Re: parse error in php backend sqlite
| From: | Doug Thompson | Date: | Wed, 20 Oct 2004 23:28:38 +0000 |
| Subject: | Re: parse error in php backend sqlite | ||
| References: | 1 | Groups: | php.db |
| Request: | Send a blank email to php-db+get-36947@lists.php.net to get a copy of this message | ||
Aravalli Sai wrote:
hi this is php code for inventory management system which is done using SQLite.....this is an user interface where an user wil giv values to the text boxes and it should be saved in the backend database which is sqlite.. but when i run this code on browser it is giving an error ... Parse error: parse error in /home/saravall/.HTML/inv.php on line 63 <HTML> <HEAD> </HEAD> <body> <?php $db ="/home/saravall/office.db"; $handle = sqlite_open($db) or die("could not open database"); if(isset($_POST['submit'])) {Your braces don't match up. The problem starts at the ELSE statement in line 33. Dougif (!empty($_post['tagno']) && !empty($_post['desc']) && !empty ($_post['acqdt']) && !empty ($_post['manufacturer']) && !empty($_post['model']) && !empty($_post['serialid']) &&!empty( $_post ['custloc']) && !empty($_post['totcost'])) { $insquery = "insert into inventory (tagno,desc,acqdt,manufacturer,model, serial id,custloc,totcost) values (\"".sqlite_escape_string( $_post['tagno'])."\",\"".sqlite_escape_string($_post['desc'])."\",\"" .sqlite_escape_string($_post['serialid'])."\",\"".sqlite_escape_string($_post ['custloc'])."\",\"".sqlite_escape_string($_post['totcost'])."\")"; $insresult = sqlite_query($handle,$insquery) or die("error:".sqlite_error_string(sqlite_last_error($handle))); echo "<i>successfully inserted!</i><p />"; } else { echo "<i>incomplete from input!</i><p />"; $query = "SELECT * FROM inventory "; $result = sqlite_query($handle,$query) or die("err:".sqlite_error_string(sqlite_last_error($handle))); if (sqlite_num_rows($result) > 0) { echo "<table cellpadding=20 border=1>"; while ($row = sqlite_fetch_array($result)) { echo "<tr>"; echo "<td>".$row[0]."</td>"; echo "<td>".$row[1]."</td>"; echo "<td>".$row[2]."</td>"; echo "<td>".$row[3]."</td>"; echo "<td>".$row[4]."</td>"; echo "<td>".$row[5]."</td>"; echo "<td>".$row[6]."</td>"; echo "<td>".$row[7]."</td>"; echo "<td>".$row[8]."</td>"; echo "<td>".$row[9]."</td>"; echo "</tr>"; } echo "</table>"; } sqlite_close($handle); ?> </body> </html> i would appreciate if you can help me in correcting this error.. thanks in advance.. sai