我對PHP非常陌生。我不想創建json服務,它將負責保存數據庫中的註冊表單值。問題是,它會在數據庫中保存空值。問題是我不明白如何從JSON獲取值,並將其保存在數據庫PHP - 數據庫中的JSON空值
PHP代碼
<?php
// Include confi.php
include_once('confi.php');
if($_SERVER['REQUEST_METHOD'] == "POST"){
// Get data
$name = isset($_POST['name']) ? mysql_real_escape_string($_POST['name']) :'';
$email = isset($_POST['email']) ? mysql_real_escape_string($_POST['email']) : '';
$password = isset($_POST['pwd']) ? mysql_real_escape_string($_POST['pwd']) : '';
$status = isset($_POST['status']) ? mysql_real_escape_string($_POST['status']) : '';
// Insert data into data base
$sql = "INSERT INTO `test`.`users` (ID,`name`, `email`, `password`, `status`) VALUES (NULL,'$name', '$email', '$password', '$status');";
$qur = mysql_query($sql);
if($qur){
$json = array("status" => 1, "msg" => "Done User added!");
}else{
$json = array("status" => 0, "msg" => "Error adding user!");
}
}else{
$json = array("status" => 0, "msg" => "Request method not accepted");
}
@mysql_close($conn);
/* Output header */
header('Content-type: application/json');
echo json_encode($json);
發送值與POST方法
{
"name":"aamir",
"email":"[email protected]",
"pwd":"12345678",
"status":"yes"
}
This is the result 和我跟隨此鏈接 https://trinitytuts.com/build-first-web-service-php/
1)請使用'mysql_ *'功能,因爲他們從PHP開始7.0版本中刪除避免。改爲使用'mysqli'或'PDO'。 2)如果你'echo $ sql;',它顯示正確的值嗎? –
@Jiri Hrazdil tihis line如果我在冒號後用單引號表示價值,它可以節省價值。像這條線在數據庫中保存香蕉... $ name = isset($ _ POST ['name'])? mysql_real_escape_string($ _ POST ['name']):'Banana'; – Aamir
這意味着'$ _POST ['name']'沒有設置。問題出在您用POST方法將值發送到您的腳本的部分... –