我一生都找不到正确的语法来返回 JSON 数组中的元素
该数组是
{
"_total": 1,
"values": [{
"isCommentable": true,
"isLikable": true,
"isLiked": false,
"numLikes": 0,
"timestamp": 1453718959851,
"updateComments": {"_total": 0},
"updateContent": {
"company": {
"id": 2691316,
"name": "Rising 5th Web Design"
},
"companyStatusUpdate": {"share": {
"comment": "This is a test update for testing the jQuery REST API",
"id": "s6097339248095035392",
"source": {
"serviceProvider": {"name": "LINKEDIN"},
"serviceProviderShareId": "s6097339248095035392"
},
"timestamp": 1453718959851,
"visibility": {"code": "anyone"}
}}
},
"updateKey": "UPDATE-c2691316-6097339248082460672",
"updateType": "CMPY"
}]
}
我试图达到的元素是
"comment": "这是用于测试 jQuery REST API 的测试更新"
但是没有充分的理由,我无法弄清楚如何将PHP带到这个元素的语法。
任何帮助将不胜感激。
如果它是一个静态 json,那么您可以使用 json_decode
$content = json_decode($string, true);
echo $content['values'][0]['updateContent']['companyStatusUpdate']['share']['comment'];
解释:
首先解码您的json
并获得正确的索引。
请注意,在此示例中,我使用函数json_decode()
第二个参数作为true
如果您忽略此参数,您将以OBJECT
形式获得解码结果。
尝试将 json 放入变量中,并使用 json_decode() 在 php 中解码 json。
//json variable.
$json ='{
"_total": 1,
"values": [{
"isCommentable": true,
"isLikable": true,
"isLiked": false,
"numLikes": 0,
"timestamp": 1453718959851,
"updateComments": {"_total": 0},
"updateContent": {
"company": {
"id": 2691316,
"name": "Rising 5th Web Design"
},
"companyStatusUpdate": {"share": {
"comment": "This is a test update for testing the jQuery REST API",
"id": "s6097339248095035392",
"source": {
"serviceProvider": {"name": "LINKEDIN"},
"serviceProviderShareId": "s6097339248095035392"
},
"timestamp": 1453718959851,
"visibility": {"code": "anyone"}
}}
},
"updateKey": "UPDATE-c2691316-6097339248082460672",
"updateType": "CMPY"
}]
}';
// decode json
$c = json_decode($json, true);
// get your element
echo $c['values'][0]['updateContent']['companyStatusUpdate']['share']['comment'];