solution here

Preview 0