Question

我有这个JSON对象，包含时间戳和单词以及使用Amazon Transcribe制作的转录的其他元素：

[  
    {  
        "start_time":"29.420",
        "end_time":"40.500",
        "alternatives":[  
            {  
                "confidence":"0.4332",
                "content":"Foo"
            }
        ],
        "type":"pronunciation"
    },
    {  
        "start_time":"156.087",
        "end_time":"156.567",
        "alternatives":[  
            {  
                "confidence":"0.8786",
                "content":"Hello"
            }
        ],
        "type":"pronunciation"
    },
    {  
        "start_time":"156.597",
        "end_time":"157.737",
        "alternatives":[  
            {  
                "confidence":"0.9439",
                "content":"how"
            }
        ],
        "type":"pronunciation"
    },
    {  
        "start_time":"157.737",
        "end_time":"157.917",
        "alternatives":[  
            {  
                "confidence":"0.9916",
                "content":"are"
            }
        ],
        "type":"pronunciation"
    },
    {  
        "start_time":"157.917",
        "end_time":"158.287",
        "alternatives":[  
            {  
                "confidence":"0.6898",
                "content":"you"
            }
        ],
        "type":"pronunciation"
    },
    {  
        "start_time":"158.717",
        "end_time":"160.637",
        "alternatives":[  
            {  
                "confidence":"0.6357",
                "content":"Bar"
            }
        ],
        "type":"pronunciation"
    }
]

我需要在start_time = 100和end_time = 160之间返回内容：

你好，你好吗

我尝试使用json_decode将其转换为数组，然后使用ksort，implode，slice，等进行操作，但未成功。

Answer 1

json_decode()获取数组的JSON，然后循环遍历记录，只连接（或存储或任何你需要的）$start_date > 100 && $end_time < 160。

<?php
$data = json_decode($json, true);
$endResult = "";
foreach ($data as $res) {
    if ($res["start_time"] > 100 && $res["end_time"] < 160) {
        $endResult .= $res["alternatives"][0]["content"]." ";
    }
}
var_dump($endResult); // Hello how are you

Demo

或者您可以array_filter()具有相同条件的数组并获得仅包含您需要的元素的数组，以便以后根据需要进一步处理：

<?php
$data = json_decode($json, true);
$endResult = array_filter($data, function ($el) {
    return $el["start_time"] > 100 && $el["end_time"] < 160;
});
var_dump($endResult); // only elements within that time

Demo

Answer 2

也许尝试这样的事情？而不是第二次进入foreach循环，你可以使用数组来满足你的需求。

$array = json_decode($json);

$data = [];

foreach($array as $el) {
    $object = (object) $el;
    $start = $object->start_time;
    $stop  = $object->end_time;
    $content = $object->alternatives[0]->content;
    $data[] =  (array) [
        'start' => (float) $start,
        'stop'  => (float) $stop,
        'content' => (string) $content,
    ];
}

$between = [100,160];

foreach($data as $el) {

 if($el['start'] >= $between[0] && $el['stop'] <= $between[1]) {
     echo $el['content'] . ' ';
 }

}

Answer 3

您可以使用json_decode函数将JSON编码的字符串$json转换为PHP变量。然后，您将能够迭代它并应用条件来构建您的内容：

<?php
$json = '[  
    {  
        "start_time":"29.420",
        "end_time":"40.500",
        "alternatives":[  
            {  
                "confidence":"0.4332",
                "content":"Foo"
            }
        ],
        "type":"pronunciation"
    },
    {  
        "start_time":"156.087",
        "end_time":"156.567",
        "alternatives":[  
            {  
                "confidence":"0.8786",
                "content":"Hello"
            }
        ],
        "type":"pronunciation"
    },
    {  
        "start_time":"156.597",
        "end_time":"157.737",
        "alternatives":[  
            {  
                "confidence":"0.9439",
                "content":"how"
            }
        ],
        "type":"pronunciation"
    },
    {  
        "start_time":"157.737",
        "end_time":"157.917",
        "alternatives":[  
            {  
                "confidence":"0.9916",
                "content":"are"
            }
        ],
        "type":"pronunciation"
    },
    {  
        "start_time":"157.917",
        "end_time":"158.287",
        "alternatives":[  
            {  
                "confidence":"0.6898",
                "content":"you"
            }
        ],
        "type":"pronunciation"
    },
    {  
        "start_time":"158.717",
        "end_time":"160.637",
        "alternatives":[  
            {  
                "confidence":"0.6357",
                "content":"Bar"
            }
        ],
        "type":"pronunciation"
    }
]';
$json = json_decode($json, true);
$content = "";
foreach ($json as $item) {
    if ($item["start_time"] >= 100 && $item["end_time"] <= 160) {
        $content .= $item["alternatives"][0]["content"] . " ";
    }
}
echo trim($content);

需要使用PHP内存一片JSON对象

3 个答案: