顯示名稱而不是ID

Question

數據庫設置：

表：客戶

id | name | address | zipcode | city | phone | email | active

表：待辦事項

id | customerid | description | information | active


$sql = "
SELECT * 
  FROM todo
  ORDER 
    BY `customerid` ASC
     , `description` ASC
";

顯示結果：

echo $row['customerid'] $row['description'] $row['information'];

輸出：

customerid description information

所需的輸出：

customername (from table customers) description information

我一直在閱讀這個論壇，我發現我應該使用INNER JOIN，但我無法使其正常工作。
有人可以幫我嗎？

Answer 1

首先，您需要使用join來獲取客戶名值

SELECT t.description,t.information,c.name 
    FROM todo t JOIN customers c ON c.id=t.customerid 
ORDER BY `t.customerid` ASC, `description` ASC

然后嘗試以下方法：

echo $row['name'] $row['description'] $row['information'];

Answer 2

問題是，當您使用“ ORDER BY”時，不能使用“ *”。 這是代碼的運行方式：

<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}

$sql = "SELECT customers.customerid,name, description, information FROM  customers INNER JOIN todo ON customers.customerid=todo.customerid ORDER 
BY customers.customerid ASC
 , description ASC";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
    echo "<br>". $row["name"]." ".$row["description"]. " " . $row["information"] . "<br>";
 }
 } else {
echo "0 results";
 }

 $conn->close();
 ?>

這應該做。