使用PHP和MySQL生成JSON格式的表

Question

I am trying to generate a JSON formatted table using data from my database. When doing so, I get the error PHP Fatal error: Call to a member function fetch() on a non-object in G:\\PleskVhosts\\mywebsite.com\\httpdocs\\getData.php on line 23

I'm not sure why it's giving this error as I did something very similar on another page and it works pefectly fine. Here is my code.

<?

$db = new PDO("mysql:host=".$servername.";dbname=".$dbname.";", $username, $password);

$query = $db->query("SELECT 'MAJOR' as 'Major' SUM(IF(MAJOR = 'Computer Science',1,0)) as CS, SUM(IF(MAJOR = 'Computer Information Systems',1,0)) as CIS, SUM(IF(MAJOR = 'Other',1,0)) as Other FROM ages");



$table = array();
$table['cols'] = array(
    array('label' => 'Major', 'type' => 'string'),
    array('label' => 'CS', 'type' => 'number'),
    array('label' => 'CIS', 'type' => 'number'),
    array('label' => 'Other', 'type' => 'number')
);

$rows = array();
while($r = $query->fetch()){
    $temp = array();
    $temp[] = array('v' => $r['Major']);
    $temp[] = array('v' => (int) $r['CS']);
    $temp[] = array('v' => (int) $r['CIS']);
    $temp[] = array('v' => (int) $r['Other']);

    $rows[] = array('c' => $temp);
}

$table['rows'] = $rows;

$jsonTable = json_encode($table);

echo $jsonTable;


?>

Answer 1

If you think it is not a back tick issue, as well as an error reporting issue, then run the below. And change the select back ticks to single quotes. And see what you get.

Use the below for a template for how to setup error reporting and a try/catch block.

Schema

create table basic_info2
(   id int auto_increment primary key,
    n_id int not null,
    name varchar(20) not null,
    date_released date not null,
    genres varchar(20) not null
);

insert basic_info2(n_id,name,date_released,genres) values (1,'a','2015-04-11','b');

PHP

<?php
error_reporting(E_ALL);
ini_set("display_errors", 1);
$servername = "host";
$username = "dbuser";
$password = "pwd";
$database = "dbname";

try {
    //Make your connection handler to your database
    $conn = new PDO("mysql:host=".$servername.";dbname=".$database, $username, $password, array(PDO::ATTR_ERRMODE => PDO::ERRMODE_WARNING));

    $sql = "SELECT `N_ID`, `NAME`, `DATE_RELEASED`, `GENRES` FROM basic_info2 ORDER BY N_ID DESC LIMIT 10";
    $stmt = $conn->prepare($sql);
    //Execute the query
    $stmt->execute();
    $result = $stmt->fetchAll();
    //Fetch the results
    foreach ($result as $row) {
        echo '<p>'.$row['NAME'].'</p>';
    }

} catch(PDOException $e) {
    echo $e->getMessage();
    die();
}
?>

Edit

concerning you comment below, this is what you had

SELECT MAJOR as 'MAJOR'
SUM(IF(MAJOR = 'Computer Science',1,0)) as CS, 
SUM(IF(MAJOR = 'Computer Information Systems',1,0)) as CIS, 
SUM(IF(MAJOR = 'Other',1,0)) as Other 
FROM ages

this is what you need (just a missing comma line 1)

SELECT MAJOR as 'MAJOR',
SUM(IF(MAJOR = 'Computer Science',1,0)) as CS, 
SUM(IF(MAJOR = 'Computer Information Systems',1,0)) as CIS, 
SUM(IF(MAJOR = 'Other',1,0)) as Other 
FROM ages