Php 如何以JSON格式发送请求以发布数据
我有一个Api端点来处理代码Php 如何以JSON格式发送请求以发布数据,php,json,Php,Json,我有一个Api端点来处理代码 <?php $name= $_POST['name']; $email = $_POST['email']; $password= $_POST['password']; $gender= $_POST['gender']; $con = mysqli_connect("localhost", "root", "qwerty", "db"); $query= mysqli_query($con, "INSER
<?php
$name= $_POST['name'];
$email = $_POST['email'];
$password= $_POST['password'];
$gender= $_POST['gender'];
$con = mysqli_connect("localhost", "root", "qwerty", "db");
$query= mysqli_query($con, "INSERT INTO users(name,email, password, gender) VALUES('$name','$email', '$password', '$gender')");
if($query){
echo "You are sucessfully Registered";
}
else{
echo "your details could not be registered";
}
mysqli_close($con);
?>
我是php Api开发新手您必须像这样构建一个数组:
$data = array(
'name' => $_POST['name'],
'email' => $_POST['email'],
'password' => $_POST['password']
);
然后你必须对它进行编码:$encoded=json\u encode($data)
使用此方法,您可以在post中发送请求您的代码易受SQL注入攻击,您需要修复此问题。可能会重复将值插入数组并使用json_编码PHPT中的数组。这不是答案,因为没有上下文。OP是否提供了上下文?不是真的。他问我如何获得JSON数组,这就是我的答案。
$data = array(
'name' => $_POST['name'],
'email' => $_POST['email'],
'password' => $_POST['password']
);
$data = array(
'userID' => 'a7664093-502e-4d2b-bf30-25a2b26d6021',
'itemKind' => 0,
'value' => 1,
'description' => 'Boa saudaÁ„o.',
'itemID' => '03e76d0a-8bab-11e0-8250-000c29b481aa'
);
$json = json_encode($data);
$client = new Zend_Http_Client($uri);
$client->setRawData($json, 'application/json')->request('POST');