parse error in php backend sqlite
| From: | Aravalli Sai | Date: | Wed, 20 Oct 2004 23:12:38 +0000 |
| Subject: | parse error in php backend sqlite | ||
| Groups: | php.db | ||
| Request: | Send a blank email to php-db+get-36945@lists.php.net to get a copy of this message | ||
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($_POST['submit'])) {
if (!empty($_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
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