mysql query data insertion through php

Question

<?php
$servername = "localhost";
$username = "root";
$password = "";
$database = "test";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $database);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
echo "Connected successfully";
$value1=$_POST['txtname'];
$value2=$_POST['cellnumber'];
$value3=$_POST['dist'];
$value4=$_POST['specialization'];
$value5=$_POST['membername'];
$value6=$_POST['date'];
$sql = "INSERT INTO students (StudentName, CellNumber, District, Specialization, PromotionMember, Date)
VALUES ('$value1', '$value2', '$value3', '$value4', '$value5', '$value6')";
if (!mysqli_query($sql)) { 
die ('Error: ' . mysql_error());
}
else
{ 
echo ("معلومات ارایه شده شما ثبت شد");
header("Location: register.html");
}
mysqli_close();
?>

connection is successfull but data insertion is getting error on line 23 which is (if (!mysqli_query($sql)) { ) enter image description here

Did you try to find a manual for `mysqli_query`? Here, I did it for you http://php.net/manual/en/mysqli.query.php — u_mulder, Mar 08 '17 at 08:00
Your code is vulnerable to [*SQL injection*](http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php?rq=1). — MrDarkLynx, Mar 08 '17 at 08:00
Stackoverflow is not a programming-service. Did you checked your error-logs? — Oliver, Mar 08 '17 at 08:01

score 1 · Answer 1 · answered Mar 08 '17 at 08:03

1

Your query doesn't run because you are using the MySQLi function wich need 2 parameters to execute your query.

So instead of

mysqli_query($sql)

you have to do:

mysqli_query($conn, $sql)

Your code also looks vulnerable to SQL Injections, so you want to know how to escape the strings in MySQLi. I recommend you to use prepared statements. I hope this will help!

answered Mar 08 '17 at 08:03

node_modules

4,790
6
21
37

Thank you I make changes and that problem solved, but still its not inserting the data Connected successfully Error: – Ghufran Ataie Mar 08 '17 at 08:17
My error is exactly (Connected Successfully error:) and its on line 23 which is (die ('Error: ' . mysql_error());) – Ghufran Ataie Mar 08 '17 at 08:24
@GhufranAtaie Change `mysql_error()` to `mysqli_error($conn)` – node_modules Mar 08 '17 at 09:11

score 0 · Answer 2 · answered Mar 08 '17 at 08:49

Try this code...

$sql = "INSERT INTO students (StudentName,CellNumber, District, Specialization, PromotionMember,      Date)VALUES ('$value1', '$value2', '$value3', '$value4', '$value5', '$value6')";
$q=mysqli_query($conn,$sql);
if (!$q) { 
die ('Error: ' . mysqli_error($conn));
}else{header("Location: register.html");}

score 0 · Answer 3 · answered Jul 13 '17 at 13:50

0

Try This code..

$sql = "INSERT INTO students (StudentName, CellNumber, District, Specialization, PromotionMember, Date)
VALUES ('$value1', '$value2', '$value3', '$value4', '$value5', '$value6')";
$qw=mysqli_query($conn,$sql);

if($qw)

{

header("Location: register.html");

}

answered Jul 13 '17 at 13:50

Ragul N

9
2

1

Why should the OP "try this code"? A **good answer** will always have an explanation of what was done and why it was done in such a manner, not only for the OP but for future visitors to SO. – B001ᛦ Jul 13 '17 at 14:27

mysql query data insertion through php

3 Answers3