contra10 Posted January 31, 2009 Share Posted January 31, 2009 this script for some reason doesnt work <script type="text/javascript"> <?php echo "var commentsbook={\n"; //Dynamically output javascript variable $commentids=mysql_query("SELECT * FROM post_profile WHERE userid = '$idp' ORDER BY ppid DESC") or die(mysql_error()); //get IDs to last 5 comment pages $idarray=array(); while ($theid=mysql_fetch_array($commentids)){ array_push($idarray, "'http://localhost/mypost/index.php?user='".$idp."'&id=" . $theid[id] . "'"); } echo "pages: [" . implode(",", $idarray) . "],\n"; //output: pages: echo "selectedpage: 0\n}"; ?> var comments=new ajaxpageclass.createBook(commentsbook, "bookdiv", ["paginate-top", "paginate-bottom"]) </script> i get no mysql errors Quote Link to comment Share on other sites More sharing options...
DeanWhitehouse Posted January 31, 2009 Share Posted January 31, 2009 what exactly isn't it doing? Quote Link to comment Share on other sites More sharing options...
contra10 Posted January 31, 2009 Author Share Posted January 31, 2009 its suppose to show the posts that a user left...I'm trying to make the pagination dynamic so that a page doesn't reload when clicking onto anothe page i uused this before as pagination <?php if(is_numeric($_GET['user'])){ $idp = $_GET['user']; $insert3= "SELECT * FROM post_profile WHERE userid = '$idp' ORDER BY ppid DESC"; $topic3 = mysql_query($insert3) or die(mysql_error()); } //This checks to see if there is a page number. If not, it will set it to page 1 if (!isset($pagenum)) { $pagenum = (isset($_GET['pagenum'])) ? $_GET['pagenum'] : 1; } //Here we count the number of results //Edit $data to be your query $data = mysql_query("SELECT * FROM post_profile WHERE userid = '$idp' ORDER BY ppid DESC") or die(mysql_error()); $rows = mysql_num_rows($data); //This is the number of results displayed per page $page_rows = 4; //This tells us the page number of our last page $last = ceil($rows/$page_rows); //this makes sure the page number isn't below one, or more than our maximum pages if ($pagenum < 1) { $pagenum = 1; } elseif ($pagenum > $last) { $pagenum = $last; } //This sets the range to display in our query $max = 'limit ' .($pagenum - 1) * $page_rows .',' .$page_rows; //This is your query again, the same one... the only difference is we add $max into it $data_p = "SELECT * FROM post_profile WHERE userid = '$idp' ORDER BY ppid DESC LIMIT 10"; $posts = mysql_query($data_p) or die(mysql_error()); //This is where you display your query results while($info = mysql_fetch_array($posts)) { $userpost= "{$info['post']}"; $usernamep= "{$info['postingusername']}"; $userdate= "{$info['datepost']}"; echo "<table border='0' align='center'>"; echo "<tr>"; echo"<td width= '500' align='center' bgcolor='black'><FONT FACE='ariel' SIZE='2' color='#0094f7'> $userpost</td>"; echo "</tr>"; echo "<tr>"; echo "<td align='right'><FONT FACE='ariel' SIZE='2' color='#0094f7'>Posted by $usernamep on $userdate</td>"; echo "<tr>"; echo"</table>"; } echo "<p>"; // This shows the user what page they are on, and the total number of pages echo " --Page $pagenum of $last-- <p>"; // First we check if we are on page one. If we are then we don't need a link to the previous page or the first page so we do nothing. If we aren't then we generate links to the first page, and to the previous page. if ($pagenum == 1) { } else { echo " <a href='http://localhost/mypost/index.php?user=$id&pagenum=1'> <<-First</a> "; echo " "; $previous = $pagenum-1; echo " <a href='http://localhost/mypost/index.php?user=$id&pagenum=$previous'> <-Previous</a> "; } //just a spacer echo " ---- "; //This does the same as above, only checking if we are on the last page, and then generating the Next and Last links if ($pagenum == $last) { } else { $next = $pagenum+1; echo " <a href='http://localhost/mypost/index.php?user=$id&pagenum=$next'>Next -></a> "; echo " "; echo " <a href='http://localhost/mypost/index.php?user=$id&pagenum=$last'>Last ->></a> "; } ?> Quote Link to comment Share on other sites More sharing options...
DeanWhitehouse Posted January 31, 2009 Share Posted January 31, 2009 Wow, i asked one question and amazingly you failed to anwser Quote Link to comment Share on other sites More sharing options...
contra10 Posted January 31, 2009 Author Share Posted January 31, 2009 not showing results Quote Link to comment Share on other sites More sharing options...
DeanWhitehouse Posted January 31, 2009 Share Posted January 31, 2009 Ok first off make sure you have connected to the database, to check add if(mysql_ping()) { echo "Connected"; } else echo "not"; put that above the query, then after the query put echo mysql_num_rows($commentids); And tell me the output Quote Link to comment Share on other sites More sharing options...
contra10 Posted January 31, 2009 Author Share Posted January 31, 2009 <div id="paginate-top"> </div> <div id="bookcontent"> </div> <div id="paginate-bottom"> </div> <p><a href="javascript:mybookinstance.selectpage(1)">Select 2nd page dynamically</a></p> <script type="text/javascript"> <?php mysql_connect("localhost", "root", "") or die(mysql_error()); mysql_select_db("registration") or die(mysql_error()); if(mysql_ping()) { echo "Connected"; } else{ echo "not"; } echo "var commentsbook={\n"; //Dynamically output javascript variable $commentids=mysql_query("SELECT `ppid` from `post_profile` limit 5"); //get IDs to last 5 comment pages echo mysql_num_rows($commentids); $idarray=array(); while ($theid=mysql_fetch_array($commentids)){ array_push($idarray, "'http://www.localhost.com/test.php?id=" . $theid[id] . "'"); } echo "pages: [" . implode(",", $idarray) . "],\n"; //output: pages: [filepath1, filepath2, filepath3, etc], echo "selectedpage: 0\n}"; ?> var comments=new ajaxpageclass.createBook(commentsbook, "bookdiv", ["paginate-top", "paginate-bottom"]) </script> </body> nothing echoes Quote Link to comment Share on other sites More sharing options...
DarkSuperHero Posted January 31, 2009 Share Posted January 31, 2009 are you looking at your pages source when you test out the page ? your php is inside <script> tags </script> that might be preventing it from showing anythign to the browser, but only visible when you see source.... just a thought...i could be wrong.. :-) Quote Link to comment Share on other sites More sharing options...
contra10 Posted January 31, 2009 Author Share Posted January 31, 2009 i c what your saying... perhaps this <div id="paginate-top"> </div> <div id="bookcontent"> <?php mysql_connect("localhost", "root", "") or die(mysql_error()); mysql_select_db("registration") or die(mysql_error()); if(mysql_ping()) { echo "Connected"; } else{ echo "not"; } echo "var commentsbook={\n"; //Dynamically output javascript variable $commentids=mysql_query("SELECT `ppid` from `post_profile` limit 5"); //get IDs to last 5 comment pages echo mysql_num_rows($commentids); $idarray=array(); while ($theid=mysql_fetch_array($commentids)){ array_push($idarray, "'http://www.localhost.com/test.php?id=" . $theid[id] . "'"); } echo "pages: [" . implode(",", $idarray) . "],\n"; //output: pages: [filepath1, filepath2, filepath3, etc], echo "selectedpage: 0\n}"; ?> </div> <div id="paginate-bottom"> </div> <script type="text/javascript"> var comments=new ajaxpageclass.createBook(commentsbook, "bookdiv", ["paginate-top", "paginate-bottom"]) </script> i get error Connectedvar commentsbook={ 2pages: ['http://www.localhost.com/test.php?id=1','http://www.localhost.com/test.php?id=2'], selectedpage: 0 } Quote Link to comment Share on other sites More sharing options...
DeanWhitehouse Posted January 31, 2009 Share Posted January 31, 2009 No you don't that is what you are telling it to print, btw i might be able to save you some hassle with my kinda ajax pagination using mysql http://djw-webdesign.awardspace.com/code.php?snippet=11 Quote Link to comment Share on other sites More sharing options...
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