【发布时间】:2020-08-16 15:38:01
【问题描述】:
//创建mysql表
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
//邮递员发帖请求
{
"name":"krithi",
"age":15,
"gender":"female",
"phone":1234567890,
"designation":"volunteer",
"address":"chennai",
"pincode":90,
"password":"ac",
"cpassword":"ac",
"email":"abc@gmail.com"
}
// 在邮递员中输出
{"code":1010,"message":"Data integrity violation"}
但它适用于获取请求。任何帮助将不胜感激。
I changed my table into a simpler one
创建表d (
name varchar(45) 非空
) 引擎=InnoDB 默认字符集=utf8mb4 整理=utf8mb4_0900_ai_ci
我的json是`
{
"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"
}
【问题讨论】:
-
显示表定义(
SHOW CREATE TABLE {tablename}),你在执行什么代码?生成的具体命令是什么?欢迎来到 SO。请预先提供模式详情。 -
你必须在更简单的表上指定一个主键。