If condition for moving to another page

newphpcoder

Good day!

I want to know what is wrong in my code that's why when i choose in select option it did not go to another page...Like when I select "Incoming" nothing happen also when I choose "Outgoing" nothing also happen.

Here is my code:

<?php
include ("config.php");

$call_type = $_POST['call_type'];

$query=mysql_query("SELECT `call_type` FROM `tbl_calltype` WHERE `call_type` = '{$call_type}'") or die(mysql_error());
$result = mysql_num_rows($query);

if ($result == 1){

if($call_type == 'Incoming'){
	header ('Location:incoming.php');
}
elseif($call_type == 'Outgoing'){
	header ('Location:outgoing.php');
}
else{
	header('Location:index.php');
}
}
?>
<html>
<body>
<form id="form1" name="form1" method="post" action="">
  <select name="call_type">
    <option value="Select Call Type">Select Call Type</option>
    <option value="Incoming" <?php if($_POST['call_type'] == 'Incoming') echo "selected='selected'"; ?>>Incoming</option>
    <option value="Outgoing" <?php if($_POST['call_type'] == 'Outgoing') echo "selected='selected'"; ?>>Outgoing</option>
  </select>
</form>
</body>
</html>

Thank you

bradgrafelman

You have to submit the form if you expect PHP to be able to process the data.

EDIT: Also note that user-supplied data should never be placed directly into a SQL query else your code will be vulnerable to SQL injection attacks and/or just plain SQL errors. Instead, you must first sanitize the data with a function such as [man]mysql_real_escape_string/man (for string data).

newphpcoder

bradgrafelman;10976101 wrote:
You have to submit the form if you expect PHP to be able to process the data.

EDIT: Also note that user-supplied data should never be placed directly into a SQL query else your code will be vulnerable to SQL injection attacks and/or just plain SQL errors. Instead, you must first sanitize the data with a function such as [man]mysql_real_escape_string/man (for string data).

How can I submit the form?

Thank you

bradgrafelman

Add a submit button, or use the 'onchange' property of the SELECT entity to automatically submit the form when the selection changes.

newphpcoder

Can you give me an idea or code for onchange?
I tried this code:

<?php
include ("config.php");

$call_type = mysql_real_escape_string($_POST['call_type']);  

$query=mysql_query("SELECT `call_type` FROM `tbl_calltype` WHERE `call_type` = '{$call_type}'") or die(mysql_error());
$result = mysql_num_rows($query);

if ($result == 1){

if($call_type == 'Incoming'){
	header ('Location:incoming.php');
}
elseif($call_type == 'Outgoing'){
	header ('Location:outgoing.php');
}
else{
	header('Location:index.php');
}
}
?>
<html>


<script type="text/javascript">                

function handleEnter (field, event) {
        var keyCode = event.keyCode ? event.keyCode : event.which ? event.which : event.charCode;
        if (keyCode == 13) {
            var i;
            for (i = 0; i < field.form.elements.length; i++)
                if (field == field.form.elements[i])
                    break;
            i = (i + 1) % field.form.elements.length;
            field.form.elements[i].focus();
            return false;
        } 
        else
        return true;
    }      

</script>

<body onload="document.form1.call_type.focus();">
<form id="form1" name="form1" method="post" action="">
  <select name="call_type" onkeypress="return handleEnter(this, event)">
    <option value="Select Call Type">Select Call Type</option>
    <option value="Incoming" <?php if($_POST['call_type'] == 'Incoming') echo "selected='selected'"; ?>>Incoming</option>
    <option value="Outgoing" <?php if($_POST['call_type'] == 'Outgoing') echo "selected='selected'"; ?>>Outgoing</option>
  </select>
</form>
</body>
</html>

But still it did not go to another page

newphpcoder

John;295547 wrote:
You need an <input type="submit" /> before you close your </form>, so users can submit the form.

I tried this code:

<?php
include ("config.php");

//ob_start();
//header ('Location:incoming.php'); 
if (isset($_POST['call_type'])) { // Check if form has been submitted
$call_type = mysql_real_escape_string($_POST['call_type']); // SECURE THE FECKING CONTENT!!!!!!!!!!!!!!!!!!!!!!

echo $call_type;

$query=mysql_query("SELECT `call_type` FROM `tbl_calltype` WHERE `call_type` = '{$call_type}'") or die(mysql_error());
$result = mysql_num_rows($query);
echo '<br />' . $result;
if ($result == 1){

if($call_type == 'Incoming'){
    header ('Location:incoming.php');
}
elseif($call_type == 'Outgoing'){
    header ('Location:outgoing.php');
}
else{
    header('Location:index.php');
}
}
}

?>
<html>



<body>
<form id="form1" name="form1" method="post" action="">
  <select name="call_type">
    <option value="Select Call Type">Select Call Type</option>
    <option value="Incoming" <?php if($_POST['call_type'] == 'Incoming') echo "selected='selected'"; ?>>Incoming</option>
    <option value="Outgoing" <?php if($_POST['call_type'] == 'Outgoing') echo "selected='selected'"; ?>>Outgoing</option>
  </select>

  <input type="submit" name = "Submit" value="Submit">
</form>
</body>
</html>

And when I choose Incoming and I press submit button the output is:
Incoming
1
Warning: Cannot modify header information - headers already sent by (output started at D:\xampp\htdocs\OJT\mae_ann\index.php:9) in D:\xampp\htdocs\OJT\mae_ann\index.php on line 17

Thank you

bradgrafelman

You can't output anything before calling [man]header/man.

newphpcoder

bradgrafelman;10976142 wrote:
You can't output anything before calling [man]header/man.

I deleted this code:

if ($result == 1){

if($call_type == 'Incoming'){
    header ('Location:incoming.php');
}
elseif($call_type == 'Outgoing'){
    header ('Location:outgoing.php');
}
else{
    header('Location:index.php');
}
}

and the warning was remove and the output is incoming 1 and it was displayed in the same page...Instead of moving to Incoming page