如何在PHP中检查while循环中的if-else语句？

我调用了一个函数，该函数从数据库中的表中检索数据&我想添加一个if-else语句来检查该表，以查看某些指定字段是否为空。因此，为了做到这一点，我写了以下代码：

    <?php 
function post_pages(){
global $con;
$get_settings = "select * from site_settings;";
$run_settings = mysqli_query($con,$get_settings);
while($row_settings = mysqli_fetch_array($run_settings)){
    $addpost_id = $row_settings['id'];
    $addpost_name = $row_settings['sitename'];
    $addpost_url = $row_settings['siteurl'];
    $addpost_tags = $row_settings['sitetags'];
    $addpost_pp1= $row_settings['postpageone'];
    $addpost_pp2= $row_settings['postpagetwo'];
    $addpost_pp3= $row_settings['postpagethree'];
    $addpost_imgG= $row_settings['imagegalleryone'];
    $addpost_imgG= $row_settings['imagegallerytwo'];
    $addpost_vG= $row_settings['videogallery'];

if (empty($addpost_pp1) || (empty($addpost_pp2)) || (empty($addpost_pp3))){
    echo "<h4 class='widgettitle'>Please adjust your site settings first!</h4>";
}else{
        echo "
            <form action='' method='POST'>
                <select name='pages'>
                    <option value='home'>home page</option>
                    <option value='blogs'>blogs page</option>
                    <option value='aboutus'>aboutus page</option>
                </select></br>
                <input type='submit' name='submit' value='Submit'/>
            </form>
        ";
       }        
     }
   }
            ?>

---

如果表为空，则while循环中的行将不会执行。因为没有什么可重复的，这就是你没有看到你给出的信息的原因。解决方案是在while循环之前设置if条件。大概

if (mysqli_num_rows($run_settings) == 0) {
    echo "No records found";
} else {
    // Your while logic
}

您可能希望查看num_行：并检查是否从db返回任何结果。如@Rizier123所述，尝试添加额外的检查，以检查是否存在行，并相应地处理该情况。