如何在PHP中解析ElasticSearch JSON

Question

我以前解析过基本的JSON文件，没有遇到任何问题，但这个文件的结构（来自ElasticSearch）让我完全困惑

{
  "took": 7,
  "timed_out": false,
  "_shards": {
    "total": 2,
    "successful": 2,
    "failed": 0
  },
  "hits": {
    "total": 1017,
    "max_score": 2.8167849,
    "hits": [
      {
        "_index": "myindex",
        "_type": "mytype",
        "_id": "119479",
        "_score": 2.8167849,
        "_source": {
          "title": "my title",
          "url": "my url",
          "company": "my company",
          "location": "my location",
          "description": "my description",
          "industry": "my industry"
        }
      },
      {
        "_index": "myindex",
        "_type": "mytype",
        "_id": "119480",
        "_score": 2.8167849,
        "_source": {
          "title": "my title",
          "url": "my url",
          "company": "my company",
          "location": "my location",
          "description": "my description",
          "industry": "my industry"
        }
      }
    ]
  }
}

现在，假设我想得到两个结果结果的"title"值。我尝试了很多不同的事情，但都没有成功。例如：

//json_decode works fine. I have verified with a var_dump();
$myobj = json_decode($json);
//this is where I'm not sure what to do:
foreach($myobj->hits->hits->_source as $result) {
    echo $result->title;
}

我尝试了很多不同的变体，但我只是不确定如何解析这个结构。任何帮助都将不胜感激。

Answer 1

正如Marc B所说，var_dump（$myobj）将为您提供json对象的结构

要迭代对象的属性，请使用以下方法：

foreach($myobj->hits->hits->_source as $key => $val) {
    if($key == 'title') {
        echo $val;
    }
}

Answer 2

我得到了。下面是稍后可能会发现它有帮助的人的代码：

foreach ($myobj->hits->hits as $result) {
echo $result->_source->title;
}