有时会丢失来自Google API查询的数据(例如输入无效地址时),当发生这种情况时,会出现一个未知密钥的丑陋错误。为了避免这个丑陋的错误,我将调用包装成条件,但似乎无法让它一直工作,因为我的面向对象编程技能是不存在的。以下是我所拥有的和一些被注意的尝试,所以我做错了什么?我真的很在乎$ dataset-> results [0]是否有效,如果是的话,那将是有效的。
$url = "https://maps.googleapis.com/maps/api/geocode/json?address=$Address&key=$googlekey";
// Retrieve the URL contents
$c = curl_init();
curl_setopt($c, CURLOPT_SSL_VERIFYPEER, false);
curl_setopt($c, CURLOPT_RETURNTRANSFER, true);
curl_setopt($c, CURLOPT_FRESH_CONNECT, true);
curl_setopt($c, CURLOPT_URL, $url);
$jsonResponse = curl_exec($c);
curl_close($c);
$dataset = json_decode($jsonResponse);
if (isset($dataset->results[0])) :
//if (isset($dataset->results[0]->geometry->location)) :
//if (!empty($dataset)) :
//if (!empty($dataset) && json_last_error() === 0) :
$insertedVal = 1;
$latitude = $dataset->results[0]->geometry->location->lat;
$longitude = $dataset->results[0]->geometry->location->lng;
return "$latitude,$longitude";
endif;
您应该知道,Geocoding API Web服务还会在响应中返回状态。状态指示响应中是否存在有效项目或出现错误且您没有任何结果。
看看文档https://developers.google.com/maps/documentation/geocoding/intro#StatusCodes,你会发现有以下可能的状态
因此,在您尝试访问$dataset->results[0]
之前,请先检查$dataset->status
的值。如果它是“OK”,您可以安全地获得结果,否则正确处理错误代码。
代码段可能是
$dataset = json_decode($jsonResponse);
if ($dataset->status == "OK") {
if (isset($dataset->results[0])) {
$latitude = $dataset->results[0]->geometry->location->lat;
$longitude = $dataset->results[0]->geometry->location->lng;
}
} elseif ($dataset->status == "ZERO_RESULTS") {
//TODO: process zero results response
} elseif ($dataset->status == "OVER_DAILY_LIMIT" {
//TODO: process over daily quota
} elseif ($dataset->status == "OVER_QUERY_LIMIT" {
//TODO: process over QPS quota
} elseif ($dataset->status == "REQUEST_DENIED" {
//TODO: process request denied
} elseif ($dataset->status == "INVALID_REQUEST" {
//TODO: process invalid request response
} elseif ($dataset->status == "UNKNOWN_ERROR" {
//TODO: process unknown error response
}
我希望这有帮助!