使用foreach遍历JSON
使用foreach遍历JSON
提问于 2018-05-31 01:25:21
我一直在遵循这个指南，使用两个不同的app创建一个应用程序，但该指南太旧了，因此其中一个app的工作方式与指南中的不同。我试图从谷歌地理编码API中获取坐标，并将其粘贴到Web的位置中。我是PHP新手，所以我按照指南中的示例遍历了一个JSON对象，但我整晚都在尝试让它正常工作。这是来自place search API的JSON对象
{
"html_attributions":[ ],
"results":[
{
"geometry":{ },
"icon":"https://maps.gstatic.com/mapfiles/place_api/icons/restaurant-71.png",
"id":"d4b0fb0f7bf5b2ea7df896a0c120a68efae039cf",
"name":"Guadalajara Mexican Grill & Cantina",
"opening_hours":{ },
"photos":[
{
"height":2952,
"html_attributions":[ ],
"photo_reference":"CmRaAAAAfO4JKUaO8vCFM2dcu5LMu4mA4_HXQGJ1FyAnyJUre_kD6VOWiQj7tBEECx4AAct5AORIKipSYWg-Zprjlf8o-SFd7mBRGMXMVMwodFZ5KMLwPYPUhBnTTehGPkb9275pEhCkAqMwfmK29vYenk1wdwFvGhSIHR8ch6FONc99tGn4rVnesbuteg",
"width":5248
}
],
"place_id":"ChIJ27es4SWa3IARcvjmt3xL2Aw",
"price_level":2,
"rating":4.4,
"reference":"CmRRAAAA7Rx-l7juDX-1or5dfpK6qFcZ0trZ9cUNEUtKP2ziqHb2MhOE6egs-msJ2OdFKEuHhuNe-3Yk6yxUYwxCBqhDT3ci8pYZI4xYhPGyyDgDenbEU_8k84JiEtCGvj4bdIR0EhDR2Pqte5_kDUcCC9PJFVknGhQomvD4d7NBIhCFxI4i2iEc0w9UiA",
"scope":"GOOGLE",
"types":[ ],
"vicinity":"105 North Main Street, Lake Elsinore"
},
{ },
{ },
{ },
{ },
{ },
{ },
{ },
{ },
{ },
{ },
{ },
{ },
{ }
],
"status":"OK"
}我正在尝试将所有的照片引用抓取到一个数组中，然后将它们插入google的Place Photos API中。这是我在这方面的尝试：
更新
<?php
if(!empty($_GET["location"])){
//$API_key = "";
$maps_url = 'https://' .
'maps.googleapis.com/' .
'maps/api/geocode/json' .
'?address=' . urlencode($_GET['location']) .
'&key=';
$maps_json = file_get_contents($maps_url);
$maps_array = json_decode($maps_json, true);
$lat = $maps_array['results'][0]['geometry']['location']['lat'];
$lng = $maps_array['results'][0]['geometry']['location']['lng'];
$places_url = 'https://maps.googleapis.com/maps/api/place/nearbysearch/json?' .
'location=$lat,$lng' .
'&radius=1500' .
'&rankby=distance' .
'&key=';
$places_json = file_get_contents($places_url);
$places_array = json_decode($places_json, true);
if (!empty($places_array)) {
foreach ($places_array as $item) {
var_dump($places_array );
}
}
}
?>
<!DOCTYPE html>
<html lang="en">
<head>
<title>What is Here?</title>
</head>
<body>
<h1>Type in a location</h1>
<p>This program will display pictures of places to go in that area</p>
<form action ="">
<input type ="text" name ="location"/>
<button type ="submit">Go!</button>
</form>
<br/>
<?php
echo "$lat $lng";
?>似乎不能让foreach循环做任何事情
回答 2
Stack Overflow用户
回答已采纳
发布于 2018-05-31 02:17:28
无效的请求意味着错误的url或错误的参数如果$lat和$lng是变量，那么插值将不适用于单引号尝试使用双引号，如"location=$lat，$lng“
$places_url = 'https://maps.googleapis.com/maps/api/place/nearbysearch/json?' .
"location=$lat,$lng" .
'&rankby=distance' .
'&key=mykey';您应该移除半径或距离，因为在docs https://developers.google.com/places/web-service/search?hl=en-419上无法同时获取它们
这是我在本地主机上工作的修改过的代码，请注意$contextOptions你不应该把它复制到你的代码上这是一个使file_get_contents在我的机器上工作的变通方法
之后，foreach应该很简单，因为它只是一个查看代码的数组
$thelocation = "1600+Amphitheatre+Parkway,+Mountain+View,+CA";
$thekey = "someapikey";
$maps_url = 'https://' .
'maps.googleapis.com/' .
'maps/api/geocode/json' .
'?address=' . urlencode($thelocation) .
'&key=' . $thekey;
$contextOptions = array(
"ssl" => array(
"verify_peer" => false,
"verify_peer_name" => false,
),
);
$maps_json = file_get_contents($maps_url, 0, stream_context_create($contextOptions));// file_get_contents($maps_url);
$maps_array = json_decode($maps_json, true);
$lat = $maps_array['results'][0]['geometry']['location']['lat'];
$lng = $maps_array['results'][0]['geometry']['location']['lng'];
$places_url = 'https://maps.googleapis.com/maps/api/place/nearbysearch/json?' .
"location=$lat,$lng" .
'&rankby=distance' .
'&key='.$thekey;
//https://maps.googleapis.com/maps/api/place/nearbysearch/json?location=-33.8670522,151.1957362&rankby=distance&key=
$places_json = file_get_contents($places_url,0, stream_context_create($contextOptions));
$places_array = json_decode($places_json, true);
if (!empty($places_array)) {
foreach ($places_array["results"] as $item) {
echo $item["name"]."<br>";
}
}此prints....easy
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AdExperts
Optical Spectroscopy and Nanomaterials GroupStack Overflow用户
发布于 2018-05-31 01:30:11
$lat,$lng变量或API调用是您的第一个问题，foreach循环是第二个问题。
json_decode($someJSON, true);从您的json创建了一个关联数组，因此您不能使用->箭头，这些箭头是用于对象的。More about this。没有$item->photo_reference，请使用：
$results = $places_array["results"];
foreach ($results as $item) {
echo $item["photos"]["photo_reference"];
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/50610394
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