如何从另一个表中选择并插入数据到另一个表PHP mysql？ [英] How to select and insert data from another table into another table PHP mysql?

问题描述

我正在为我的学习做最后的项目。现在我的fyp是裁缝管理系统出了问题。



我的问题是我有两个表（对于数据库phpmyadmin）和表单输入，即Customer_detail （cid，customername，address，phoneno）和customer_order（orderid，customername，order_date，pickup_date）..我想选择数据customername并插入客户订单表格（名称：必须选择选项）这是数据必须从数据库获取来自customer_detail表的列customername。我不知道如何在php和mysql中创建代码。



如何为order_date&创建php代码必须插入phpmyadmin数据库的pickup_date。



i希望有人可以帮助我:)



我尝试过：



客户名称：



connect_error）{

die（连接失败：。$ conn-> connect_error）;

}

mysqli_select_db（$ conn，tailorsystem）; //设置数据库名称



$ menu =;



$ sql =SELECT cuname FROM customer; //选择查询

$ rs = mysqli_query（$ conn，$ sql）; // odbc_exec（$ conn，$ sql）;





if（mysqli_num_rows（$ rs）> 0）{

//每行输出数据

while（$ row = mysqli_fetch_assoc （$ rs））{

$ menu。=。 $行[ cuname]。 ;

}

}



echo $ menu;



mysqli_close（$ conn）;



？>

hi , i am middle of doing my final project for my studies . Now i'm having trouble for my fyp which is tailor management system .

my problem is i have two table (for database phpmyadmin) and form input which is ,Customer_detail(cid,customername,address,phoneno) and customer_order(orderid,customername,order_date,pickup_date) ..i want to select data customername and insert into customer order form (name :must be select option) which is the data must get from database column customername from customer_detail table .. i dont know how to create the code in php and mysql .

also how to create php code for order_date & pickup_date that must be insert into phpmyadmin database .

i hope anyone can help me :)

What I have tried:

for customer name :

connect_error) {
die("Connection failed: " . $conn->connect_error);
}
mysqli_select_db($conn,"tailorsystem"); //set the database name

$menu=" ";

$sql="SELECT cuname FROM customer"; //selection query
$rs = mysqli_query($conn, $sql);//odbc_exec($conn,$sql);


if (mysqli_num_rows($rs) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($rs)) {
$menu .= "" . $row['cuname']. "";
}
}

echo $menu;

mysqli_close($conn);

?>

推荐答案

conn-> connect_error）;

}

mysqli_select_db（

conn->connect_error);
}
mysqli_select_db(

conn，tailorsystem）; //设置数据库名称

conn,"tailorsystem"); //set the database name

menu =;

menu=" ";

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