MySQLi Select to PHP variable

Question

I wrote a PHP script with a database connection, which appears to work properly. The problem I'm having is in using a SELECT statement to fetch values from my database and assign those values to PHP variables.

This is my table:

If I copy the SQL into PHPMyAdmin, I get the right result, so I trust the query itself.

I get an error, if I write:

echo $result;

What do I do wrong?

This is my full code:

<?php
$servername = "XXXXXX";
$username = "XXXXXX";
$password = "XXXXX";
$dbname = "XXXXXX";


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

$test_code = $_GET['code'];
$product_name = $_GET['product'];


// sql to create table
$sql = "CREATE TABLE IF NOT EXISTS code_ever (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY, 
produkt VARCHAR(50) NOT NULL,
code VARCHAR(30) NOT NULL
)";

if ($conn->query($sql) === TRUE) {

} else {

}

// sql to create table
$sql = "CREATE TABLE IF NOT EXISTS code_scanned (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY, 
produkt VARCHAR(50) NOT NULL,
code VARCHAR(30) NOT NULL
)";

if ($conn->query($sql) === TRUE) {

} else {

}

$query  = "SELECT code FROM code_ever WHERE code = 'XKBKNABJK'";
$result = mysqli_query($conn, $query);

if($result)
{

echo $result;

} 
else
{

echo("Sorry:".mysqli_error($conn));

}

$conn->close();
?>

Dharman · Accepted Answer · 2021-05-10 09:04:24Z

4

You need to fetch the results.

$query = "SELECT code FROM code_ever WHERE code = 'XKBKNABJK'";

$result = mysqli_query($conn, $query);
    
/* fetch associative array */
while ($row = mysqli_fetch_assoc($result)) {
    echo $row["code"];
}

http://php.net/manual/en/mysqli-result.fetch-assoc.php

Aside from fetching the results you should redesign the table. You shouldn't have more than one value per 'table cell'. For instance you have comma separated values for the size/color etc of the product. You should have those as separate attributes in the table or maybe even a separate table.

Example:

edited May 10, 2021 at 9:04

Dharman♦

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answered Dec 23, 2015 at 20:17

pajamas

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1 Comment

Devil089 Over a year ago

Hi Thanks for your answer it works now :). I see syntax like this $configs = $db->QueryFetchArrayAll("SELECT config_name,config_value FROM site_config"); for example. How can i use this. this look much shorter to use a query and fetch data.

Dharman · Accepted Answer · 2021-05-10 09:04:52Z

0

Without fetching record you cant print the result

Use

$result = mysqli_query($con, $query);
while ($final = mysqli_fetch_array($result)) {
    echo $final['code'];
}

edited May 10, 2021 at 9:04

Dharman♦

34k27 gold badges106 silver badges157 bronze badges

answered Dec 23, 2015 at 20:21

devpro

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