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Answered Snippet output - how to output html correctly

    • 52857
    • 8 Posts
    sf23103 Reply #1, 7 years, 6 months ago
    Hello,

    I've been reading the documentation and I have learned that it is not best practice to output html from a snippet. Can you help identify best practice for this scenario?

    I have a snippet that queries a MySQL database on my server. It finds the 10 latest db entries (they happen to be news headlines), and outputs the list. My first thought is that I should change the echo line to '$output =', and then call that $output on my page. Is that correct? If so, I know how to call the snippet, but how do I print the $output variable on my page?

    Thanks for your help. Loving modx.

    <?php
include "/path/to/dblogin/details.php";
$db = new mysqli('localhost', $username, $password, $database);

if($db->connect_errno > 0){
    die('Unable to connect to database [' . $db->connect_error . ']');
}

$sql = <<<SQL
    SELECT *
    FROM `database_16`
    WHERE `col_11` = 'Include'
ORDER BY submission_id DESC LIMIT 10 
SQL;

if(!$result = $db->query($sql)){
    die('There was an error running the query [' . $db->error . ']');
}

while($row = mysqli_fetch_array($result))
  {
 echo "<a href=\"/news-template/full.php?id=". $row['submission_id'] . "\">" . $row['col_5'] . "</a><br /><br />";
  }

$db->close();


    This question has been answered by BobRay. See the first response.

    • discuss.answer
      • 3749
      • 24,544 Posts
      BobRay Reply #2, 7 years, 6 months ago
      Just return $output in the snippet. The value returned will replace the snippet tag on the page.

      while($row = mysqli_fetch_array($result))
  {
 $output = "<a href=\"/news-template/full.php?id=". $row['submission_id'] . "\">" . $row['col_5'] . "</a><br /><br />";
  }
 
$db->close();

return $output;

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        • 52857
        • 8 Posts
        sf23103 Reply #3, 7 years, 6 months ago
        Ok, that makes sense. Thank you!

        Alex

        Quote from: BobRay at Oct 14, 2016, 07:06 AM
        Just return $output in the snippet. The value returned will replace the snippet tag on the page.

        while($row = mysqli_fetch_array($result))
  {
 $output = "<a href="\"/news-template/full.php?id="." $row['submission_id']="" .="" "\"="">" . $row['col_5'] . "</a>

";
  }
 
$db->close();

return $output;