Question

我收到此错误注意：未定义的索引：第17行的C：\ xampp \ htdocs \ RAU \ courses.php中的pharmacy.Name注意：未定义的索引：第19行的C：\ xampp \ htdocs \ RAU \ courses.php中的delivery.Phone注意：未定义的索引：第21行的C：\ xampp \ htdocs \ RAU \ courses.php中的delivery.Address

 <link rel="stylesheet" href="style1.css">
    <?php  
    $servername = "localhost";
        $userrname = "root";
        $password = "";
        $dbname = "bless";
        $conn = new mysqli($servername, $userrname, $password ,$dbname);
        if ($conn->connect_error) {
            die("Connection failed: " . $conn->connect_error);
        }
    $sqlq="SELECT pharmacy.name,delivery.phone,delivery.address FROM delivery INNER JOIN pharmacy ON delivery.Pharmacy_id = pharmacy._id";
    $rslt=$conn->query($sqlq) ;
    $roww = $rslt->fetch_assoc();

    echo "<br>";
    echo "<br>";
    echo "<td>" . $roww['pharmacy.name'] ;
    echo "<br>";
        echo"<td>". $roww['delivery.phone'];
    echo "<br>";
        echo"<td>". $roww['delivery.address'];
    echo "<br>";
    ?>

Answer 1

在MySQL查询中选择表时，列名称前面的名称是表：

'表名'。'列名'

它返回列名，但它告诉MySQL哪个表在不明确的情况下从中获取列数据。对于连接是一种安全的做法，但是当仅从一个表中进行选择时，它默认为该表。

使用

$roww['name']
$roww['phone']
$roww['address']

我如何修复未定义的变量

1 个答案: