hi all,
i have written the program for deleting multiple records using php.
it is displaying the error as " Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING, expecting ‘,’ or ‘;’ in C:\xampp\htdocs\multi.php on line 33
"
dont no what went wrong…
below ids the code…
“multi.php”
<?php
$host = "localhost";
$user = "root";
$pass = "";
$con = 0;
$dbs = 0;
$re = 0;
$se = 0;
$C = mysql_connect($host,$user,$pass);
if($C)
{
$con = 1;
}
$D = mysql_select_db("test",$C);
if($D)
{
$dbs = 1;
}
$Q = mysql_query("SELECT * FROM emp");
if($Q)
{
$re = 1;
}
if(mysql_num_rows($Q) > 0)
{
$se = 1;
}
if($con == 1 && $dbs == 1 && $re == 1 && $se == 1)
{
echo "<form method=\\"post\\" action=\\"p.php\\">";
while(($dat = mysql_fetch_assoc($Q)) !== false)
{
echo "<input type=\\"checkbox\\" name=\\"id[]\\" value=\\"$dat[empno]\\" />" . $dat['empno'] . ": " . $dat['empname'] . ": " . $dat['desig'] "<br />";
}
echo "<input type=\\"submit\\" value=\\"delete\\" />";
echo "</form>";
}
?>
below is “p.php”
<?php
$uv = $_POST['id'];
$cs = 0;
$host = "localhost";
$user = "root";
$pass = "";
$con = 0;
$dbs = 0;
$ct = 0;
$ac = 0;
$C = mysql_connect($host,$user,$pass);
if($C)
{
$con = 1;
}
$D = mysql_select_db("test",$C);
if($D)
{
$dbs = 1;
}
if($uv)
{
$cs = 1;
}
if($cs == 1)
{
if(is_array($uv))
{
$ac = count($uv);
if($con == 1 && $dbs == 1)
{
foreach($uv as $k)
{
$Q = mysql_query("DELETE FROM emp WHERE empno = '".$k."'");
if($Q)
{
$ct++;
}
}
echo "You sent " . $ac . " items to be deleted. " . $ct . " items where deleted!";
}
else
{
echo "not connected to server and no database selected";
}
}
else
{
echo "not an array";
}
}
?>
please tell me what is the error…