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Question

我想为我的数据库选择最后一行，并将标题为s1和s2的列回显到网页正文，以下是我的代码。这会产生错误。

<html>
<body>
<?php

$servername = "localhost";
    $username = "root";
    $password = "";
    $dbname = "mydb2";

            $conn = new mysqli($servername, $username, $password, $dbname);
            if ($conn->connect_error) {die("Connection failed: " . $conn->connect_error);
            } 
$sql = "SELECT s1, s2 from reading ORDER BY id DESC LIMIT 1";
$row = array();
$row = mysqli_query($conn,$sql);
echo " cup 1". $row["s1"]. "CUP 2". $row["s2"];
 ?>



</body>
</html>

Answer 1

您必须使用mysqli_fetch_assoc（将遍历结果集），如下所示：

if ($result = mysqli_query($conn, $sql)) {

    while ($row = mysqli_fetch_assoc($result)) {
        echo " cup 1". $row["s1"]. "CUP 2". $row["s2"];
    }

    /* free result set */
    mysqli_free_result($result);
}

没有循环：

if ($result = mysqli_query($conn, $sql)) {

    $row = mysqli_fetch_assoc($result);
    if($row)
    {
       echo " cup 1". $row["s1"]. "CUP 2". $row["s2"];
    }

    /* free result set */
    mysqli_free_result($result);
}

Answer 2

您的解决方案

<html>
<body>
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "mydb2";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
    die("MySQL Connection Error"); // Try not to output SQL error messages on the front-end, look into error_reporting()
}

$sqlQuery = "SELECT s1, s2 from reading ORDER BY id DESC LIMIT 1";
$results = mysqli_fetch_assoc(mysqli_query($conn, $sqlQuery));
echo " cup 1". $results["s1"]. "CUP 2". $results["s2"];
?>
</body>
</html>

除此之外，我建议不要在程序风格和面向对象的风格中进行SQL操作，因为这可能导致将来出现许多复杂情况。

Answer 3

使用mysqli_fetch_assoc将结果提取到数组

$sql = "SELECT s1, s2 from reading ORDER BY id DESC LIMIT 1";
if ($result = mysqli_query($conn, $sql)) {
    $row = mysqli_fetch_assoc($result);
    echo " cup 1". $row["s1"]. "CUP 2". $row["s2"];
}

不能在php Mysql中使用mysqli_result类型的对象作为数组

3 个答案:

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