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我想验证一个持有 json 的变量只有一个级别的简单格式。有人可以提供一个仅验证一个级别而不是多级验证的示例吗？

   <?php
$json = '{"a":1,"b":2,"c":3,"d":4,"e":5}';
isValidOnlayer($json); // True
$json = '{
    "id": "0001",
    "type": "donut",
    "name": "Cake",
    "ppu": 0.55,
    "batters":
        {
            "batter":
                [
                    { "id": "1001", "type": "Regular" },
                    { "id": "1002", "type": "Chocolate" },
                    { "id": "1003", "type": "Blueberry" },
                    { "id": "1004", "type": "Devil's Food" }
                ]
        },
    "topping":
        [
            { "id": "5001", "type": "None" },
            { "id": "5002", "type": "Glazed" },
            { "id": "5005", "type": "Sugar" },
            { "id": "5007", "type": "Powdered Sugar" },
            { "id": "5006", "type": "Chocolate with Sprinkles" },
            { "id": "5003", "type": "Chocolate" },
            { "id": "5004", "type": "Maple" }
        ]
    }'
    isValidOnlayer($json); // False
?>
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2 回答 2

2

您可以将其用作您的功能的结果：

count(array_filter(json_decode($json, 1), 'is_array'))
于 2013-08-06T13:10:33.790 回答
0 
$tmp = json_decode($json, true, 2); // depth=2, the array "itself" is level 1, its elements are level 2

如果 $tmpnull发生错误。如果深度限制是问题json_last_error()将返回JSON_ERROR_DEPTH

于 2013-08-06T13:14:34.787 回答