mnybud Posted November 3, 2007 Share Posted November 3, 2007 Hey guys I have this script that will check a sites indexed pages in Google using the Google API. Right now it requires that you enter one domain in a form and hit submit. Is there a way for me to change this so it loads a list of domains from a text file instead? I am guessing yes but I have been having a tough time getting it working right. Here is the working script that is using the form...If anyone could help me out I would REALLY appreciate it <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.1//EN" "http://www.w3.org/TR/xhtml11/DTD/xhtml11.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en"> <head> <title>Google API DEMO- Number of indexed pages from a site</title> <META NAME="description" CONTENT=" This demo will output the number of links indexed by Google in a site"> <META NAME="keywords" CONTENT="Google API DEMO"> <meta http-equiv="content-type" content="text/html; charset=iso-8859-1" /> <!-- *** This simple demo is put together to illustrate the use of Google API (to query the number of indexed pages) by KC TAN at http://www.kchut.com. No linkback is required but would appreciate if the publisher can leave this notice intact. Thank you. *** --> </head> <body> <?php // include the nusoap class, can be downloaded from http://sourceforge.net/projects/nusoap/ require_once('lib/nusoap.php'); if(isset($_POST['submit'])) { //trim the user's url $qurl=trim($_POST['url']); $mq='site:'.$qurl; $parameters = array( 'key'=> 'Your Google API Key', 'q'=> $mq, 'start'=> '0', 'maxResults'=>'10', 'filter'=> 'false', 'restrict'=>'', 'safeSearch'=>'false', 'lr'=>'', 'ie'=>'', 'oe'=>'' ); //Create a new soap client, feeding it googlesearch.wsdl $soapclient=new soapclient('http://api.google.com/GoogleSearch.wsdl','wsdl'); //save the results into $results array $results = $soapclient->call('doGoogleSearch',$parameters); //save the total number of pages indexed $no_pages=$results['estimatedTotalResultsCount']; // test if any results were returned if(is_array($results['resultElements'])) $enable=1; else $enable=2; } ?> <div align="center"> <form name="demoform" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> <table> <tr><td>Site URL:</td><td><input type="text" name="url" size='45' value="<?php echo $_POST['url']; ?>"> (Eg. http://www.useseo.com)</td></tr> <tr align="center"><td colspan="2"><input type="submit" name="submit" value="Search"></td></tr></table> </form> <?php if($enable==1) echo'<p style="margin-top:15px;">Google has indexed <b>'.number_format($no_pages).'</b> pages from the above domain.</p>'; elseif ($enable==2) echo'<p style="margin-top:10px;color:red;">There are no results returned.</p>'; ?> </body> </html> Quote Link to comment Share on other sites More sharing options...
ignace Posted November 3, 2007 Share Posted November 3, 2007 if(is_array($results['resultElements'])) $enable=1; else $enable=2; // should be if(is_array($results['resultElements'])) { $enable=1; } else { $enable=2; } // or if you want to keep short form if(is_array($results['resultElements'])): $enable=1; else: $enable=2; // but it can be that it is interpreted the wrong way Quote Link to comment Share on other sites More sharing options...
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