PHP:在字符串中的特定单词后提取特定单词？

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country_name来自该字符串的值.

所以我试过这个:

$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;

但是没有任何回应 $result

有人可以就这个问题提出建议吗？

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;

你有什么JSON字符串.

JSON代表JavaScript Object Notation.PHP可以通过json_decode($ string,FALSE)将其解码为数组或对象;

默认情况下,第二个参数为FALSE,这意味着它会将字符串转换为对象,然后您可以按照上面的显示进行访问.