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Valentina
Asked: 2021-10-22 12:36:09 +0800 CST 2021-10-22 12:36:09 +0800 CST

How to avoid repeating data when using a PHP concatenation from a MYSQL query?

  • 772

I have this CSS styles layout, as you can see there are two different containers col-1andcol-grid

 <div class="col-1">
      <div class="article-two">
        <img src="https://i.imgur.com/fNiFRdd.jpg" />
      </div>
    </div>

    <div class="col-grid">
      <div class="article-two">
        <img src="https://i.imgur.com/fNiFRss.jpg" />
      </div>
      <div class="article-two">
        <img src="https://i.imgur.com/fNiFRfN.jpg" />
      </div>
      <div class="article-two">
        <img src="https://i.imgur.com/fNiFRkN.jpg" />
      </div>
    </div>

To obtain the same design I have used the PHP concatenation -> .=but the data is repeated and the same design of the structure is not obtained HTMLwhere only one image goes in the first container and more than one image can go in the second container, given to the error what changes should be added to the code to get the same layout of the structureHTML

    $DivCol = '<div class="col-1">';
    $DivColGrid = '<div class="col-grid">';
    while ($stmt->fetch()) {
        $DivCol .= '<div class="article-two">
                        <img src="'.$cover_page.'">
                    </div>';
        $DivColGrid .= '<div class="article-two">
                        <img src="'.$cover_page.'">
                    </div>';
    }
    $DivColGrid .= '</div>';
    $DivCol .= '</div>';

This is the error I get as a result, the data printout was with a limit de 3in the queryMYSQL

<div class="col-1">
      <div class="article-two">
        <img src="https://i.imgur.com/QCcU9VB.jpg" />
      </div>
      <div class="article-two">
        <img src="https://i.imgur.com/Sy5jQrQ.jpg" />
      </div>
      <div class="article-two">
        <img src="https://i.imgur.com/t6MofoQ.jpg" />
      </div>
    </div>

    <div class="col-grid">
      <div class="article-two">
        <div class="article-two">
          <img src="https://i.imgur.com/QCcU9VB.jpg" />
        </div>
        <div class="article-two">
          <img src="https://i.imgur.com/Sy5jQrQ.jpg" />
        </div>
        <div class="article-two">
          <img src="https://i.imgur.com/t6MofoQ.jpg" />
        </div>
      </div>
    </div>
php

1 Answers

  1. Best Answer

    I think you can try the following:

    $DivCol = '<div class="col-1">';
$DivColGrid = '<div class="col-grid">';
$first = true;
while ($stmt->fetch()) {
    if ($first) {
        $DivCol .= '<div class="article-two">
                        <img src="' . $cover_page . '">
                    </div>';
        $first = false;
    } else {
        $DivColGrid .= '<div class="article-two">
                        <img src="' . $cover_page . '">
                    </div>';
    }
}
$DivColGrid .= '</div>';
$DivCol .= '</div>';

    in this case I have added the variable $firstwith value true: The first time it will add the first record to the first $DivColand the rest of the times to $DivColGrid.

    I hope it works for you.

    • 1