Create a Simple API In PHP with MySQL Database

Great tutorial. Simple and ez to understand. Many tutorial out there are using classes. It make me confused 😵

This code gives me an Internal Server 500 Error so I updated the code if anyone getting same error use my code - <?php
$con = mysqli_connect("localhost", "root", "", "{database_name_isert_here}");
$response = array();

if ($con) {
$sql = "SELECT * FROM {table_name_insert_here}";
$result = mysqli_query($con, $sql);

if ($result) {
// Set the correct Content-Type header
header("Content-Type: application/json");

$rows = mysqli_fetch_all($result, MYSQLI_ASSOC);
$response['data'] = $rows; // Store the data in an associative array
echo json_encode($response, JSON_PRETTY_PRINT);
} else {
// Handle query execution error
$errorResponse = array('error' => 'Query execution failed');
echo json_encode($errorResponse);
}

// Close the database connection
mysqli_close($con);
} else {
// Handle database connection error
$errorResponse = array('error' => 'Database connection failed');
echo json_encode($errorResponse);
}
?>

please make a video on information about MySql, Sqlite what is the need and difference between
and what is php plugins?

where is the code

how to do a post rquest?

do you have an actual tutorial that connects to a thrid party api

aahhh thanks you guyz completely wrecked my 2 hour , in the line no 5 there `data` should bein back thick 😤😤😤😤

thank you bro

Bro this is not a API its just a Database query

This is good but how to protect of other people, every guy who know how to use fetch in javascript can get data from db using this method. This isnt secured bro.

Good Tutorial Bro.

Is there a way that instead of calling all the data. We can only get one we required. Creating an API that somehow asks to enter the name and will only return data containing that name

Thanks

This is not an api... lots of paid commenters.

Very informative & to the point
Helped me a lot in making an API that I want to use with ReactJS

Is this API???? you've just connected mySql. Api is different :)

mslqi code

<?php
$mysqli = new mysqli("localhost","root","","API_DATA");
$response = array();
if ($mysqli->ping())
{
$db = "select * from data";
$result = mysqli_query($mysqli,$db);
if($result){
header("Content-Type: JSON");
$i=0;
while($row = mysqli_fetch_assoc($result)){
$response[$i]['id'] = $row['id'];
$response[$i]['name'] = $row['name'];
$response[$i]['age'] = $row['age'];
$response[$i]['email'] = $row['email'];

$i++;
}
echo json_encode($response,JSON_PRETTY_PRINT);
}
}
?>

I love you

Really Easy to understand, good work

when I run the file, it says "Not found". why?

Thanks dear sir

Thanks for your tutorial very easy to understand.