[RESOLVED] how to insert 3 values into a single database field

[lisa007]

i have 3 drop down which the user can only select one each has they own name like select 1 select 2 and select 3 but i want whichever the user select to go into the same field on the database.. at the moment i named the field select3 but that only store value of the drop down named SELECt3 how do i get if user chose select2 insert the value on same field as if chose select 1 or 3

i tried

$select3 = $_POST['select1 || select2 || select3'];


$query="INSERT INTO photos values ('$select3')"; 

this is the drop down

$select3 = $_POST['select1 && select2 && select3'];





$query="INSERT INTO photos values ('$select3')";





<select id="in-campu" name="select1">

 <option name="hot"  value="default">--Select  Music Type--</option>

<option name="how" value="Concerts">Concerts</option>

<option name="hot" value="Clubs" >Clubs</option>

<option name="hot" value="Festival">Festival</option>

<option name="hot" value="Opera">Opera</option>



</select>



<select id="off-campus" class="item" name="select2" style="display: none;">

 <option name="hot" value="default"> -- Select Sport Type -- </option>

<option name="hot" value="Formula 1">Formula 1</option>

<option name="hot" value="Footbal">Footbal</option>

<option name="hot" value="Basketball">Basketball</option>

<option name="hot" value="Rugby">Rugby</option>

<option name="hot" value="Cricket">Cricket</option>

</select>



<select id="one" class="item" name="select3" style="display: none;">

<option name="one" value="default"> -- Select Art & Theatre Type -- </option>

<option name="one" value="Comedy">Comedy</option>

<option name="one"  value=" Drama">Drama</option>

<option name="one" value="Museus">Museus</option>

</select> 

[bradgrafelman]

Something like..

if(!empty($_POST['select1']) && $_POST['select1'] != 'default')
    $value = mysql_real_escape_string($_POST['select1']);
elseif(!empty($_POST['select2']) && $_POST['select2'] != 'default')
    $value = mysql_real_escape_string($_POST['select2']);
// etc.
else {
    $value = '';
    // user didn't select anything... display error message here, perhaps?
}

// use $value in SQL query

[lisa007]

this is what i tried

$name = strip_tags ($POST['name']);
$phone = strip_tags ($POST['phone']);
$email = $POST['email'];
$image = $POST['image'];
$campustype = $POST['campustype'];
$select3 = $POST['select3'];
$select3 = $POST['select2'];
$select3 = $POST['select1'];

$query="INSERT INTO photos values ('$image','$name','$email','$phone','$campustype','$select3')"; //into database
mysql_query( $query );

if(!empty($POST['select1']) && $POST['select1'] != 'default')
$value = mysql_real_escape_string($POST['select1']);
elseif(!empty($POST['select2']) && $POST['select2'] != 'default')
$value = mysql_real_escape_string($POST['select2']);
elseif(!empty($POST['select3']) && $POST['select3'] != 'default')
$value = mysql_real_escape_string($_POST['select3']);

// etc.
else {
$value = '';
// user didn't select anything... display error message
here, perhaps?
}

[bradgrafelman]

The point of using my if() statements was so that the final value of $value is what you would use in your SQL query. Not only did you put the if() statements after you already executed the query (making them completely pointless), but instead of using $value you used $select3, a variable which you overwrite twice after defining it:

$select3 = $_POST['select3'];
$select3 = $_POST['select2'];
$select3 = $_POST['select1'];

// values are overwritten, so the above might as well have been written as:
$select3 = $_POST['select1'];

[lisa007]

sorry about the previous post i'm just learning but yeah i fix the way you told me works wonders thank you very much