turpentyne Posted April 22, 2011 Share Posted April 22, 2011 It's been a while since I sat down to build some pages and teach myself php. So now that I've started back up, I'm at a loss for what I've done. I deleted a file, and have to rebuild from an old broken version: I have a form that submits a query to the database, but the results pages is giving me this error: Oops, my query failed. The query is: SELECT COUNT 'descriptors'.* ,'plantae'.* FROM 'descriptors' LEFT JOIN 'plantae' ON ('descriptors'.'plant_id' = 'plantae'.'plant_name') WHERE 'leaf_shape' LIKE '%auriculate%' AND 'leaf_venation' LIKE '%%' AND 'leaf_margin' LIKE '%%' The error is: 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 '.* ,'plantae'.* FROM ' at line 2 But I'm not seeing what the syntax error is. Here's the code: <?php require ('connection.php'); $display = 2; // it's intentionally only 2 for the moment if (isset($_GET['np'])) { $num_pages = $_GET['np']; } else { $data = "SELECT COUNT 'descriptors'.* ,'plantae'.* FROM 'descriptors' LEFT JOIN 'plantae' ON ('descriptors'.'plant_id' = 'plantae'.'plant_name') WHERE 'leaf_shape' LIKE '%$s1%' AND 'leaf_venation' LIKE '%$s3%' AND 'leaf_margin' LIKE '%$s4%'"; $result = mysql_query ($data); if (!$result) { die("Oops, my query failed. The query is: <br>$data<br>The error is:<br>".mysql_error()); } $row = mysql_fetch_array($result, MYSQL_NUM); $num_records = $row[0]; if ($num_records > $display) { $num_pages = ceil ($num_records/$display); } else { $num_pages = 1; } } if (isset($_GET['s'])) { $start = $_GET['s']; } else { $start = 0; } if(isset($_POST[submitted])) { // Now collect all info into $item variable $shape = $_POST['s1']; $color = $_POST['s2']; $vein = $_POST['s3']; $margin = $_POST['s4']; // This will take all info from database where row tutorial is $item and collects it into $data variable $data = mysql_query("SELECT 'descriptors'.* ,'plantae'.* FROM 'descriptors' LEFT JOIN 'plantae' ON ('descriptors'.'plant_id' = 'plantae'.'plant_name') WHERE 'leaf_shape` LIKE '%$s1%' AND 'leaf_venation' LIKE '%$s3%' AND 'leaf_margin' LIKE '%$s4%' ORDER BY 'plantae'.'scientific_name` ASC LIMIT $start, $display"); //chs added this in... echo '<table align="center" cellspacing="0" cellpading-"5"> <tr> <td align="left"><b></b></td> <td align="left"><b></b></td> <td align="left"><b>Leaf margin</b></td> <td align="left"><b>Leaf venation</b></td> </tr> '; //end something chs added in // This creates a loop which will repeat itself until there are no more rows to select from the database. We getting the field names and storing them in the $row variable. This makes it easier to echo each field. while($row = mysql_fetch_array($data)){ echo '<tr> <td align="left"> <a href="link.php">View plant</a> </td> <td align="left"> <a href="link.php">unknown link</a> </td> <td align="left">' . $row['scientific_name'] . '</td> <td align="left">' . $row['common_name'] . '</td> <td align="left">' . $row['leaf_shape'] . '</td> </tr>'; } echo '</table>'; // row 95 } if ($num_pages > 1) { echo '<br /><p>'; $current_page = ($start/$display) + 1; // row 100 if ($current_page != 1) { echo '<a href="leafsearch2a.php?s=' . ($start - $display) . '&np=;' . $num_pages . '">Previous</a> '; } for ($i = 1; $i <= $num_pages; $i++) { if($i != $current_page) { echo '<a href="leafsearch2a.php?s=' . (($display * ($i - 1))) . '$np=' . $num_pages . '">' . $i . '</a>'; } else { echo $i . ' '; } } if ($current_page != $num_pages) { echo '<a href="leafsearch2a.php?s=' . ($start + $display) . '$np=' . $num_pages . '"> Next</a>'; } } //added curly ?> Quote Link to comment https://forums.phpfreaks.com/topic/234485-syntax-error-on-left-join-query/ Share on other sites More sharing options...
analog Posted April 22, 2011 Share Posted April 22, 2011 The single quotes around the table names and columns should be backticks (`). Also should it be COUNT(*)? This should work: SELECT COUNT(*), `descriptors`.*, `plantae`.* FROM `descriptors` LEFT JOIN `plantae` ON (`descriptors`.`plant_id` = `plantae`.`plant_name`) WHERE `leaf_shape` LIKE '%$s1%' AND `leaf_venation` LIKE '%$s3%' AND `leaf_margin` LIKE '%$s4%' If the variables $s1, $s3 & $s4 are coming from user input make sure to use mysql_real_escape_string to avoid SQL Injection attacks. Quote Link to comment https://forums.phpfreaks.com/topic/234485-syntax-error-on-left-join-query/#findComment-1205101 Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.