Question

我可以执行var_dump，但在尝试访问这些值时，我收到的错误是找不到值。

    {
  "metrics": {
    "timers": [
      {
        "name": "com.android.timer.launchtime",
        "startTime": 1232138988989,
        "duration_ms": 1900
      },
      {
        "name": "com.android.timer.preroll-load-time",
        "startTime": 1232138988989,
        "duration_ms": 1000
      }
    ]
  }
}

到目前为止，我使用以下内容尝试解析它。

$json_file = file_get_contents('test.json'); 
$json_a = json_decode($json_file,true);


var_dump(json_decode($json_file)); //This works

echo $json_a['name']; //I want to print the name of each (from timers).

Answer 1

尝试：

$yourDecodedJSON = json_decode($yourJson)

echo $yourDecodedJSON->metrics->timers[0]->name;

或者你可以：

$yourDecodedJSON = json_decode($yourJson, true); // forces array

echo $yourDecodedJSON['metrics']['timers'][0]->name;

在您的情况下，您可能想要..

foreach($yourDecodedJSON['metrics']['timers'] as $timer){

    echo $timer['name']; echo $timer['duration_ms']; // etc

}

如果出现问题，请使用：

echo json_last_error_msg()

进一步排查

Answer 2

您需要按以下方式进行： -

<?php

 $data =  '{
  "metrics": {
    "timers": [
      {
        "name": "com.android.timer.launchtime",
        "startTime": 1232138988989,
        "duration_ms": 1900
      },
      {
        "name": "com.android.timer.preroll-load-time",
        "startTime": 1232138988989,
        "duration_ms": 1000
      }
    ]
  }
}';
$new_array = json_decode($data); //convert json data into array
echo "<pre/>";print_r($new_array); //print array
foreach ($new_array->metrics->timers as $new_arr){ // iterate through array
    echo $new_arr->name.'<br/>'; // rest you can do also same   
}
?>

输出： - https://eval.in/407418

如何用PHP解析JSON？

2 个答案: