如何在PHP中回显INNER join

Question

我已经在MySQL中测试了我的代码，并且可以正常工作。 我唯一的问题是我不明白如何回应INNER JOIN（以前从未使用过），而且我似乎无法在网上找到清晰的示例。

我需要将代码回显到与数据库的<table>连接中（有效） ：

include 'db_connection1.php';

$conn = OpenCon();

echo "Connected Successfully";

数据库连接将转到：

<?php
    function OpenCon()
    {
        $dbhost = "localhost";
        $dbuser = "root";
        $dbpass = "admin";
        $db = "theDBname";

        $conn = new mysqli($dbhost, $dbuser, $dbpass,$db) or
            die("Connect failed: %s\n". $conn -> error);

        return $conn;
    }

    function CloseCon($conn)
    {
        $conn->close();
    }
?>

码：

$sql = "SELECT orders.Order_ID AS OrderID,
            customer.First_Name AS FirstName,
            customer.Last_Name AS LastName,
            orders.Order_Date AS OrderDate
            FROM Orders
            INNER JOIN customer ON orders.Customer_Customer_ID=customer.Customer_ID";

Answer 1

根据我的评论，如果您想在内容中显示查询，只需执行；

echo $sql;

（我知道这并不是一个完整的答案-这是写下来的一种形式）

解释后编辑

根据您的评论，您希望结果在表格中吗？
所以...

<?php

$table = "<table><tr><th>Order ID</th><th>First Name</th><th>Last Name</th><th>Order Date</th></tr>";

// Set up DB connection
$conn = new MySqli("db_hostname", "db_user", "db_pass", "db_name");
// Excecute the query
$result = $conn->query("SELECT orders.Order_ID AS OrderID,
    customer.First_Name AS FirstName, 
    customer.Last_Name AS LastName,
    orders.Order_Date AS OrderDate
    FROM Orders
    INNER JOIN customer ON orders.Customer_Customer_ID=customer.Customer_ID");
// For each row, add the results to a table string using concatenate (.=)
while ($row = $result->fetch_assoc())
{
    $table .= "<tr>";
    $table .= "<td>{$row['OrderID']}</td>";
    $table .= "<td>{$row['FirstName']}</td>";
    $table .= "<td>{$row['LastName']}</td>";
    $table .= "<td>{$row['OrderDate']}</td>";
    $table .= "</tr>";
}

$table .= "</table";
print $table;

Answer 2

您的订单表似乎没有名为Customer_Customer_ID的列。 因此，我不知道您如何声明查询的工作原理。

但是，添加一列Customer_ID ，并将最后一行调整为：

INNER JOIN customer ON orders.Customer_ID=customer.Customer_ID";

Answer 3

我不明白我如何回应内在的加入

这里的重要概念是INNER JOIN（或与此相关的任何联接）都返回相同的结果集。 您可以获得许多行，但这不是通过使用联接来确定的。 您可以简单地遍历结果集。 每次迭代为1行。

<?php
...
$results = $conn->query("SELECT orders.Order_ID AS OrderID,
    customer.First_Name AS FirstName, 
    customer.Last_Name AS LastName,
    orders.Order_Date AS OrderDate
    FROM Orders
    INNER JOIN customer ON orders.Customer_Customer_ID = 
    customer.Customer_ID");

print '<table border="1">';
while($row = $results->fetch_assoc()) {
    print '<tr>';
    print '<td>'.$row["OrderID"].'</td>';
    print '<td>'.$row["FirstName"].'</td>';
    print '<td>'.$row["LastName"].'</td>';
    print '<td>'.$row["OrderDate"].'</td>';
    print '</tr>';
}  
print '</table>';