php - Why does this XML not render in the browser the way I want? -

-

i want use php create rss mysql. can see content in page source code. can not see item part in web browser(ie, firefox or opera). web browser show

`your rss feed name or website name

a description of feed or site.` 

<?php require_once ('mysql_connect.php');      //set xml header     header('content-type: text/xml');      //construct rss feed headers     $output = '<rss version="2.0">';     $output .= '<channel>';     $output .= '<title>your rss feed name or website name</title>';     $output .= '<description>a description of feed or site.</description>';     $output .= '<link>http://www.yoursite.com/</link>';     $output .= '<copyright>your copyright details</copyright>';      //body of rss feed     mysql_select_db("rss", $db);     $result = mysql_query("select * rss limit 15",$db);         while ($row = mysql_fetch_array($result)) {             $output .= '<item>';                 $output .= '<title>'. $row['title'] .'</title>';                 $output .= '<description>'. $row['content'] .'</description>';                 $output .= '<link>'. $row['link'] .'</link>';                 $output .= '<pubdate></pubdate>';             $output .= '</item> ';         }     mysql_close($db);      //close rss feed     $output .= '</channel>';     $output .= '</rss>';      //send complete rss feed browser     echo($output);  ?>

the xml source looks like:

<rss version="2.0"> <channel> <title>your rss feed name or website name</title> <description>a description of feed or site.</description> <link>http://www.yoursite.com/</link> <copyright>your copyright details</copyright>   <item>     <title>milan, italy - september 26: models walk runway @ emi&hellip</title>            <description>date: sep 26, 2009 7:45 pmnumber of comments on photo:0view photo&hellip;</description>     <link>http://picasaweb.google.com/roxyluvtony/kendraspears#5551895410815389042</link>     <pubdate></pubdate>   </item>   ... </channel> </rss>

moving comment answer...

it's not valid xml because of &hellip; entity in <title> , <description> nodes.

wrap strings <![cdata[tag]]> , try again.


Comments

Post a Comment

Popular posts from this blog

java - SNMP4J General Variable Binding Error -

-
i trying use snmp4j snmp bulkget. when ever use snmp4j make call returned responseevent error says "general variable binding error" , data returned equal null. to debug: print out, console, exact oid , version number using in snmp4j. use printed out data snmpwalk on command line. valid results. know sending snmp4j correct pdu correct oid, version number, max repititions, etc. i have used snmp4j , code wrote succesffuly monitor other devices years. don't know different time. leaving me stumped. why "general variable binding error?" causes error? ideas debug? can reproduced? if so, first action should using wireshark or microsoft network monitor capture network packets. my guess agent gives generr response, not surprising, http://www.ietf.org/rfc/rfc1157.txt
Read more

windows - Python Service Installation - "Could not find PythonClass entry" -

-
i working on building python service. inducing error, not sure inducing it. ps c:\...> python .\file.py debug debugging service myservice - press ctrl+c stop. error 0xc00000f4 - not find service's pythonclass entry in registry error 1814 - specified resource name cannot found in image file. error 0xc0000080 - not locate module name in python class string (ie, no '.') this error coming pythonservicemessages.mc in pythonwin32. it turns out error result caused exception being thrown , python script dying off.
Read more

Determine if a XmlNode is empty or null in C#? -

-
the following code takes xmlnode data type , populates dataset object xmlnode content. write dataset's content file. public void populatedataset(xmlnode node) { xmlnodereader reader = new xmlnodereader(node); dataset ds = new dataset(); ds.readxml(reader); system.guid guid = system.guid.newguid(); string name = string.format("{0}{1}_{2}.xml", utility.xmloutputpath, utility.xmloutputfilename, guid.tostring()); //need write "node empty" file if xmlnode object empty of null ds.writexml(name, xmlwritemode.ignoreschema); } the problem encountered 1 scenario not write content file. how determine if xmlnode object null or empty? you can check if node parameter null or has innertext or innerxml properties null or empty, when enter method before creating xmlnodereader .
Read more