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
这篇关于如何从另一个表中选择并插入数据到另一个表PHP mysql?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!