我一直在試圖將PHP放入PHP中一段時間,但我永遠無法使它工作。相反,我不得不使用這個PHP內部的HTML不能正常工作
<?php //code in here
while($record = mysql_fetch_array($myData)){
?>
//HTML code
<?php
}
?>
它已經工作了一段時間,但我現在推行的網頁搜索功能,我需要所有的HTML動態地生成,或頁面會顯示兩次。因此,我需要一些幫助。這是我目前擁有的新舊代碼。
<?php
$sql = "SELECT * FROM `LISTINGS` ORDER BY YA DESC $limit";
$myData = mysql_query($sql,$con);
while($record = mysql_fetch_array($myData)){
//this is what i have manage to get done
echo '<div class="cards-row">';
echo '<div class="card-row">';
echo '<div class="card-row-inner">';
echo '<div class="card-row-image" data-background-image="http://domains/$record["IMAGENAME"].jpg">';
echo '</div>';
echo '<div class="card-row-body">';
echo '<h2 class="card-row-title">';
echo '<a href="http://domain/listing-detail.php?ID='; $record['ID']echo'">'
echo '</a>';
echo '</h2>';
echo '</div>';
echo '</div>';
echo '</div>';
echo '</div>';
?>
靜態/動態代碼
<div class="card-row">
<div class="card-row-inner">
<div class="card-row-image" data-background-image="http://www.domain/<?php echo $record['IMAGENAME'];?>.jpg" width="150" height="180" ">
</div><!-- /.card-row-image -->
<div class="card-row-body">
<h2 class="card-row-title">
<a href="http://www.domain/listing-detail.php?ID=<?php echo $record['ID'];?>"> <?php custom_echo ($record['TITLE'], 80); ?><?php if($record['STREET']===' '){ echo "Not provided" ;}?>
</a>
</h2>
<div class="card-row-content"><?php custom_echo ($record['DESCRIPTION'], 250); ?><?php if($record['STREET']===' '){ echo "Not provided" ;}?></div><!-- /.card-row-content -->
</div><!-- /.card-row-body -->
<div class="card-row-properties">
<dl>
<dd></dd><dt><a href="<?php echo $record['WEBSITE'];?>">Visit Website</a></dt>
<dd></dd><dt><a href="http://www.domain/listing-detail.php?ID=<?php echo $record['ID'];?>">More Info</a></dt>
<dd>Added</dd><dt><?php echo $record['YA'];?>/<?php echo $record['MU'];?>/<?php echo $record['DU'];?></dt>
<dd>Viewed</dd><dt>Visited</dt>
</dl>
</div><!-- /.card-row-properties -->
</div><!-- /.card-row-inner -->
</div><!-- /.card-row -->
</div><!-- /.cards-row -->
<br/>
<?php
}
?>
我的目標是把所有上面的代碼中
while($record = mysql_fetch_array($myData)){
請[停止使用'mysql_ *'功能](http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php)。 [這些擴展](http://php.net/manual/en/migration70.removed-exts-sapis.php)已在PHP 7中刪除。瞭解[編寫](http://en.wikipedia.org/ wiki/Prepared_statement)語句[PDO](http://php.net/manual/en/pdo.prepared-statements.php)和[MySQLi](http://php.net/manual/en/mysqli.quickstart .prepared-statements.php)並考慮使用PDO,[這真的很簡單](http://jayblanchard.net/demystifying_php_pdo.html)。 –
[您的腳本存在SQL注入攻擊風險。](http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php) –
您的目標是將代碼循環內?所以把它放在循環中。我真的不知道你有什麼問題。你目前的產出和期望的產出是多少?第二和第三代碼塊如何組合在一起? – Mike