Question

我有以下代码。我试图从请求的Feed中获取城市名称，并将其添加到新阵列中。任何人都可以给我一个指示。

    $city = $_GET['city'];
$json = @file_get_contents('http://ws.geonames.org/searchJSON?country=GB&maxRows=10&name_startsWith=$city');
$json = utf8_encode($json);

$city_suggest = json_decode($json, true);
foreach($city_suggest['geonames'] as $city){
    $cities = $city['geonames']['name'];    
}
print_r ($cities);

编辑 - json响应的1行

{"totalResultsCount":323,"geonames":[{"countryName":"United Kingdom","adminCode1":"ENG","fclName":"city, village,...","countryCode":"GB","lng":-0.12883186340332,"fcodeName":"capital of a political entity","toponymName":"London","fcl":"P","name":"London","fcode":"PPLC","geonameId":2643743,"lat":51.5005149421307,"adminName1":"England","population":7556900},

编辑 - var_dump的响应

array(2) { ["totalResultsCount"]=>  int(0) ["geonames"]=>  array(0) { } }

Answer 1

你已经在这个城市的地名部分，所以你不需要有 $city['geonames']['name']，只是$city['name']。

Answer 2

在$city_suggest上执行var_dump以查看此变量的结构。有了这些信息，提取所需的数据应该相当容易但是，试试这个：

 $cities[] = $city->name;

Answer 3

$cities = array();
foreach($city_suggest['geonames'] as $city){
    $cities[] = $city['name'];    
}

Answer 4

$city = $_GET['city'];
$json = file_get_contents("http://ws.geonames.org/searchJSON?country=GB&maxRows=10&name_startsWith=" . rawurlencode($city));
$json = utf8_encode($json);
$city_suggest = json_decode($json, true);
foreach($city_suggest['geonames'] as $city){
    print $city['name'];
    // there are other available variables too
    // print $city['countryName'];
    // print $city['adminCode1'];
    // print $city['fclName'];
    // print $city['countryCode'];
    // print $city['lng'];
    // print $city['fcodeName'];
    // print $city['toponymName'];
    // print $city['fcl'];
    // print $city['name'];
    // print $city['fcode'];
    // print $city['geonameId'];
    // print $city['lat'];
    // print $city['adminName1'];
    // print $city['population'];
}

另外，请注意您有一行：

$json = @file_get_contents('http://ws.geonames.org/searchJSON?country=GB&maxRows=10&name_startsWith=$city');

除非您将字符串括在双引号中，否则不会解释

$city，如下所示：

$json = @file_get_contents("http://ws.geonames.org/searchJSON?country=GB&maxRows=10&name_startsWith=$city");

从json响应中获取一系列城市

4 个答案: