Issue passing variables in PHP with $_POST - php

Question

I have a searchresult.php file where I pull all the menus names out of my database and print them with a button close to them where the button has a input type hidden where I associated the name of the menu through the variable $nameofmenu . Something like
Menu1 --> Button (See Menu)
Menu2 --> Button (See Menu)
Menu3 --> Button (See Menu)
<?PHP
while($myrow = pg_fetch_assoc($result)) {
$nameofmenu = $myrow[name];
echo $nameofmenu; //It prints correctly the name of the menu
echo '<form name"formname" method="post" action="resultsmenu.php">';
echo "<input type='hidden' name='menuname' value=' $nameofmenu ' />";
echo "<input type='submit' name='submit' value='See Menu' />";
}
?>
Then I have my resultsmenu.php file that will open when I click in any of the buttons.
<?php
$nameofmenu = $_POST['menuname'];
$db = pg_connect('host=localhost dbname=test user=myuser password=mypass');
$query = "SELECT * FROM menu where name='$nameofmenu'";
$result = pg_query($query);
if (!$result) {
echo "Problem with query " . $query . "<br/>";
echo pg_last_error();
exit();
}
$myrow = pg_fetch_assoc($result);
$description = $myrow[description];
?>
I have 2 Issues:
First issue: in the searchresult.php, even it prints out correctly all the different menu names, whenever I click any of the buttons, all the buttons return my the same menu (The last one, in the example: Menu 3 in the $nameofmenu = $_POST['menuname']; of the resultsmenu.php file.
Second issue: this issue happend in the resultsmenu.php file. Even for Menu 3, it seams that the recognized the value in the variable. Meaning I don't get back any result or value from the query: $query = "SELECT * FROM menu where name='$nameofmenu'";
However, if I set up the value Manually like $nameofmenu = "Menu3"; then It works.
I even tried this code to see if there is any different in the value passed from $_POST['menuname'] and the value typed in manually and it seams to be the same value because it prints "Both variables are the same"
$nameofmenu2 = "Menu3";
$nameofmenu = $_POST['menuname']; //where the value in menuname is Menu3
if ($nameofmenu2 = $nameofmenu){echo "Both variables are the same";}
Thank you so much

Answer 1

You overwrite the same input field (=name) over and over again, so only the "last" overwrite is returned/available. To avoid this, put the data as POST-ARRAYs like this (also switch your quotes):
echo '<input type="hidden" name="menuname[]" value="'.$nameofmenu.'" />';
After submitting, you now will see that $_POST['menuname'] is an array! e.g.
var_dump($_POST['menuname']);

Answer 2

As mentioned by David, your are assigning nameofmenu2 to nameofmenu. But furthermore, when you echoed:
echo "<input type='hidden' name='menuname' value=' $nameofmenu ' />";
you included whitespace on either side of $nameofmenu. This could cause problems when you go to make your query. Try using:
var_dump($_POST['menuname']);
to check your expected post data.