我正在读
,
它使用MySQL作为时态表连接中的查找表,如
-- Customers is backed by the JDBC connector and can be used for lookup joins
CREATE TEMPORARY TABLE Customers (
id INT,
name STRING,
country STRING,
zip STRING
) WITH (
'connector' = 'jdbc',
'url' = 'jdbc:mysql://mysqlhost:3306/c
我有这个码头文件
services:
db:
# We use a mariadb image which supports both amd64 & arm64 architecture
image: mariadb:10.6.4-focal
# If you really want to use MySQL, uncomment the following line
#image: mysql:8.0.27
command: '--default-authentication-plugin=mysql_native_passwor
我将创建下表
$sql[] = "CREATE TABLE IF NOT EXISTS #__GmQuestions(
QnID int(11) NOT NULL AUTO_INCREMENT,
Question text COLLATE utf8_unicode_ci NOT NULL,
Answer text COLLATE utf8_unicode_ci NOT NULL,
QnLevel int(11) NOT NULL,
QnPrize text COLLATE utf8_unicode_ci,
QnPoints DECIMAL( 10, 2 )
首先,对不起,我是一个真正的初学者。
我有点小问题
我有一个从我的用户中选择数据的mysql查询
the code looks like this
$r = mysql_query(" SELECT reader FROM users WHERE username = 'Tom' ") or die (mysql_error());
$result = mysql_fetch_array($r);
$vv = $result['reader'];
echo '<div id="reader">'.
我得到以下错误:致命错误:在我的php脚本中找不到类'mysqli‘我已经安装了php5.3.3-7,当我运行apt-cache show php5-mysql时得到这个错误
Description: MySQL module for php5
This package provides modules for MySQL database connections directly from
PHP scripts. It includes the generic "mysql" module which can be used
to connect to al
我在MySQL中使用实体框架,每次我尝试插入数据时,它都会给我一个NullReferenceException。
我可以通过创建一个命令直接插入数据,但是当我使用实体框架时,它就会爆炸掉。
实体框架将从表或更新表中选择,所以这可能与主键有关
从SaveChanges()方法引发以下异常。
failed: System.NullReferenceException : Object reference not set to an instance of an object.
at MySql.Data.Entity.ListFragment.WriteSql(String