i have a problem when inserting data to my database. but when i'm echoing it doesn't duplicate. here's my code:
foreach($_POST['purpose'] as $i => $a){
echo $a."-".$i."<br />";
if($_POST['purpose'][$i] == "recieved"){
$purpose[$i] = "'1','0','0','0','0'";
} elseif($_POST['purpose'][$i] == "released"){
$purpose[$i] = "'0','1','0','0','0'";
} elseif($_POST['purpose'][$i] == "recalled"){
$purpose[$i] = "'0','0','1','0','0'";
} elseif($_POST['purpose'][$i] == "charges"){
$purpose[$i] = "'0','0','0','1','0'";
} elseif($_POST['purpose'][$i] == "adjustments"){
$purpose[$i] = "'0','0','0','0','1'";
}
$query = "INSERT INTO responsibility_center VALUES(
'',
'".$_POST['or_number']."',
'".$_POST['response'][$i]."',
$purpose[$i])";
echo $query."<br />";
$result1 = mysql_query($query) or die(mysql_error()." inserting responsibility center");
$responseid = mysql_insert_id();
$query2 = "INSERT INTO particulars VALUES (
'',
'".$responseid."',
'".$_POST['details'][$i]."')";
echo $query2."<br />";
$result2 = mysql_query($query) or die(mysql_error()." inserting particulars");
}
i dont know where in my code is wrong cause it doesn't give me errors. :/
$result2 = mysql_query($query)
that should be
$result2 = mysql_query($query2)
This is why you should name your variables something more meaningful than $query
. Please also read through this: http://bobby-tables.com/php.html