Display drowdown values after user has selected 3 option - php

Question

I have 3 drop downs I want to display whatever the user will select after he/she has selected a function or a scrpt will do but it must be within the script
<?php
$resource_names = mysql_query("SELECT DISTINCT NAME FROM selections ORDER BY id ASC");
$names = array();
while($row = mysql_fetch_row($resource_names)){
$names[] = $row[0]
}
$resource_surnames = mysql_query("SELECT DISTINCT SURNAME FROM selections ORDER BY id ASC");
$surnames = array();
while($row = mysql_fetch_row($resource_surnames)){
$surnames[] = $row[0];
}
$resource_emails = mysql_query("SELECT DISTINCT EMAIL FROM selections ORDER BY id ASC");
$emails = array();
while($row = mysql_fetch_row($resource_emails)){
$emails[] = $row[0];
}
if(count($emails) <= 0 || count($surnames) <= 0 || count($emails) <= 0){
echo 'No results have been found.';
} else {
// Display form
echo '<form name="form" method="post" action="test.php">';
//Names dropdown:
echo '<select name="id" id="names">';
foreach($names as $name) echo "<option id='$name'>$name</option>";
echo '</select>';
//Surnames dropdown
echo '<select name="id" id="surnames">';
foreach($surnames as $surname) echo "<option id='$surname'>$surname</option>";
echo '</select>';
//Emails dropdown
echo '<select name="id" id="emails">';
foreach($emails as $email) echo "<option id='$email'>$email</option>";
echo '</select>';
echo "<button id='write_in_div'>Click me!</button>";
echo '</form>';
}
?>
Something that will call the write_in_div When Click me! button is press or any other method that can be used to display 3 selection user selected
The Output should be something like You select 1) Name 2)Surname and Email

Answer 1

You have an error in your html selects each select has the same name "id" they each need to be unique so you can detect then.
You need to detect if the user has submitted the form
if(isset($_POST["select_name"])) {
echo $_POST["select_name"];
}

Answer 2

There is a big mistake on you form.
In a form, each select and input MUST have a unique name. You need this name to retrieve the submitted value back in your php script.
I suppose you have this:
<?php
$resource_names = mysql_query("SELECT DISTINCT NAME FROM selections ORDER BY id ASC");
$names = array();
while($row = mysql_fetch_row($resource_names)){
$names[] = $row[0]
}
$resource_surnames = mysql_query("SELECT DISTINCT SURNAME FROM selections ORDER BY id ASC");
$surnames = array();
while($row = mysql_fetch_row($resource_surnames)){
$surnames[] = $row[0];
}
$resource_emails = mysql_query("SELECT DISTINCT EMAIL FROM selections ORDER BY id ASC");
$emails = array();
while($row = mysql_fetch_row($resource_emails)){
$emails[] = $row[0];
}
if(count($emails) <= 0 || count($surnames) <= 0 || count($emails) <= 0){
echo 'No results have been found.';
} else {
// Display form
echo '<form method="post" action="test.php">';
//Names dropdown:
echo '<select name="names">';
foreach($names as $name) echo "<option id='$name'>$name</option>";
echo '</select>';
//Surnames dropdown
echo '<select name="surnames">';
foreach($surnames as $surname) echo "<option id='$surname'>$surname</option>";
echo '</select>';
//Emails dropdown
echo '<select name="emails">';
foreach($emails as $email) echo "<option id='$email'>$email</option>";
echo '</select>';
echo '<button id="write_in_div">Click me!</button>';
echo '</form>';
}
When the form is submitted, test.php will have the posted data: $_REQUEST['names'], $_REQUEST['surnames'] and $_REQUEST['emails'].
You just have to check the content of thoses vars and print them if not null.
Note1: ?> is useless at the end of a file.
Note2: be carefull about ' and " when writing an html file. The value of an html attribute is written with ", not '.