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Sware

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  1. Is there a way to make that PHP rss feed into a Javascript Variable then?
  2. I don't think it works, it doesn't display anything.. How do I fix it?
  3. I changed to this.. <?php echo "var pausecontent=new Array();"; echo "var rsstxt=" . $rssticker->recent-post->subject . ";"; echo "pausecontent[0]='rsstxt';"; echo "pausecontent[1]='<a href='#'>New Divisions Launched</a> - <i>7 Hours Ago</i>';"; echo "pausecontent[2]='<a href='#'>Welcome New Leaders</a> - <i>20 Hours Ago</i>';"; ?> Now i'm getting "Parse error: syntax error, unexpected T_OBJECT_OPERATOR, expecting ',' or ';'"
  4. Bascially I have a Javascript script that has to have PHP elements in it. When I try to echo it all it doesn't work, and when I put PHP in the variable it still doesnt work.. Any luck? This is what I currently have.. <?php echo "var pausecontent=new Array()"; echo "pausecontent[0]='$rssticker->recent-post->subject'"; echo "pausecontent[1]='<a href='#'>New Divisions Launched</a> - <i>7 Hours Ago</i>'"; echo "pausecontent[2]='<a href='#'>Welcome New Leaders</a> - <i>20 Hours Ago</i>'"; ?> I tried it this way also. pausecontent[0]='<?php $rssticker->recent-post->subject ?>'; No worky. Any suggestions?
  5. I need help I don't know if it's quite possible in PHP but I need it when an image is clicked it changes into a new image. So the peusdo way is like Image Click (Default: img1.png) Image Click --> Change to img2.png Image Click --> Change to img3.png Image Click --> Change to img4.png Then it raps around to Img1.png again. Anyone know a good way about doing this?
  6. Paying 5$ per request and 10$ overall if someone can help me or do themselves (I'm not THAT great with PHP coding). Basically take this PHP script http://www.hotscripts.com/Detailed/70852.html and having an option to format it as HTML when sent out as email. If someone can do that I'll pay $5. also another $5 for whoever can make it so you can text box to put in multi emails address in it for like a mailing list of some sort. Not sure if its possible. I can only pay via paypal.
  7. [code]<?PHP                //MYSQL CONNECTION $dbh=mysql_connect ("localhost", "XXXX", "XXXXXX") or die ('I cannot connect to the database because: ' . mysql_error()); mysql_select_db ("sware_phpb2"); $query = "SELECT phpbb_users.username, phpbb_posts_text.post_subject, phpbb_posts_text.post_text, phpbb_posts.post_id, phpbb_posts.topic_id, phpbb_posts.forum_id FROM phpbb_users JOIN phpbb_posts ON phpbb_posts.poster_id = phpbb_users.user_id JOIN phpbb_posts_text ON phpbb_posts.post_id = phpbb_posts_text.post_id ORDER BY post_id DESC"; $result = mysql_query($query) or die(mysql_error()); // Contents while($row = mysql_fetch_array($result)){ echo "<br>";     echo "<table border="0" cellpadding="0" cellspacing="2" width="510" class="pageoutline">";           echo "<tr>";             echo "<td width="504" height="19" valign="top" bgcolor="#A3C159"><div align="center" class="navtitle">";               echo "<div align="center">". $row['post_subject'] ."</div>";             echo "</div></td>";             echo  "</tr>";           echo "<tr>";             echo "<td height="45" valign="top">""; echo "Posted By:". $row['username'] .".";               echo "<p align="center">". $row['post_text'] .";                 echo "<br>";             echo  "</tr>";           echo "</table>";   } ?>[/code] It says [code]Parse error: syntax error, unexpected T_LNUMBER, expecting ',' or ';' in /home/sware/public_html/test.php on line 13[/code] Line 13 is [code]    echo "<table border="0" cellpadding="0" cellspacing="2" width="510" class="pageoutline">";[/code]
  8. How do I add another join onto that [code]SELECT phpbb_posts.poster_id, phpbb_users.username FROM phpbb_posts JOIN phpbb_users ON phpbb_posts.poster_id = phpbb_users.poster_id (+ ANOTHER JOIN)[/code] or can you not do that?
  9. $query = "SELECT phpbb_posts.poster_id, phpbb_users.username FROM phpbb_posts, phpbb_users WHERE phpbb_posts.poster_id = phpbb_users.poster_id";  something like that? yes.
  10. Bump - Looked at once, not cool.
  11. I was doing fine and all of a sudden.. Bam! Not unique table/alias: 'phpbb_posts' I'm new to MySQL and trying to create a portal with phpBB. $query = "SELECT phpbb_posts.poster_id, phpbb_users.username ". "FROM phpbb_posts, phpbb_users". "WHERE phpbb_posts.poster_id = phpbb_users.poster_id"; $result = mysql_query($query) or die(mysql_error()); // Print out the contents of each row into a table while($row = mysql_fetch_array($result)){ echo $row['poster_id']; echo $row['username']; echo "<br />"; } Just a little test but it shows that error, ANY help will be thanked for.
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