CODEWITHSUNDEEP
phpmysqljson
user3493797
user3493797

Reputation: 63

Return MySQL SELECT as JSON

I have this php code to build the result of my MYSQL into JSON:

<?php
$mysql_db_hostname = "localhost";
$mysql_db_user = "root";
$mysql_db_password = "";
$mysql_db_database = "essensplan";


$con = @mysqli_connect($mysql_db_hostname, $mysql_db_user, $mysql_db_password, $mysql_db_database);

if (!$con) {
 trigger_error('Could not connect to MySQL: ' . mysqli_connect_error());
}
$var = array();
 $sql = "SELECT * FROM beispiel";
$result = mysqli_query($con, $sql);

while($obj = mysqli_fetch_object($result)) { $var[] = $obj; }

echo '{"users":'.json_encode($var).'}';
?>

The PHP script work on my Apache and the PHP Version is 5.5.11. But when I call the script, there only shows {users:} and the json data from the mysql data base is missed. Have anybody an idea?

EDIT: I found a solution. After connect to the database set this: 

mysqli_set_charset($con, 'utf8');

Upvotes: 0

Views: 296

Answers (1)

Gabor
Gabor

Reputation: 560

You should modify the part after you've checked for the connection error as so:

$var = array();
$sql = "SELECT * FROM beispiel";
$result = mysqli_query($con, $sql);

while($obj = mysqli_fetch_object($result)) { 
    $var['users'][] = $obj; 
}

echo json_encode($var);

If for example your returned results are arrays as so

array(
    'name' => 'person1',
    'age' =>  25
),
array(
    'name' => 'person2',
    'age' => 30
),
array(
    'name' => 'person3',
    'age' => 22
)

Then your returned JSON response will be as so:

{"user":[{"name":"person1","age":25},{"name":"person2","age":30},{"name":"person3","age":22}]}

Hope that solved your problem, if any modifications are required, please provide more info.

Upvotes: 1

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