Undefined variable: php sql

Question

This might be a dumb problem. My php code gives me below error.

Notice: Undefined variable: country in line 19

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

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

if(isset($_POST['country'])){ $country = $_POST['country']; }
if(isset($_POST['name'])){ $name = $_POST['name']; }
if(isset($_POST['tel'])){ $name = $_POST['tel']; }

$sql = "INSERT INTO orders (country, name, tel)
VALUES ($country, $name, $tel)";

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

$conn->close();

?>

Thanks in advance.

don't forget to harden against sql injection. Note that it's far better to do it the first time than to believe, likely wrongly, that you'll get to it before it's discovered. — erik258, Dec 11 '18 at 18:00

score -1 · Answer 1 · answered Dec 11 '18 at 18:01

-1

After isset $_POST[tel] why it is $name not $tel

answered Dec 11 '18 at 18:01

Tommy Rachmadiono

1

hifebriansyah · Accepted Answer · 2018-12-11T19:06:39.800

your question already answered here "Notice: Undefined variable", "Notice: Undefined index", and "Notice: Undefined offset" using PHP

additionally :

no it's not dumb, but to be honest it's a beginner common mistakes

you almost do it this right, by doing

if(isset($_POST['country'])){ $country = $_POST['country']; }

however it will not completely solve your problem. there are two ways to solve it, depend on your preferences.

first solutions is you can save it using default value

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

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

//default values, coz you will used it whetter isset true or not in line 25
$country = '';
$name = '';
$tel = '';

if(isset($_POST['country'])){ $country = $_POST['country']; }
if(isset($_POST['name'])){ $name = $_POST['name']; }
if(isset($_POST['tel'])){ $tel = $_POST['tel']; } //you should use $tel instead $name

// use escaped double quote
$sql = "INSERT INTO orders (country, name, tel)
VALUES (\"$country\", \"$name\", \"$tel\")";

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

$conn->close();

alternatively, the second solution is to only insert if all the data valid

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

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

//only insert if all data valid!
if(isset($_POST['country']) && isset($_POST['name']) && isset($_POST['tel'])) {
    $country = $_POST['country']; }
    $name = $_POST['name']; }
    $tel = $_POST['tel']; } //you should use $tel instead $name

    $sql = "INSERT INTO orders (country, name, tel)
    VALUES (\"$country\", \"$name\", \"$tel\")";

    if ($conn->query($sql) === TRUE) {
        echo "New record created successfully";
    } else {
        echo "Error: " . $sql . "<br>" . $conn->error;
    }
} else {
    echo "Error: data not valid";
}

$conn->close();

hopefully it can help you in some way,

have a nice day!

Hi, thanks for the quick response. I get below error after adding this.**Error: INSERT INTO orders (country, name, tel) VALUES (, , ) You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near ' , )' at line 2** — Chami M, Dec 11 '18 at 18:16
You should use prepared statements, but if not at least put quotes around the string values in the SQL statement. — Nigel Ren, Dec 11 '18 at 18:22

Undefined variable: php sql

2 Answers2