如何使用php在url中编辑json格式的数据?
这是我的url如何使用php在url中编辑json格式的数据?,php,json,Php,Json,这是我的urlhttp://localhost:4000/users json数据如下所示: "users": [ { "id": "1", "name": "Admin", "email": "admin@gmail.com", "username": "admin", "passwor
http://localhost:4000/users
json数据如下所示:
"users": [
{
"id": "1",
"name": "Admin",
"email": "admin@gmail.com",
"username": "admin",
"password": "e19d5cd5af0378da05f63f891c7467af",
"wallet_address": "4p7tg5cb2yulvmz3sjfeanoi0r8wkdx4h69q",
"balance": "1000"
},
]
我想改变平衡。我该怎么做?我尝试了以下代码:
$url = "http://localhost:4000/users";
$data = file_get_contents($url);
$decodedData = json_decode($data);
for ($i=0; $i<count($decodedData); $i++) {
if ($decodedData[$i]->username == $_SESSION['username']) {
$decodedData[$i]->balance = $decodedData[$i]->balance - $amount;
$data4 = array("name"=>$decodedData[$i]->name, "email"=>$decodedData[$i]->email,
"username"=>$decodedData[$i]->username, "password"=>$decodedData[$i]->password,
"wallet_address"=>$decodedData[$i]->wallet_address, "balance"=>$decodedData[$i]->balance);
$options4 = array(
'http' => array(
'header' => "Content-type: application/x-www-form-urlencoded\r\n",
'method' => 'PUT',
'content' => http_build_query($data4)
)
);
$context4 = stream_context_create($options4);
$result_users = file_get_contents($url, false, $context4);
}
}
我从代码中删除了最后两行,现在没有显示警告。但平衡的价值仍然没有改变
感谢您的回答。听起来URL要么不正确,要么无法从运行PHP的地方访问。您是否在承载URL的同一台计算机上运行此代码?侦听该URL的Web服务器是否已启动并正在运行?URL是否正确。我在url所在的同一台机器上运行代码。我已经成功地使用了POST方法,比如
'method'=>'POST',
我从代码中删除了最后两行,现在警告没有显示。但余额的值仍然没有改变。您是否碰巧在我的php.ini中设置了错误报告(error\u reporting)设置?error\u reporting=E\u ALL&~E\u DEPRECATED&~E\u STRICT
和display\u errors=On
Warning: file_get_contents(http://localhost:4000/users): failed to open stream: HTTP request failed! HTTP/1.1 404 Not Found in ....