php echo href='javascript:void(0);'
Question
I have the following image list on an html page that I am converting into php to be driven from a mysql database. While I thought it was a simple enough procedure -I am struggling to avoid syntax errors because of the existing use of ' in my code.
I need to convert a simple list of images into an array based on what is in the database. The html list is:
<li><a href='javascript:void(0);' rel="{gallery: 'gal1', smallimage: './imgProd/2.jpg',largeimage: './imgProd/2.jpg'}">
<img src='imgProd/2.jpg' style="width:110px; height:110px;"></a></li>
<li><a href='javascript:void(0);' rel="{gallery: 'gal1', smallimage: './imgProd/3.jpg',largeimage: './imgProd/3.jpg'}">
<img src='imgProd/3.jpg' style="width:110px; height:110px;"></a></li>
<li><a href='javascript:void(0);' rel="{gallery: 'gal1', smallimage: './imgProd/4.jpg',largeimage: './imgProd/4.jpg'}">
<img src='imgProd/4.jpg' style="width:110px; height:110px;"></a></li>
I am trying to generate this from php using:
<?php
$result = mysql_query("SELECT * FROM table WHERE id='$id'");
while($row = mysql_fetch_array($result))
{
echo '<li>';
echo '<a href='javascript:void(0);' rel="{gallery: 'gal1', smallimage: './' . $row['photo'] . '',largeimage: './' . $row['photo'] . ''}">';
echo '<img src=' . $row['photo'] . '' style="width:110px; height:110px;">';
echo '</a>';
echo '</li>';
}
?>
but am obviously getting loads of syntax errors as I am attempting to use single quotes within single quotes (I assume!?). Does anyone know how I can incoporate this list into an array??
Thanks very much in advance
JD
5 answers
solution1
1 2011-10-18 21:42:31
You're right that you can't use single quotes within a single-quoted string, you need to use double quotes, or escape them with a backslash - ie
echo '<a href=\'javascript:void(0);\' rel="{gallery: \'gal1\', smallimage: \'./' . $row['photo'] . '\',largeimage: \'./' . $row['photo'] . '\'}">';
The backslash tells PHP to treat the following quote character as a character inside the string, rather than the end of the string.
solution2
1 2011-10-18 21:43:53
You are messing with quotes. PHP doesn't know anything about JS or HTML and yoiu have to escape quotes. A way is the following:
[...]
echo '<a href=\'javascript:void(0);\' rel="{gallery: \'gal1\', smallimage: \'./' . $row['photo'] . '\',largeimage: \'./' . $row['photo'] . '\'}">';
echo '<img src=\'' . $row['photo'] . '\' style="width:110px; height:110px;">';
[...]
for more: Please see http://php.net/string
solution3
1 2011-10-18 21:45:19
Use HEREDOC s:
echo <<<EOL
<li>
<a href="javascript:void(0);" rel="{gallery: 'gal1', smallimage: './{$row['photo']}' , largeimage: './{$row['photo']}'}">
<img src="{$row['photo']}" style="width:110px; height:110px;">
</a>
</li>
EOL;
Presto magico! No more quote issues.
solution4
1 2011-10-18 21:45:30
As a general rule-of-thumb, building blocks of HTML server-side is generally a bad idea. Try using this sort of syntax instead...
while($row = mysql_fetch_array($result)) : ?>
<li>
<a href="javascript:void(0)">
<img src="<?php echo htmlspecialchars($row['photo']) ?>" style="...">
</a>
</li>
<?php endwhile ?>
solution5
1 ACCPTED 2011-10-18 21:50:54
Escaping the quotes is a bad idea.
You can either simply close and reopen your <?php ?> tags or use Heredoc syntax .
Using tags (PHP is a templating language, anything that isn't between <?php ?> tags will be send to the client exactly as it is):
<?php
$result = mysql_query("SELECT * FROM table WHERE id='$id'");
while($row = mysql_fetch_array($result))
{
?>
<li>
<a href='javascript:void(0);' rel="{gallery: 'gal1', smallimage: './<?php echo $row['photo']; ?>',largeimage: './<?php echo $row['photo']; ?>'}">
<img src='<?php echo $row['photo']; ?>' style="width:110px; height:110px;">
</a>
</li><?php
}
?>
Using Heredoc (the closing HTML; should be on its on line, unindented):
<?php
$result = mysql_query("SELECT * FROM table WHERE id='$id'");
while($row = mysql_fetch_array($result))
{
echo <<<HTML
<li>
<a href='javascript:void(0);' rel="{gallery: 'gal1', smallimage: './{$row['photo']}',largeimage: './{$row['photo']}'}">
<img src='{$row['photo']}' style="width:110px; height:110px;">
</a>
</li>
HTML;
}
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.