Question

我正在尝试从数据库中回显数据，但是，即使所有3个变量都存在，我也会收到以下错误。

注意：未定义的变量：第58行的/Applications/MAMP/htdocs/PhpProject2/testing.php中的fran_phone

对于以下内容：

Fran_phone
蠢
FB

代码

mysqli_report(MYSQLI_REPORT_INDEX);
$dbc = new mysqli("localhost", "root", "root", "One_Delivery");
$dbc->set_charset("utf8mb4");
if (isset($_GET['area'])) {

    $franc_details = $_GET['area'];
    $get_franc_dets = "SELECT * FROM Franc_dets WHERE Fran_City = '$franc_details'";
    $run_get_franc_dets = mysqli_query($dbc, $get_franc_dets);
    mysqli_stmt_execute($run_get_franc_dets);


    while ($row_get_franc_dets = mysqli_fetch_array($run_get_franc_dets)) {

        $franc_phone = $row_get_franc_dets['Fran_Contact_Num'];
        $twit = $row_get_franc_dets['Twitter'];
        $fb = $row_get_franc_dets['Fb'];
    }
}
?>

<div id='franc_div' >

    <table id='franchise_dets'>

        <tr id='frnc_tbl'>
            <td class='collapse'>
                <img src='./Images/franc_dets_phone.png'  height='50' width='50' alt='Call us'>    
            </td>

            <td class='phn_dets'>
                <p id='phn_title'>Problems ordering?</p>
                <p id='phn_numb'><?php echo $franc_phone ?></p>
            </td>

            <td class='collapse'>
                <img src='./Images/franc_dets_twitter.png'  height='50' width='50' alt='Twitter logo'>    
            </td>


            <td class='twitter_dets'>
                <p id='sm_title'>Social media</p>
                <a id='sm_twit' href='https://twitter.com/<?php echo $twit ?>'>@<?php echo $twit ?></a>
            </td>

            <td class='collapse'>
                <img src='./Images/franc_dets_fb.png'  height='50' width='50' alt='Facebook logo' >    
            </td>


            <td class='fb_dets'>

                <a id='sm_fb' href='https://www.facebook.com/<?php echo $fb; ?>'><?php echo $fb; ?></a>
            </td>
        </tr>

    </table>
</div>

我哪里出错了？我该怎么做才能解决它

Answer 1

所以有两件事是错的。

您显然正在尝试学习/切换到准备好的语句。 mysqli_stmt_execute需要声明，但您要为其提供mysqli_result个对象。

您需要使用mysqli_query。

，而不是使用mysqli_prepare

$sql = "SELECT * FROM Franc_dets WHERE Fran_City = ?;";
if ($stmt = mysqli_prepare($dbc, $sql)) {
    mysqli_stmt_bind_param($stmt, "s", $franc_details);
    mysqli_stmt_execute($stmt);

    // Check how many if any rows were returned.
    $num_rows = mysqli_stmt_num_rows($stmt);

    // Do what you were already doing,
    // and loop through each returned row.
}

请注意，如果返回0行，那么您将得到相同的错误。因为那时不会定义变量。

未定义的变量，但定义/声明变量

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