我尝试通过PHP读取JSON
键{
"data": [
{
"id": "3043252fsdgdf36360354",
"name": "name1",
"access_token": "CAAIf3VEtVSoBAHrVxHL16zt4H5OvwBmdfgs4F3auPE0NZBx5PmIujBAdqw0Cv4bZACXytT1O1y6FHEZA25E1aqQZD"
},
{
"id": "3326848fdgsdfgsdf03424168",
"name": "name2",
"access_token": "CAAIf3VEtVSoBAJinePVdfgsdfgMxuY3zaj9AimaoKx7VIO9jCqZCHC6ZBixL1n6ZC72LTMn0ZB4T8rOHD27WmzbBVgvUwgspeEZD"
}}
我尝试使用此代码
$sfgsdfg= $json_a=json_decode($read,true);
echo $json_a['data'][3043252fsdgdf36360354];
echo $json_a['3043252fsdgdf36360354'][access_token];
不使用此PHP代码,我需要帮助才能阅读它
我需要选择 3043252fsdgdf36360354的 access_token 作为回声 $json_a['data'][3043252fsdgdf36360354]['access_token'];
我只需要通过 id 读取为 . mysql 命令。 选择access_token其中 id ='3326848fdgsdfgsdf03424168';
首先,您的 JSON 缺少]
字符,将导致json_decode
返回 NULL。应首先更正 JSON 字符串。可以使用联机服务(如 jsonlint.com)来验证 JSON。将 JSON 字符串解码为关联数组后,只需遍历数组并检查它是否包含给定的 ID。如果是这样,您可以轻松获取相应的访问令牌。
我把它做成一个简短的小函数。您可以使用它按 ID 获取访问令牌:
$jsonArray = json_decode($str, TRUE);
function getAccessTokenFromID($id, $jsonArray) {
foreach ($jsonArray['data'] as $k => $elem) {
if($elem['id'] == $id) {
$access_token = $elem['access_token'];
}
}
return $access_token;
}
用法:
$myid = '3043252fsdgdf36360354';
$my_accesstoken = getAccessTokenFromID($myid, $jsonArray);
演示!
格式不正确的 json。尝试:
{
"data": [
{
"id": "3043252fsdgdf36360354",
"name": "name1",
"access_token": "CAAIf3VEtVSoBAHrVxHL16zt4H5OvwBmdfgs4F3auPE0NZBx5PmIujBAdqw0Cv4bZACXytT1O1y6FHEZA25E1aqQZD"
},
{
"id": "3326848fdgsdfgsdf03424168",
"name": "name2",
"access_token": "CAAIf3VEtVSoBAJinePVdfgsdfgMxuY3zaj9AimaoKx7VIO9jCqZCHC6ZBixL1n6ZC72LTMn0ZB4T8rOHD27WmzbBVgvUwgspeEZD"
}
]
}
和 php:
$val = json_decode($read, TRUE);
echo $val['data'][0]['id'];
更新:
function find_by_id($id, $val) {
foreach($val['data'] as $key => $obj) {
if ($obj['id'] === $id)
return $obj['access_token'];
}
}