如何从没有索引的mysqli prepare（所有匹配行）中获取json object/array

Question

我花了很多时间试图找出从 mysql 数据库输出 json array/object 的正确方法，如下所示：

expected output 
    {
        "ITEM1": 20,
        "ITEM2": 15,
        "ITEM3": 12
    }

因为每个 key/value 对对应一个 mysql table 行（不需要显示 ID 列）。

id |item| price

这是我当前在 debian 上使用 MYSQLND driver 的代码：

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

$servername = "localhost";
$username = "root";
$password = "pswd";
$dbname = "json";


$conn = new mysqli($servername, $username, $password, $dbname);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}


$itemInitArr = array();
$decharge = 0;
$stmt = $conn->prepare("SELECT item, price FROM stock WHERE total_sold = ?");
$stmt->bind_param("i", $decharge);
$stmt->execute();
$result = $stmt->get_result();


                            while ($itemInit = $result->fetch_assoc())
                            {
                                $itemInitArr[] = array (
                                            $itemInit['item'] => $itemInit['price']
                                    );
                            }
$stmt->close();                         
$final_data = json_encode($itemInitArr, JSON_PRETTY_PRINT); 
echo $final_data;

$conn->close();
?>

输出这个：

[
    {
        "ITEM1": 20
    },
    {
        "ITEM2": 15
    },
    {
        "ITEM3": 12
    }
]

那么，错误在哪里以及如何解决它以匹配预期的输出？

Answer 1

尝试使用

$idx = 1; 
while ($itemInit = $result->fetch_assoc()){
     $itemInitArr['ITEM'.$idx] =  $itemInit['price'];
     $idx++;
}
$final_data = json_encode($itemInitArr, JSON_PRETTY_PRINT);

of如果ITEM 1 ... ITEM是列item的内容

     $itemInitArr[$itemInit['item']] =  $itemInit['price'];

Answer 2

如果您将构建阵列的方式更改为...

$itemInitArr[$itemInit['item']] =  $itemInit['price'];

Answer 3

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

$servername = "localhost";
$username = "root";
$password = "pswd";
$dbname = "json";


$conn = new mysqli($servername, $username, $password, $dbname);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}


$itemInitArr = array();
$decharge = 0;
$stmt = $conn->prepare("SELECT item, price FROM stock WHERE total_sold = ?");
$stmt->bind_param("i", $decharge);
$stmt->execute();
$result = $stmt->get_result();


                            while ($itemInit = $result->fetch_assoc())
                            {
                                $itemInitArr[$itemInit['item']] = $itemInit['price'];

                            }
$stmt->close();                         
$final_data = json_encode($itemInitArr, JSON_PRETTY_PRINT); 
echo $final_data;

$conn->close();
?>

如何从没有索引的mysqli prepare（所有匹配行）中获取json object/array

How to get json object/array from mysqli prepare (all matching rows) without indexes

php

json

mysqli

prepared-statement