Passing the link's text as a value to the next page

Question

I am trying to pass the link's text as a value to the next page so I can use it to search the database for the item and retrieve the information related to the value .I have tried using the POST method but regardless the information is not passed. This is the code I tried .

<form action="DetailedMenu.php"  method = "POST" action = "<?php $_PHP_SELF ?>">

<?php

for($i=0;$i<sizeof($array);$i++) {
    if($array[$i]["Food_Category"]=="starters") {
        echo str_repeat('&nbsp;', 4); ?>

        <a  href="DetailedMenu.php" ><?php echo $array[$i]["Food_Name"];?></a>

        <?php echo "   " .str_repeat('.&nbsp;', 25). "€".$array[$i]["Food_Price"]."<br>"; ?>

        <input type="hidden" name="name" value="<?php echo $array[$i]["Food_Name"];?>">

        <?php
    }
}

?>

</form>

Answer 1

You don't need the form.

The easiest way to do what you're trying to do....

In addition to including the text in the content of the link, include it as a query string parameter.

for($i=0;$i<sizeof($array);$i++) {
    if($array[$i]["Food_Category"]=="starters") {
        ...
        <a  href="DetailedMenu.php?q=<?php echo $array[$i]["Food_Name"];?>"><?php echo $array[$i]["Food_Name"];?></a>
        ...
    }
}

I would actually recommend something more like this. I obviously don't know the names of your fields, so I've just taken a guess...

for($i=0;$i<sizeof($array);$i++) {
    if($array[$i]["Food_Category"]=="starters") {
        ...
        <a  href="DetailedMenu.php?FoodID=<?php echo $array[$i]["Food_ID"];?>"><?php echo $array[$i]["Food_Name"];?></a>
        ...
    }
}

You'll be able to access "FoodID" as a parameter within your PHP, just as you would if it had been submitted from a form.

Answer 2

You may be looking for AJAX. AJAX lets you send the form data to a back end PHP file (that can then insert data into a DB, and/or get data from the DB) without refreshing the page.

In fact, when you are using AJAX you don't even need to use a <form> structure -- simple DIVs work just fine. Then you don't need to use event.preventDefault() to suppress the built-in form refresh.

Just build a structure inside a DIV (input fields, labels, etc) and when the user is ready to submit, they can click an ordinary button:

<button id="btnSubmit">Submit</button>

jQuery:

$('#btnSubmit').click(function(){
    var fn = $('#firstname').val();
    var ln = $('#lastname').val();
    $.ajax({
        type: 'post',
         url: 'ajax_receiver.php',
        data: 'fn=' +fn+ '&ln=' +ln,
        success: function(d){
            if (d.length) alert(d);
        }
    });
});

ajax_receiver.php:

<?php
    $fn = $_POST['fn'];
    $ln = $_POST['ln'];

    //Do your stuff

Check out this post and especially its examples. Copy them onto your own system and see how they work. It's pretty simple.