$_POST in PHP, need help with the code

Could you help me with the next code:

<?php

require ("Functions.php");

#geting the destinations
$link = Connect();

$destinations_query = "SELECT * FROM destinations
WHERE Held_In_Destination > 1";

$destinations_result = mysqli_query($link, $destinations_query);

#creating a form containing a destination list.
#The form will pass on a destination ID to the next form
#that shows hotels in the area.

echo "<form action='MakeAnOrder.php' method='POST'>

<strong>Select Your Destination</strong>
<select name='Destination_Name'>\n";

while ($destinations_row = mysqli_fetch_array($destinations_result))
{

extract ($destinations_row);
echo"<option value = '$Destination_Name'>$Destination_Name\n";

}


echo"</select>\n";
echo "<input type = 'submit' value = 'Select'>
</form>\n";


#Selecting a hotel

$hotels = Hotels(isset($_POST['$Destination_ID']));

echo "<form action='MakeAnOrder.php' method='POST' >

<strong>Select a Hotel</strong>
<select name='Hotel_Name'>\n";

while ($hotels_row = mysqli_fetch_array($hotels))
{

extract ($hotels_row);
echo"<option value = '$Hotel_Name'>$Hotel_Name\n";

}

echo"</select>\n";
echo "<input type = 'submit' value = 'Select'>
</form>\n";

?>



I am trying to influence the output of the second form by the users choice in the first one. Both are based in the same page.

What do you mean by "influencing" the output? SQL-injection?

I want a user to chose a name of the destination. The program the should use an ID attached to this name to find the hotels which are based in that destination and output them in to a second dropdown list.

You probably need to change this part:

Code:

---

while ($destinations_row = mysqli_fetch_array($destinations_result))
{

extract ($destinations_row);
echo"<option value = '$Destination_Name'>$Destination_Name\n";

}

---

Not sure what the extract() function does but I'm guessing you need to change the echo on the next line to:

Code:

---

echo "<option value='$Destination_ID'>$Destination_Name</option>\n";

---

Both selects seem incomplete btw...

It should be something like:

Code:

---

<select id=myselection>
  <option value=1>Option 1</option>
  <option value=2>Option 2</option>
</select>

---

Didn't work :(

oh.... This doesn't look right either:

Code:

---

$hotels = Hotels(isset($_POST['$Destination_ID']));

---

Im guessing this should be:

Code:

---

$hotels = Hotels(isset($_POST['Destination_Name']));

---

This may help:
http://www.w3schools.com/php/php_forms.asp

OK I made some changes to the code.

<?php

require ("Functions.php");

#geting the destinations
$link = Connect();

$destinations_query = "SELECT * FROM destinations
WHERE Held_In_Destination > 1";

$destinations_result = mysqli_query($link, $destinations_query);

#creating a form containing a destination list.
#The form will pass on a destination ID to the next form
#that shows hotels in the area.

echo "<form action='MakeAnOrder.php' method='POST'>

<strong>Select Your Destination</strong>
<select name='Destination_Name'>\n";

while ($destinations_row = mysqli_fetch_array($destinations_result))
{

extract ($destinations_row);
echo"<option value = '$Destination_Name'>$Destination_Name\n";

}


echo"</select>\n";
echo"<input type = 'hidden' name = 'Destination_ID' value = '$Destination_ID'>";
echo "<input type = 'submit' value = 'Select'>
</form>\n";

mysqli_close ($link);
?>


<?php
#Selecting a hotel
$Destination_ID = SelectDestinationID(isset($_POST['Destination_Name']));

echo"$Destination_ID";

$hotels = Hotels('$Destination_ID');

echo "<form action='MakeAnOrder.php' method='POST' >

<strong>Select a Hotel</strong>
<select name='Hotel_Name'>\n";

while ($hotels_row = mysqli_fetch_array($hotels))
{

extract ($hotels_row);
echo"<option value = '$Hotel_Name'>$Hotel_Name\n";

}

echo"</select>\n";
echo "<input type = 'submit' value = 'Select'>
</form>\n";
#When both forms are created, the Destination_ID doesn't transfere to the second form
?>

When I echo a Destination_ID in the second part of the code, it turns out to be fine. But the form still doesn't display the hotel names :(

Here's the code for the Hotels Funsction

function Hotels($Destination_ID)
{
#Connect to the database
$link = Connect();

$query = "SELECT * FROM hotels
WHERE Destination_ID = $Destination_ID";

$result = mysqli_query ($link,$query);

return(object) $result;
}

Quote:

---

Originally posted here by nightcat

$hotels = Hotels(isset($_POST['$Destination_ID']));

---

Is the function Hotels() defined somewhere? I'm guessing its in your functions.php include, but without that we can't really tell what is going on. You should probably also parse the destination for the currently selected desitnation and then add a "selected" tag to that line.

Also, why are you ?>ing your php and then reopening immediately. I would suggest just leaving the tag open.

Edit: hah, I guess I should have refreshed. I was editing this for about a half hour while someone was talking to me, and looks like you answered my question in the mean time. Lemme look at it and see whats going on.

My first suggestion is, if you're not going to functionize the first sql query, why do the second one? Put all of the code into the base .php file and get it working, then you can worry about functionizing it. Also, you only need 1 $link. When you do your functions, you can pass $link to it.

zENGER No offence but I didn't understand most of the stuff you said. I've been using PHP for a couploe of weeks now, but I'm not that good with it.

Look at the example I gave (the link).

You need to get something like (watch the bolded stuff):

Code:

---

<select name=destination>
<?

fireSQLquery("SELECT Destination_ID, Destination_Name FROM destinations;");

while (there are rows)
{
  // get one row as $Dest_ID, $Dest_name
  echo "<option value='$Dest_ID'>$Dest_name</option>"
}
?>
</select>

<?

 $selected_dest_id = $_POST["destination"];
 // do something with $selected_dest_id

?>

---