问题
i want to add images dynamically in my rss feed but its only showing thumbs then it disappear see the sample here
my images are stored on my server at a folder called "allimages" and all pages are retrieved from mysql database. where em i wrong!!!?
Any help ll be appreciated thaanx below is my code
<?php
header("Content-type: text/xml");
$connection = mysql_connect( "------", "-----", "-----");
mysql_select_db("-------", $connection);
$sundo = "SELECT * FROM NEWSPAGES ORDER BY ID DESC LIMIT 20";
$query = mysql_query($sundo ) or die(mysql_error());
echo (" <rss version=\"2.0\"> ");
echo (" <channel> ");
echo (" <title>People</title>");
echo (" <link>http://tabata.com</link>");
echo (" <description>tabata</description> ");
echo " <copyright>Copyright 2013</copyright> \ n";
echo " <managingEditor>tabata(sun)</managingEditor> \ n";
echo " <webMaster>tabata.com (sun)</webMaster> \ n";
echo (" <language>IT- en</language>");
while($array = mysql_fetch_array( $query )) {
extract($array);
echo "<item>
<title>$title</title>
<link>http://tabata.com/news.php?page=$id</link>
<description><![CDATA[<img src=\"http://tabata.com/$allimages\" width=\"57\" height=\"57\" />]]>$description</description>
<image>
<link>http://tabata.com/news.php?page=$id</link>
<url>http://tabata.com/$allimages</url>
<title>$title</title>
</image>
</item>";
}
echo " </channel> </rss>" ;
?>
回答1:
First of all, your code becomes quite unreadable because you use SELECT *
and extract()
.
You could better replace them with SELECT field1, field2, etc
and I suggest avoiding the extract()
function all-time, you suggest you'd use something like $field1 = $array['field1']
.
Then it should be easier to find out if the field allimages
exists in your query result. Test each step of the process with print_r()
or var_dump()
to check if a value contains the expected result.
回答2:
Save image names/full url in you database, create a column of images let say images
. Then in you while loop you can get images like $array['images']
.
Like
<image>
...
<url>http://tabata.com/$array['images']</url>
...
</image>
Like for page http://tabelltz.com/news.php?page=31
you image column should have img name 27.jpg
and url like
<url>http://tabata.com/userfiles/$images</url>
I use $images
as you're using extract($array);
Better if you don't use extract($array);
来源:https://stackoverflow.com/questions/23115259/how-to-add-different-images-in-my-rss-feed-using-php