doudiao2335 2016-03-25 21:41
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在显示MySQL数据时需要php / html的建议

So I created this script to test out if I could output from my database. And it works fine inside the php, but when I trying extracting that data inside a div tag in HTML, it does not work. Any suggestions, thanks.

//Connection to database, working fine, top of page
<?php
require "include/mysql.php";
if (!$con) {
die('Could not connect: ' . mysql_error());
}
?>

//This is also working as intended
<?php 
$sql = "SELECT * FROM `LISTINGS` ORDER BY `LISTINGS`.`YA` DESC";
$myData = mysql_query($sql,$con);
while($record = mysql_fetch_array($myData)){

    echo $record['TITLE'];

}
?>

//Here is were the problem, in the H2 tag, i'm trying to output a title but it is not working. 
<div class="cards-row">

    <div class="card-row">
        <div class="card-row-inner">
            <div class="card-row-image" data-background-image="assets/img/tmp/product-1.jpg">


            </div><!-- /.card-row-image -->

            <div class="card-row-body">
                <h2 class="card-row-title"><?php echo $record['TITLE'];?>title</h2>
                <div class="card-row-content">Description</div><!-- /.card-row-content -->
            </div><!-- /.card-row-body -->

            <div class="card-row-properties">
                <dl>

                        <dd>Test</dd><dt>Visit Website</dt>
                        <dd>Test</dd><dt>More Info</dt>                  
                        <dd>Added</dd><dt>AddedDate</dt>
                        <dd>Visited</dd><dt>Visited</dt>

                </dl>
            </div><!-- /.card-row-properties -->
        </div><!-- /.card-row-inner -->
    </div><!-- /.card-row -->


</div><!-- /.cards-row -->

I know there is a lot of unfinished code but im just testing a few things out at the moment, sorry for bad grammar.

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2条回答 默认 最新

  • duannuo7878 2016-03-25 21:45
    关注

    You cannot access $record outside of the while loop. If there is only one record you want to access just get rid of the loop and write $record = mysql_fetch_array($myData); on its own.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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