Php 如何从嵌套的JSON数据中获取值？

以下是我的JSON代码：

{
    "query": {
        "count": 2,
        "created": "2013-04-03T09:47:03Z",
        "lang": "en-US",
        "results": {
            "yctCategories": {
                "yctCategory": {
                    "score": "0.504762",
                    "content": "Computing"
                }
            },
            "entities": {
                "entity": [
                    {
                        "score": "0.902",
                        "text": {
                            "end": "19",
                            "endchar": "19",
                            "start": "0",
                            "startchar": "0",
                            "content": "Computer programming"
                        },
                        "wiki_url": "http://en.wikipedia.com/wiki/Computer_programming"
                    },
                    {
                        "score": "0.575",
                        "text": {
                            "end": "51",
                            "endchar": "51",
                            "start": "41",
                            "startchar": "41",
                            "content": "programming"
                        }
                    }
                ]
            }
        }
    }
}

下面是我的PHP代码

$json_o = json_decode($json,true);
echo "Json result:</br>";
echo $json; // json
echo "</br></br>";
echo "Value result:</br>";
$result = array();
//$entity = $json_o['query']['results']['entities']['entity'];
foreach ($json_o['query']['results']['entities']['entity'] as $theentity)
 foreach ($theentity['text'] as $thetext){
    $result[] = $thetext['content'];
 }
print_r($result);

---

删除第二个foreach并将其替换为

$result[]=$theentity['text']['content']

使用类似的东西可能会使您更容易了解json的结构。或者，只需
var\u dump
（或
print\r
）json解码的输出使用此循环

foreach ($json_o['query']['results']['entities']['entity'] as $theentity)
{
    $result[] = $theentity['text']['content'];
}


输出数组（[0]=>计算机编程[1]=>编程）
使用简单代码：

$array = $json_o['query']['results']['entities']['entity'];

foreach($array as $v){
  echo $v['text']['content'];
}

将您的foreach更改为：

$result = array();
foreach ($json_o['query']['results']['entities']['entity'] as $theentity) {
    $result[] = $theentity['text']['content'];
}
print_r($result);

$theentity['text']
是一个键数组=>值。您只需访问key
content
，而无需遍历所有条目

另一种方法是（尽管这是一个糟糕的选择）：

我提供第二个示例来说明为什么原始代码不起作用

更新

要访问其他属性，如
yctCategories
，只需执行类似操作

$categories = array();
foreach ($json_o['query']['results']['yctCategories'] as $yctCategory) {
    $categories[] = $yctCategory['content'];
}
print_r($categories);

不必要地复制数据，没有理由在这里效率低下。@Rickburges ok，但提供了更高的可读性：）和性能editing@Akam谢谢澄清。。它给了我写代码的其他选择
foreach($theentity['text'] as $key => $value) {
    if( 'content' === $key ) {
        $result[] = $value;
    }
}

$categories = array();
foreach ($json_o['query']['results']['yctCategories'] as $yctCategory) {
    $categories[] = $yctCategory['content'];
}
print_r($categories);