檢查數據庫中是否存在一行，如果不顯示消息

Question

我無法檢查數據庫中是否存在一行，如果不顯示消息。 代碼如下。 我需要一點幫助。

<?php
    $result = mysqli_query($mysqli, "SELECT client_cv FROM formular_client WHERE id = '$id'"); 
   if (mysqli_num_rows($result) > 0){
       echo "<a href='../cv/" . $rows['client_cv'] . "' target='_blank' class='btn btn-lg btn-primary'><span class='glyphicon glyphicon-download-alt'></span> Descarca CV</a>";
   } else {
       echo "<a href='#' target='_blank' class='btn btn-lg btn-primary'><span  class='glyphicon glyphicon-download-alt'></span> Fara CV</a>";
   }
?>

Answer 1

如果要基於client_cv是否具有值顯示不同的鏈接，請嘗試此操作

<?php
    $result = mysqli_query($mysqli, "SELECT client_cv FROM formular_client WHERE id = '$id'"); 
    if (mysqli_num_rows($result) > 0){
        $rows = mysqli_fetch_array($result);
        if (strlen($rows["client_cv"]) > 0){
            echo "<a href='../cv/" . $rows['client_cv'] . "' target='_blank' class='btn btn-lg btn-primary'><span class='glyphicon glyphicon-download-alt'></span> Descarca CV</a>";
        } else {
            echo "<a href='#' target='_blank' class='btn btn-lg btn-primary'><span  class='glyphicon glyphicon-download-alt'></span> Fara CV</a>";
        }
    } else {
        echo "No record found"
    }
?>