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RSS feed error on Google Chrome (5 posts)

  1. itechnified
    Member
    Posted 3 years ago #

    Hi guys,

    I am facing problem with feed on Google Chrome. In firefox its working fine, even in IE its just showing XML file code instead of feed.

    thsi is my blog feed url
    technified.net/feed

    I am getting This error on Chrome

    This page contains the following errors:

    error on line 12 at column 6: XML declaration allowed only at the start of the document
    Below is a rendering of the page up to the first error.

    please suggest me how to resolve it.

  2. esmi
    Forum Moderator
    Posted 3 years ago #

    Have you tried:

    - deactivating all plugins to see if this resolves the problem? If this works, re-activate the plugins one by one until you find the problematic plugin(s).

    - switching to the Twenty Ten theme to rule out any theme-specific problems?

    - resetting the plugins folder by FTP or phpMyAdmin? Sometimes, an apparently inactive plugin can still cause problems.

  3. dikoro
    Member
    Posted 3 years ago #

    try it with Safari everything looks very good but G. Chrome mess my feeds all up.

  4. brasofilo
    Member
    Posted 3 years ago #

    incredibly, but looks like chrome doesn't support feeds without an extension...

    here there's a way of getting support for rss in chrome

    i adapted the technique for use in my rss button

    so if browser is chrome = true, change value of href:

    <?php
    $isChrome = eregi("chrome", $_SERVER['HTTP_USER_AGENT']);
    $rss = $MY_NORMAL_RSS_URL;
    if($isChrome) $rss="javascript:(%7BaddURL:function(aURL)%7Bvar%20url=%22http://feeds.ramisp.org/preview.php?utf=y&src=%22+encodeURIComponent(aURL);location.href=url;%7D,getFeedURL:function(e,aDocument)%7Bvar%20j=e;var%20c=aDocument.location;if(e.indexOf(%22/%22)!=0)%7Bvar%20d=c.pathname.split(%22/%22);d%5Bd.length-1%5D=e;j=d.join(%22/%22)%7Dreturn%20c.protocol+%22//%22+c.hostname+j;%7D,checkForFeeds:function()%7Bvar%20f=false;var%20m=document.getElementsByTagName(%22link%22);for(var%20g=0,a;a=m%5Bg%5D;g++)%7Bvar%20h=a.getAttribute(%22type%22);var%20i=a.getAttribute(%22rel%22);if(h&&h.match(/%5B%5C+%5C/%5Dxml%24/)&&i&&i==%22alternate%22)%7Bvar%20b=a.getAttribute(%22href%22);if(b.indexOf(%22http%22)!=0)%7Bb=this.getFeedURL(b,document);%7Dthis.addURL(b);f=true;break%7D%7Dif(!f)alert(%22Oops.%20Can't%20find%20a%20feed.%22);%7D%7D).checkForFeeds();";
    ?>
    <a href="<?php echo $rss; ?>" target="_blank"><img src="/images/rss.png" alt="RSS Feed" title="RSS Feed" /></a>

    at work here: http://www.abuzza.com.br

  5. Curtiss Grymala
    Member
    Posted 3 years ago #

    The problem is that something is inserting 11 blank lines at the top of the feed. Therefore, it is not a valid XML file.

    While most browsers, as you've found, will still render it okay, it's still not valid. The beginning of the XML declaration has to be the first character in the document; otherwise it's considered invalid.

    Are you using any kind of customized theme/template for your feeds?

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