檢索數據從一個特定的SQL命令在PHP

Question

我有一個數據庫與一些溫度數據和日期。爲了繪製數據，我需要將日期時間轉換爲ISO8601。 我發現命令來選擇正確的數據並將其轉換：

SELECT DATE_FORMAT(`datelog`, '%Y-%m-%dT%TZ') AS date_formatted FROM `tempo` ORDER BY id ASC 

在phpMyAdmin數據顯示正確，但是當我運行我的PHP腳本使用相同的命令它給我的數據，它們存儲在數據庫（例如：{「datelog」：「2016-12-18 11:54:11」，「donnee」：「16.937」}）。我該如何改變它？

<?php 
//setting header to json 
header('Content-Type: application/json'); 

//database 
define('DB_HOST', '127.0.0.1'); 
define('DB_USERNAME', 'root'); 
define('DB_PASSWORD', '******'); 
define('DB_NAME', 'tempo'); 

//get connection 
$mysqli = new mysqli(DB_HOST, DB_USERNAME, DB_PASSWORD, DB_NAME); 

if(!$mysqli){ 
    die("Connection failed: " . $mysqli->error); 
} 

//query to get data from the table 
$query = "SELECT DATE_FORMAT(`datelog`, \'%Y-%m-%dT%TZ\') AS date_formatted\n" . "FROM `tempo`\n" . "ORDER BY id ASC "; 

//execute query 
$result = $mysqli->query($query); 

//loop through the returned data 
$data = array(); 
foreach ($result as $row) { 
    $data[] = $row; 
} 

//free memory associated with result 
$result->close(); 

//close connection 
$mysqli->close(); 

//now print the data 
print json_encode($data); 

Answer 1

它可以在東西串聯：

$query = "SELECT DATE_FORMAT(`datelog`, \'%Y-%m-%dT%TZ\') AS date_formatted\n" . "FROM `tempo`\n" . "ORDER BY id ASC "; 

試試這個，如果功能：

$query = "SELECT DATE_FORMAT(`datelog`, \'%Y-%m-%dT%TZ\') AS date_formatted\n . " " . FROM `tempo`\n . " " . ORDER BY id ASC . ";