Question

我根本不理解在PHP中访问JSON数据的语法。我已经习惯了这一段时间了。我太懂了。

$patchData = $_POST['mydata'];
$encoded = json_encode($patchData,true);
$patchDataJSON = json_decode($encoded,true);


/* what my JSON object looks like

{"patch_name":"whatever","sound_type":{

 "synths":[
    {"synth_name":"synth1","xpos":"29.99999725818634","ypos":"10.000012516975403"},
    {"synth_name":"synth2","xpos":"1.999997252328634","ypos":"18.000012516975403"},

    ]
  }
} 

*/


$patchName = $patchDataJSON['patch_name'];  // works!
$soundType = $patchDataJSON['sound_type']; // trying to access innards of JSON object. Does not work

echo $soundType;   // Does not work.

Answer 1

json_decode返回与JSON结构对应的嵌套PHP数组。（当传递true作为第二个参数时，否则返回PHP对象。）

要进行打印，请使用var_dump或print_r：

$soundType = $patchDataJSON['sound_type'];
print_r($soundType);

要访问字段，请使用字符串或数字索引（取决于输入JSON）：

$xpos = $soundType['synths'][0]['xpos'];

等。

一个简单的例子：

$jsonString = '{"foo": "bar", "baz": [{"val": 5}]}';
$decoded = json_decode($jsonString, true);
print_r($decoded);

这将输出：

Array
(
    [foo] => bar
    [baz] => Array
        (
            [0] => Array
                (
                    [val] => 5
                )
        )
)

不理解在PHP中访问JSON数据的语法

1 个答案: