不显示数据库中行的特定空白列

Question

I have a working code that can insert and update information from the database and echoing it to page. but I like to hide the specific empty column while displaying all the information from the row. check this screenshot my goal is to hide the "image" column when its null/empty; so the crock image wont display. here is my code below:

 <?php
    $result = mysqli_query($mysqli, "SELECT * FROM blogs ORDER BY id DESC");
    while ($row = mysqli_fetch_array($result)) 

    if (!empty($row['image'] != "")) 

    {
      echo "<div class='row'>
        <div class='col-lg-12 box'>
           <div class='content-heading'>
             <p>
                <text>".$row['title']."</text>
             </p>
           </div>

             <p>    
                <text1 class='pull-right' >".$row['image_text']."</text1><br/>
             <img class='img-size' id='hp'src='admin/upload_images/".$row['image']."'/>

             <text>".$row['definition']."</text> 
            </p>
          </div>
       </div>";     

   }
?>

but this code if (!empty($row['image'] != "")) is hiding the entire row from my database. Can anyone have a right solution to my problem?

Answer 1

You could try using the style attribute of the <img> tag to selectively show or hide the image tag:

<img class='img-size'
     id='hp'
     style='display: '. ($row["image"] != "" ? "block" : "none") . ';'
     src='admin/upload_images/".$row['image']."'/>

Answer 2

Instead of hiding entire row, you could just hide the image tag, like this:

 <?php
    $result = mysqli_query($mysqli, "SELECT * FROM blogs ORDER BY id DESC");
    while ($row = mysqli_fetch_array($result)) 
    {
        echo "
            <div class='row'>
                <div class='col-lg-12 box'>
                    <div class='content-heading'>
                        <p>
                            <text>".$row['title']."</text>
                        </p>
                    </div>

                    <p>    
                        <text1 class='pull-right' >".$row['image_text']."</text1><br/>
        ";

        if (!empty($row['image'])) {
            echo "<img class='img-size' id='hp'src='admin/upload_images/".$row['image']."'/>";      
        }                 

        echo "
                 <text>".$row['definition']."</text> 
                </p>
              </div>
           </div>
       ";     

   }
?>

Answer 3

Your if condition will hide all other html elements as well instead you should hide only img tag when no value found for image .

Update your code as below :

while ($row = mysqli_fetch_array($result)) {
    $image = (!empty($row['image'])) ? "<img class='img-size' id='hp'src='admin/upload_images/" . $row['image'] . "'/>" : "";
    echo "<div class='row'>
            <div class='col-lg-12 box'>
               <div class='content-heading'>
                 <p>
                    <text>" . $row['title'] . "</text>
                 </p>
               </div>        
                 <p>    
                    <text1 class='pull-right' >" . $row['image_text'] . "</text1><br/>
                    " . $image . "
                    <text>" . $row['definition'] . "</text> 
                </p>
            </div>
        </div>";
}