how to get value of a <select> without <form> HTML markup - php

Question

i never put too much thought on how the data would be submitted without using <form> . I initially just focused on the design of my form.
now, im having trouble trying to send POST data without having a <form> markup.
im using bootstrap framework.
also, how do i get the values in my <select> markup to be submitted. below is my HTML and PHP code.
below HTML
<div id="box1">
<input name="fname" type="text" placeholder="name">
</div>
<div id="box2">
<select id="food" name="food">
<option value="meat" name="nmeat"> meat </option>
<option value="fruit" name="nfruit"> fruit </option>
</select>
</div>
// ^ Above code is in a JQuery Conditional Statement
// When user select meat, <select id="meat"> will display else fruit
<div id="box3">
<!--MEAT-->
<select id="meat" name="topic">
<option value="hotdog" name="nhotdog"> hotdog </option>
<option value="sausage" name="nsausage"> sausage </option>
</select>
<!--FRUIT-->
<select id="fruit" name="topic">
<option value="apple" name="napple"> apple </option>
<option value="grapes" name="ngrapes"> grapes </option>
</select>
</div>
<div id="box4">
<textarea name="desc" rows="5" class="span11" placeholder="Description"></textarea>
<button name="submit" class="btn btn-large btn-block btn-primary" type="button">Submit</button>
</div>
below PHP
Im having a hardtime making the correct syntax in PHP, specially if the values being submitted is not in a <form>
<?php
if ( isset( $_POST['submit'] ) ) {
$name = $_POST["fname"]; //box1
$gender =$_POST["gender"]; //box2
$topic =$_POST["topic"]; //box3
$desc =$_POST["desc"]; //box4
//echo $msg;
$sql = "insert into z_form (name) values ('".$msg."')";
mysql_query( $sql );
} else {
echo "error";
}
?>

Answer 1

Just use the <form> tag, really. POST method requires a <form> tag. I don't see any valid reasons to exclude the <form tags from a form.
now, im having trouble trying to send POST data without having a markup. im using bootstrap framework.
Bootstrap doesn't restrict you from using <form> tags. :)
Wrap your form in <form tags, as follows:
<form action="somefile.php" method="post" name="myForm">
<!-- your form code here -->
</form>
Also, change your button from:
<button name="submit" class="btn btn-large btn-block btn-primary" type="button">Submit</button>
to:
<input type="submit" name="submit" value="Submit" class="btn btn-large btn-block btn-primary"/>
And finally, in your PHP code, you're doing:
$gender =$_POST["gender"];
But there's no input field with the name attribute gender. Either you should create one input field with that name, or remove the definition from your script.
That should fix the issues. Cheers!

Answer 2

Why would you not use a <form> ? all other methods consist of using jquery in combination with ajax, or just ajax for sending the values. But there shouldn't be a reason to not use a <form> tag.

Answer 3

I always thought that you have to use the -tag if you want to submit data just with submit button.
An alternative is to take a button and use javascript / jquery to read data from fields and submit them via get-parameters with an url.
However it seems you would like to go the classic / easy way. So you need the and define the method and action.
In addition a little hint: Try to use a php framework with mvc-(model view controller)pattern. When you use it in the right way it makes you easier to handle your code, brings predefined functions / classes and makes your functions against database safer. An example for such a framework would be codeigniter but there are many more...

Answer 4

First, you need to include form tags:
<form action="process.php" method="post">
//form content
</form>
Change your submit button. It's also good to change the name to somthing other than "submit" to avoid namespace conflicts:
<input type="submit" name="sub" class="btn btn-large btn-block btn-primary" value ="Submit" />
Also, give your selects different names (You have two with the name "topic"):
<select id="meat" name="meat">
<option value="hotdog" name="nhotdog"> hotdog </option>
<option value="sausage" name="nsausage"> sausage </option>
</select>
And then retrieve those values from $_POST:
$meat = $_POST['meat'];

Answer 5

Please see below code:
<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
<script type="text/javascript">
$(document).ready(function()
{
$(button).click(function() {
var name=$("#name").val();
var frt=$("#fruit").val();
var gender = $("#gender").val();
var topic= $("#topic").val();
var descr= $("#description").val();
var dataString = 'name='+name+'fruit='+ frt+'gender='+gender+'topic='+topic+'desc='+descr;
$.ajax
({
type: "POST",
url: "ajax.php",
data: dataString,
cache: false,
success: function(html)
{
$("#success").html(html);
}
});
});
});
</script>
You should add id in all dropdown fields.
also add which you want to view the show success or error result
<div id="success"></div>

Answer 6

In #box3 you have 2 selects with the same name.
Try something like:
<select id="meat" name="meatSelect">
<select id="fruit" name="fruitSelect">
And then:
$meatSelect = $_POST['meatSelect'];
$fruitSelect = $_POST['fruitSelect'];