Question

如何使用PHP创建这样的JSON？我正在尝试转换动态数据并将其保存到MySQL数据库中。

{
"fbd49440-a5a1-48be-b13e-e8efddad3588": {
    "0": {
        "value": "dsfrasdf5464356dfs hdhfg dfgh"
    }
},
"0fc71cea-5609-40a7-a1d2-b78139660f8f": {
    "0": {
        "value": "50"
    }
},
"73936e70-4329-4aba-b47c-42c64ced420c": {
    "0": {
        "file": "\/components\/com_djclassifieds\/images\/item\/25_juliet_ibrahim.jpg",
        "uniqid": "59a352b96773325",
        "title": "",
        "file2": "",
        "overlay_effect": "",
        "caption": "",
        "width": "",
        "height": ""
    },
"ac00b95e-9eeb-4035-bf4a-ff206319b2d6": {
    "0": {
        "value": "members-in-good-standing-2014",
        "text": "",
        "target": "0",
        "custom_title": "",
        "rel": ""
    }
},
"69072346-fe4c-489e-8e2b-5a7d7409fd44": {
    "0": {
        "value": "34"
    }
}
}

我尝试了下面的代码，但它没有给我我想要的结果。

$json = (
"fbd49440-a5a1-48be-b13e-e8efddad3588"=> (
    $array1
),
"0fc71cea-5609-40a7-a1d2-b78139660f8f"=> (
    $array2
),
"73936e70-4329-4aba-b47c-42c64ced420c"=> (
    $array3
    ),
"ac00b95e-9eeb-4035-bf4a-ff206319b2d6"=> (
    $array4
),
"69072346-fe4c-489e-8e2b-5a7d7409fd44"=> (
    $array5
)
)

echo json_encode($json);

如果有人能帮助我，我会很高兴，谢谢你

Answer 1

json_encode将获取多种类型的有效数据，并尝试将它们编码为json表示形式。它确实要求输入有效。

您粘贴的代码有多处语法错误，我们无法分辨您$array1. $array2, $array3, $array4, $array5中的内容。但是，一些更改（并假设您的$数组是实际数组）使您的代码正常工作。

还有JSON Constants形式的位掩码选项，用于定义如何存储JSON。

$array = array( "data" => "value");  // dummy array to show data working

$json = array( //defined the following as an array
    "fbd49440-a5a1-48be-b13e-e8efddad3588"=> array( //notice each of these are now defined as arrays
        $array
    ),
    "0fc71cea-5609-40a7-a1d2-b78139660f8f"=> array(
        $array
    ),
    "73936e70-4329-4aba-b47c-42c64ced420c"=> array(
        $array
    ),
    "ac00b95e-9eeb-4035-bf4a-ff206319b2d6"=> array(
        $array
    ),
    "69072346-fe4c-489e-8e2b-5a7d7409fd44"=> array(
        $array
    )
); //added semicolon to end the declaration

echo json_encode($json, ( JSON_FORCE_OBJECT + JSON_PRETTY_PRINT ) ); 
// added JSON_FORCE_OBJECT and JSON_PRETTY_PRINT 
// As bitmask options, they return a constant to give `json_encode` instructions
// JSON_FORCE_OBJECT => 16, JSON_PRETTY_PRINT => 128 
// JSON_FORCE_OBJECT + JSON_PRETTY_PRINT = 16 + 128 = 144
// json_encode( $array, ( JSON_FORCE_OBJECT + JSON_PRETTY_PRINT ) = json_encode ($array, 144);

返回

{"fbd49440-a5a1-48be-b13e-e8efddad3588":{"0":{"data":"value"}},"0fc71cea-5609-40a7-a1d2-b78139660f8f":{"0":{"data":"value"}},"73936e70-4329-4aba-b47c-42c64ced420c":{"0":{"data":"value"}},"ac00b95e-9eeb-4035-bf4a-ff206319b2d6":{"0":{"data":"value"}},"69072346-fe4c-489e-8e2b-5a7d7409fd44":{"0":{"data":"value"}}}

json数据中的数组数组。

如果您希望它打印每个结果的索引，那么也需要将其视为数组。足够简单，可以理解当结果中存在多个对时，结果是一对时不那么直观。

$array1 = array( "data" => "value");
echo json_encode($array1, JSON_FORCE_OBJECT );
//would return
//{"data":"value"}

$array2 = array( 
    array( "data" => "value" ),
    array( "data" => "value" )
    );
echo json_encode($array2, JSON_FORCE_OBJECT );
// would return
// {"0":{"data":"value"},"1":{"data":"value"}}

$array3 = array( 
    array( "data" => "value" )
    );
echo json_encode($array3, JSON_FORCE_OBJECT );
// would return
// {"0":{"data":"value"}}

如何使用php创建嵌套的JSon

1 个答案: