Undefined variables at insert into value

Question

0.00/5 (No votes)

See more:

How to fix that? i have checked all spelling and all same.

What I have tried:

<?php

require 'db.php';

if(isset($_POST['submit'])){





$DepName = $_POST['DepName'];
$Address = $_POST['Address'];
$PIC = $_POST['PIC'];
$Contact = $_POST['Contact'];
$RecordVisit = $_POST['RecordVisit'];
$Outlet = $_POST['Outlet'];
$JobTitle = $_POST['JobTitle'];
$Sales = $_POST['Sales'];
$JobDate = $_POST['JobDate'];
$Note = $_POST['Note'];
}

$result = $mysqli->query("SELECT * FROM villadata WHERE VillaID='VillaID'") or die($mysqli->error());

if ( $result->num_rows > 0 ) {
    
    $_SESSION['message'] = 'User already exists! Please try other username!';
    echo "User exist";
}

else{
$sql="INSERT INTO villadata(DepName,Address,PIC,Contact,RecordVisit,Outlet,JobTitle,Sales,JobDate,Note)
VALUES('$DepName','$Address','$PIC','$Contact','$RecordVisit','$Outlet','$JobTitle','$Sales','$JobDate','$Note')";

 if ( $mysqli->query($sql) ){

        $_SESSION['logged_in'] = true; 
        header("location: tables-data.php");

    }

  else {
        $_SESSION['message'] = 'Registration failed!';
        echo "Registration failed.";
    }

}
?>

Posted 14-Jan-19 17:50pm

Member 14118137

Updated 18-Jan-19 10:54am

Add a Solution

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)

OriginalGriff · Answer 1 · 2019-01-14T20:06:00

Don't do it like that! Never concatenate strings to build a SQL command. It leaves you wide open to accidental or deliberate SQL Injection attack which can destroy your entire database. Always use Parameterized queries instead.

When you concatenate strings, you cause problems because SQL receives commands like:

SQL

SELECT * FROM MyTable WHERE StreetAddress = 'Baker's Wood'

The quote the user added terminates the string as far as SQL is concerned and you get problems. But it could be worse. If I come along and type this instead: "x';DROP TABLE MyTable;--" Then SQL receives a very different command:

SQL

SELECT * FROM MyTable WHERE StreetAddress = 'x';DROP TABLE MyTable;--'

Which SQL sees as three separate commands:

SQL

SELECT * FROM MyTable WHERE StreetAddress = 'x';

A perfectly valid SELECT

SQL

DROP TABLE MyTable;

A perfectly valid "delete the table" command

SQL

--'

And everything else is a comment.
So it does: selects any matching rows, deletes the table from the DB, and ignores anything else.

So ALWAYS use parameterized queries! Or be prepared to restore your DB from backup frequently. You do take backups regularly, don't you?

Once you have gone through your whole app fixing this, you may well find the problem you noticed has gone away ...

Patrice T · Answer 2 · 2019-01-18T10:54:00

PHP

$sql="INSERT INTO villadata(DepName,Address,PIC,Contact,RecordVisit,Outlet,JobTitle,Sales,JobDate,Note)
VALUES('$DepName','$Address','$PIC','$Contact','$RecordVisit','$Outlet','$JobTitle','$Sales','$JobDate','$Note')";

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[^]

Undefined variables at insert into value

2 solutions

Solution 1

Solution 2

Add your solution here

Preview 0