There is a syntex error in line 12.how to solve it

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PHP

<?php
include_once('connection.php');
if(isset($_POST['submit']))
{
$name= $_POST['name'];
$email=$_POST['email'];
$add=$_POST['address'];
$adm=$_POST['admission'];
if(mysql_query("insert into student_record(name,email,address,joining_date) VALUES

('$name','$email','$add','$adm',)")
{
	echo "values has been submitted";
}
}
?>


<!DOCTYPE html>

<title>niraj php

<p>Name:<br></p>
<p>Email:<br></p>
<p>Address : <br> </p>
<p>Admission Date: <br></p>
<p>
</p>

What I have tried:

I have tried to remove the curly bracket but it didn't work.

Posted 28-Aug-17 22:09pm

niraj kr

Updated 28-Aug-17 23:06pm

Graeme_Grant

v2

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Thomas Daniels · Answer 1 · 2017-08-28T22:14:00

The problem is:

PHP

if(mysql_query("insert into student_record(name,email,address,joining_date) VALUES ('$name','$email','$add','$adm',)")
{

You have one ) after your string, and that closes the ( after mysql_query. However, your ( of the if statement is still open, so you need another ) to close that as well:

PHP

if(mysql_query("insert into student_record(name,email,address,joining_date) VALUES ('$name','$email','$add','$adm',)"))
{

That aside, I'm not sure if that comma after '$adm' is supposed to be there; I think you have to remove that as well (although that would be a MySQL syntax error, not a PHP one)

Also, this way of executing a query is very bad. You are simply putting all POST values into your query; you have an SQL Injection[^] vulnerability. Learn about parameterized queries to overcome this.

Patrice T · Answer 2 · 2017-08-28T23:06:00

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[^]

There is a syntex error in line 12.how to solve it

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