如何修改PHP代码以输出符合要求的JSON Schema？

目标JSON结构 #

需要输出的JSON结构如下：

{"lots":[{"Id":32932,"Type":"G","x":1,"y":2,"z":3},{"Id":32933,"Type":"R","x":4,"y":5,"z":6}]}

当前错误的PHP代码及输出 #

错误代码 #

$lots = array("lots");

$lots[] = array("Id" => 32932, "Type" => "G", "x" => 1, "y" => 2, "z" => 3);
$lots[] = array("Id" => 32933, "Type" => "R", "x" => 4, "y" => 5, "z" => 6);

$json = json_encode($lots);

echo $json;

错误输出 #

["lots",{"Id":32932,"Type":"G","x":1,"y":2,"z":3},{"Id":32933,"Type":"R","x":4,"y":5,"z":6}]

报错信息 #

ArgumentException: JSON must represent an object type.

解决方案 #

当前代码生成的是索引数组（对应JSON数组），但目标结构是JSON对象——外层以lots为键，对应的值是包含两个元素的数组。修改PHP代码如下：

// 定义lots对应的数组内容
$lotItems = [
    ["Id" => 32932, "Type" => "G", "x" => 1, "y" => 2, "z" => 3],
    ["Id" => 32933, "Type" => "R", "x" => 4, "y" => 5, "z" => 6]
];

// 外层构造关联数组，键为lots
$result = ["lots" => $lotItems];

// 编码并输出JSON
echo json_encode($result);

也可以写成更简洁的形式：

echo json_encode([
    "lots" => [
        ["Id" => 32932, "Type" => "G", "x" => 1, "y" => 2, "z" => 3],
        ["Id" => 32933, "Type" => "R", "x" => 4, "y" => 5, "z" => 6]
    ]
]);

修改后生成的JSON会和目标结构完全一致，满足C# Unity项目的格式要求，解决报错问题。

内容的提问来源于stack exchange，提问作者domagogon