having issues on inserting data into database from table in php

Question

Having issues on inserting data into database from table in php, shows data inserted successfully but data not shown on phpmyadmin. I'm using wamp server on this project . please help me overcome this problem guys....

CODE:-

<?php  
$host = 'localhost:3306';  
$user = 'root';  
$pass = '';  
$conn = mysqli_connect($host, $user, $pass);    
?>  
<html>
<head></head>
<body>
<center>
<form method="POST">
<label> Firstname </label> 
            <input type="text" name="name" placeholder= "Name" size="15" required /> </br></br>
<label> Salary </label> 
            <input type="number" name="salary" placeholder= "salary" size="15" required /> </br></br>
            <center>
                        <button type="submit"  name="submit" id="submit" class="submit"><b>Submit</b></button>
                </center>

</form>
</center>


<?php
if(isset($_POST['submit']))
{

$name=$_POST['name'];
{
echo $name;
}
$salary=$_POST['salary'];
{
    echo $salary;
}
$sql = 'INSERT INTO emp5(name,emp_salary) VALUES ("$name", "$salary")';
    

}

?>
</body>
</html>

I dont see, where your SQL query actually gets executed. Like with mysqli_query. — Daniel Heinrich, Mar 24 '21 at 06:00

score 1 · Answer 1 · answered Mar 24 '21 at 06:29

First you need to add one more line in top of the page like :

<?php
    $host = 'localhost:3306';
    $user = 'root';
    $pass = '';
    $database_name = <Your database name>;
    $conn = mysqli_connect($host, $user, $pass, $database_name); 
?>

Then you need to execute the query like :

after the $sql line add this :

$insert_result = mysqli_query($conn, $sql);

score 1 · Answer 2 · answered Mar 24 '21 at 06:38

1

As Daniel pointed out you're missing your query execution, as is you're just saving your query as a variable, and you're missing one parameter in your connection.

I would add the following:

Connection:

$db_name = "your_db";
$conn = mysqli_connect($host, $user, $pass, $db_name);

Query:

// Check connection
if ($conn->connect_error) {
  die("Connection failed: " . $conn->connect_error);
} 
else {
// If connection is succesful, execute query.
mysqli_query($conn, $sql);
}

answered Mar 24 '21 at 06:38

Mackno

11
1

Additionally, avoid using string interpolation to construct SQL queries and prefer using prepared statements and parameter binding instead to avoid SQL injection attacks. (See [discussion](https://stackoverflow.com/questions/8263371/how-can-prepared-statements-protect-from-sql-injection-attacks) and [documentation](https://www.php.net/manual/en/mysqli.prepare.php)). – superhawk610 Mar 24 '21 at 07:26
One should also avoid printing mysql errors as it can potentially leak sensitive data. See [mysqli or die, does it have to die?](https://stackoverflow.com/questions/15318368/mysqli-or-die-does-it-have-to-die). – El_Vanja Mar 24 '21 at 10:18
Oh okay, thanks for the advice, i'll check out those posts. – Mackno Mar 25 '21 at 06:00

having issues on inserting data into database from table in php

2 Answers2