Im using PHP & MySQL, the Problem im facing to INSERT data using "mysql_fetch_array"
.
this is my my connection to Mysql and my query to display data.
There are 2 table in this. 1 table for display. 1 more for insert data.
<?php
$host="localhost"; // Host name
$username="root"; // Mysql username
$password=""; // Mysql password
$db_name="skpj"; // Database name
// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM `student` WHERE cls_id = '13' ";
$result=mysql_query($sql);
// Count table rows
$count=mysql_num_rows($result);
?>
<form name="form1" method="post" action="">
<?php while($rows=mysql_fetch_array($result)){ ?>
<?php echo $rows['s_no']; ?>
<?php echo $rows['name']; ?>
<input name="s_no[]" type="hidden" id="name"
value="<?php echo $rows['s_no']; ?>">
<?php echo $rows['ic']; ?>
<?php echo $rows['cls_id']; ?>
<input name="class_n[]" type="hidden" id="cls_id"
value="<?php echo $rows['cls_id']; ?>">
<select name="att[]" id="att" style=" width:80px" >
<option value="1">Atten</option>
<option value="2">Absend</option>
<option value="3">MC</option>
</select>
<input name="tmp[]" type="hidden" id="name" value="1">
<?php } ?>
<input type="submit" name="submit" value="submit"></td>
</form>
<?php
if($submit){
for($i=0;$i<$count;$i++){
$sql1="INSERT INTO attendance (s_no, class_n, att, tmp)
VALUE '$s_no[i]','$class_n[i]','$att[i]','$tmp[i]' ";
$result1=mysql_query($sql1);
}
}
if($result1){
header("location:att2.php");
}
mysql_close();
?>
below is the notice foe my error.
SCREAM: Error suppression ignored for
Notice: Undefined variable: submit in C:\wamp\www\att\att2.php on line 67
Notice: Undefined variable: result1 in C:\wamp\www\att\att2.php on line 74
furthermore i cannot insert my data