// mysql создание таблицы
CREATE TABLE `users` (
`id` int NOT NULL,
`name` varchar(45) NOT NULL,
`gender` varchar(45) NOT NULL,
`designation` varchar(45) NOT NULL,
`address` varchar(45) NOT NULL,
`email` varchar(45) NOT NULL,
`password` varchar(45) NOT NULL,
`cpassword` varchar(45) NOT NULL,
`age` int NOT NULL,
`phone` int NOT NULL,
`pincode` int NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_0900_ai_ci
// почтовый запрос
{
"name":"krithi",
"age":15,
"gender":"female",
"phone":1234567890,
"designation":"volunteer",
"address":"chennai",
"pincode":90,
"password":"ac",
"cpassword":"ac",
"email":"abc@gmail.com"
}
// вывод в почтальоне
{"code":1010,"message":"Data integrity violation"}
Но он хорошо работает для запроса на получение. Любая помощь будет принята с благодарностью.
I changed my table into a simpler one
CREATE TABLE d
(name
varchar (45) NOT NULL) ENGINE = InnoDB DEFAULT CHARSET = utf8mb4 COLLATE = utf8mb4_0900_ai_ci и мой json is` {
"name" : "abc"
}
*and the response is*
{
"code": 9999,
"message": "Argument 1 passed to Tqdev\\PhpCrudApi\\Database\\ColumnsBuilder::quoteColumnName() must be an instance of Tqdev\\PhpCrudApi\\Column\\Reflection\\ReflectedColumn, null given, called in C:\\Users\\vimy9\\OneDrive\\Desktop\\clone\\php-crud-api\\api.php on line 4892"
}