php mysql查詢返回多行，將數據保存到對象

Question

我試圖從返回多行的mysql查詢中提取數據。 該代碼當前已編碼為僅處理一行。 我需要怎么做才能將多行數據保存到對象Date的實例變量中？

<?php
session_start();
require 'database.php';

class Date {
    private $id = '';
    private $titl = '';
    private $tim = '';
    private $dat = '';
    private $usern = '';

}

if ($_SESSION['loginstatus']!=1){
    die;
}

$username=$_SESSION['username'];
$date=$_POST['dateChosen'];

$stmt = $mysqli ->prepare("select ID, title, time from events where date=? and creator=?");    
if (!$stmt){
    $error = $mysqli->error;
    $string="Query Prep Failed:" . $error; 
    echo json_encode(array(
        "message"=> $string));
    exit;
}


$stmt -> bind_param('ss',$date,$username);
$stmt -> execute();
$stmt ->bind_result($ID, $title,$time);
$count = 0;
while ($stmt->fetch()){
    $count += 1;

    echo json_encode(array("id"=>$ID,
        "title"=> $title, "time"=>$time));
}
if ($count == 0){
    echo json_encode(array(
        "title"=> "NOT SET", "time"=>"NOT SET"));
}
$stmt->close();
header("Content-Type: application/json");
?>

Answer 1

您正在創建多個單獨的/獨立的JSON字符串，這是不正確的。 您需要在循環中構建一個數組，然后在循環完成后對數組ONCE進行編碼。

例如

$data = array();
while($stmt->fetch()) {
   $data[] = array('id' => $ID, 'title' => etc....);
}
echo json_encode($data);

現在你正在生產

{"id":1,"title":"foo",...}{"id":2,"title":"bar",...}{etc...}

這是語法錯誤。 它應該是

[{"id":1,"title":"foo",...},{"id":2,"title":"bar",...},{etc...}]

Answer 2

更改while循環：

while($res = $stmt->fetch())
{
    //do whatever, data is in the $res variable
}