Question

对于这个json api响应，我能够用json_decode解析：

 $string = '{"name": "Google",
     "permalink": "google",
     "homepage_url": "http://google.com",
     "blog_url": "http://googleblog.blogspot.com",
     "blog_feed_url": "http://googleblog.blogspot.com/feeds/posts/default?alt=rss",
     "twitter_username": "google",
     "category_code": "search",
     "number_of_employees": 20000,
     "founded_year": 1998,
     "founded_month": 9,
     "founded_day": 7,// bla bla.....}';

$obj=json_decode($string);
echo $obj->number_of_employees."<br>";// 20000
echo $obj->founded_year; //1998

我得到了上面的结果但是得到了空白的结果，下面是一个：

$string =  '{"offices":
  [{"description": "Google Headquarters",
    "address1": "1600 Amphitheatre Parkway",
    "address2": "",
    "zip_code": "",
    "city": "Mountain View",
    "state_code": "CA",
    "country_code": "USA",//blah blah }]//blah blah...}';

$obj=json_decode($string);
echo $obj->address1."<br>";// ""
echo $obj->city; //""

我知道address1又在另一个数组或循环中，但不知道如何检索它......任何想法？

Answer 1

您将需要以下内容：

foreach($obj->offices as $office) {
    echo $office->address1; // The first would be '1600 Amphitheatre Parkway'
}

要查看已解码的json的内容，请执行以下操作：

echo "<pre>";
print_r($obj);
echo "</pre>";

在你的情况下，这会给你：

stdClass Object
(
    [offices] => Array
        (
            [0] => stdClass Object
                (
                    [description] => Google Headquarters
                    [address1] => 1600 Amphitheatre Parkway
                    [address2] => 
                    [zip_code] => 
                    [city] => Mountain View
                    [state_code] => CA
                    [country_code] => USA
                )

        )

)

为什么json_decode不能用于某些代码？

1 个答案: