PHP:在字符串中的特定单词之后提取特定单词？

Question

我有一个看起来像这样的字符串:

{"ip":"XX.XX.XX","country_code":"IE","country_name":"Ireland","region_code":"L","region_name":"Leinster","city":"Dublin","zip_code":"","time_zone":"Europe/Dublin","latitude":53.333,"longitude":-6.249,"metro_code":0}

我只需要该字符串中 country_name 的值。

所以我试了一下:

$country = '{"ip":"XX.XX.XX","country_code":"IE","country_name":"Ireland","region_code":"L","region_name":"Leinster","city":"Dublin","zip_code":"","time_zone":"Europe/Dublin","latitude":53.333,"longitude":-6.249,"metro_code":0}';

if (preg_match('#^country_name: ([^\s]+)#m', $country, $match)) {
    $result = $match[1];
}

echo $result;

但$result

中没有任何回显

有人可以就此问题提出建议吗？

Answer 1

$country = json_decode('{"ip":"XX.XX.XX","country_code":"IE","country_name":"Ireland","region_code":"L","region_name":"Leinster","city":"Dublin","zip_code":"","time_zone":"Europe/Dublin","latitude":53.333,"longitude":-6.249,"metro_code":0}');

echo $country->country_name;

你有一个 JSON 字符串。

JSON 代表 JavaScript 对象表示法。 PHP 可以通过 json_decode($string, FALSE); 将其解码为数组或对象；

默认情况下，第二个参数是 FALSE，这意味着它将把字符串转换成一个对象，然后您可以像我上面展示的那样访问它。