Innovati0n Posted December 20, 2010 Share Posted December 20, 2010 I’m new to PHP so I’m just to get to grips with a few things. It's giving me the error shown below. You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '-10, 10' at line 1 What I'm trying to make it so is give me some text when nothing is in the MySQL table, such as "No articles just yet". I also don't want any of the buttons to show. <?php $table = "news"; $prows = 10; $spread = 6; $page = $_GET['p']; $query = mysql_query("SELECT * FROM $table ORDER BY id ASC"); $rows = mysql_num_rows($query); $lastp = ceil($rows / $prows); // ### if (!isset($page)) { $page = 1; } if ($page < 1) { $page = 1; } if ($page > $lastp) { $page = $lastp; } // ### $page = isset($_GET['p']) ? $_GET['p'] : 1; $pn = ($page * $prows) - $prows; // ALL MY CODE IS HERE!!!!!!!!!!!! } echo("\t<div class=\"pagination_wrapper\">\n"); echo("\t <div class=\"pagination\">"); // ### if ($page == 1) { echo(""); } else { echo("<a href='index.php?p=1' class=\"pagination\">First</a>"); } // ### if ($page > 1) { $prev = $page - 1; echo("<a href='index.php?p=$prev' class=\"pagination\">Previous</a>"); } else { echo(""); } // ### $min = $page - $spread; $max = $page + $spread; if ($min < 1) { $dif = 1 - $min; $min = 1; $max = $max + $dif; } if ($max > $lastp) { $dif = $lastp - $max; $max = $lastp; $min = $min + $dif; } if ($min < 1) { $min = 1; } // ### for ($i = $min; $i <= $max; $i++) { if ($i == $page) { echo("<a href='index.php?p=$i' class=\"pagination selected\">$i</a>"); } else { echo("<a href='index.php?p=$i' class=\"pagination\">$i</a>"); } } // ### if ($page < $lastp) { $next = $page + 1; echo("<a href='index.php?p=$next' class=\"pagination\">Next</a>"); } else { echo(""); } // ### if ($page == $lastp) { echo(""); } else { echo("<a href='index.php?p=$lastp' class=\"pagination\">Last</a>"); } echo("</div>\n"); echo("\t</div>\n"); // ### ?> If you could help then it would be amazing, thank you. Quote Link to comment Share on other sites More sharing options...
BlueSkyIS Posted December 20, 2010 Share Posted December 20, 2010 take a look at your SQL to see what's wrong. i always do this with my queries: $sql = "SELECT * FROM $table ORDER BY id ASC"; $result = mysql_query($sql) or die (mysql_error() . " IN $sql); but i don't think that error is coming from the code you posted. Quote Link to comment Share on other sites More sharing options...
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