How to echo a MySQLi row in escaped PHP code, inside an HTML block

Question

I'm wondering what is the appropriate syntax is to echo a row from MySQLi code in a block of HTML text. I'm working on a page that uses PHP code at the start to determine if a session is started and to post comments that user has posted, after pulling said comments from a MySQLi database. The interesting and confusing part is, I've accomplished what I'm trying to do in one HTML div, but I can't seem to get it to work in the next.

    <?php
    $page_title = "store";
    require_once('connectvars.php');
    require_once('startsession.php');
    require_once('header.php');

    // Connect to the DB
    $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);

    // Grab location data from the DB
    // Grab post data
    $query2 = "SELECT sp.post_id, sp.admin_id, sp.post, sp.date, sa.admin_id, s.store_name, s.store_city, s.store_state, s.store_zip,s.store_phone, s.store_email, s.store_address FROM store_posts sp, store_admin sa, stores s WHERE sp.admin_id='$admin_id' and sa.admin_id='$admin_id' and s.admin_id='$admin_id'ORDER BY date DESC";
    $data = mysqli_query($dbc, $query2);
    if (mysqli_num_rows($data) == 0) {
        $blankwall = "Empty feed? '<a href=manage.php><u>Click here</u></a> to manage posts and communicate with your customers!";
    }
?>
<body>
<div id="content">
<!-- BANNER & CONTENT-->
    <h2>Recent Posts</h2>
  <div id="store_wrap">
        <div id="left_col">
        <br />
        <?php
            **// Loop through posts and show them
            if (mysqli_num_rows($data) == 0) {
                echo $blankwall;
            }
                else {
            while ($row = mysqli_fetch_array($data)) {
                // Show the posts
                echo '' . $row['post'] . ' | ';
                echo date('M j, Y g:i A', strtotime($row['date']));
                echo '<br /><hr />';
            }**
            }
        ?>
        </div><!-- closes left_col -->

So all the above code is there to query the DB to grab the correct array and then show $row['posts'] in the HTML div below the PHP code, titled left_col. I am trying to do the exact same thing in the next div but instead of echoing $row['posts'], I want to echo rows such as $row['store_city'] to have the page display the store's location after pulling it out of the previously selected array. Here's my non-functioning code for that part:

<div id="right_col">
            <div id="store_picture">
          <img src="images/store.jpg" style="width:325px;"/>
            </div><!-- closes picture --><br />
            <div id="store_address">
                <br /><br /><h2>Location Info</h2>
                <?php
                if (mysqli_num_rows($data) == 1) {
                    while ($row = mysqli_fetch_array($data)) {
                    echo '<p>' . $row['store_city']. '</p>';
                    }
            }
                mysqli_close($dbc);
                ?>

            </div><!-- closes store_address -->
            <div id="store_info">
                <p></p>
            </div><!-- closes store_info -->
        </div><!-- closes right_col -->
        <div class="clear"></div>
    </div><!-- closes store_wrap -->
</div><!-- closes content -->

For whatever reason, the second time, when I try to echo data from that array, I just have empty space within that div. I don't get any errors. I just don't get...anything. Therefore, I think this is a syntax issue. I've tried exactly the same thing I did with the section where I echo $row['post'] and it isn't working.

Answer 1

The chief issue you're facing is that you make a second attempt at fetching rows from the $data MySQLi result resource object after already having fetched them once. This won't work as intended, as MySQL keeps an internal recordset pointer which advances every time mysqli_fetch_array() is called. So when the end of the first loop is reached, the pointer is at the end of the recordset and a subsequent call will return FALSE .

Therefore, your second loop gets nowhere because its first call to mysqli_fetch_array() returns FALSE , exiting the loop. You have two options.

The quickest fix is to just rewind the record pointer using mysqli_data_seek() . Called right before the second loop, it will set the pointer back to the first record, allowing you to fetch them all again.

if (mysqli_num_rows($data) == 1) {
  // Rewind the record pointer back to the first record
  mysqli_data_seek($data, 0);

  while ($row = mysqli_fetch_array($data)) {
    // note addition of htmlspecialchars() to escape the string for HTML!
    echo '<p>' .htmlspecialchars($row['store_city']). '</p>';
  }
}

Perhaps a better option if your recordset is small is to fetch all rows into an array once, then use that array with a foreach loop for both your subsequent output loops:

// To hold all rows
$rows = array();
// Do the query
$data = mysqli_query($dbc, $query2);
// (don't forget to check for errors)
if ($data) {
  while ($row = mysqli_fetch_array($data)) {
    // Append the row onto the $rows array
    $rows[] = $row;
  }
}
else{
  // handle the error somehow...
}

Later, instead of using $data again, you will loop over $rows with a foreach loop.

// use count($rows)
if (count($rows) == 0) {
  echo $blankwall;
}
else {
  foreach ($rows as $row) {
    // Show the posts
    echo '' . $row['post'] . ' | ';
    echo date('M j, Y g:i A', strtotime($row['date']));
    echo '<br /><hr />';
  }
}

...And do the same thing again for your second loop later in the code. It's recommended to use htmlspecialchars() on the string fields output from the query where appropriate.

How to echo a MySQLi row in escaped PHP code, inside an HTML block

Question

1 answers

solution1
0 ACCPTED 2015-01-04 03:27:07

How to echo a MySQLi row in escaped PHP code, inside an HTML block

Question

1 answers

solution1 0 ACCPTED 2015-01-04 03:27:07

solution1
0 ACCPTED 2015-01-04 03:27:07