jQuery AJAX表单没有发布结果

Question

I have been having problems returning results from my database using AJAX. I have tried to echo $diskspace and $price , both of which are returning undefined.

index.php

<form id="form" method="POST">
    Diskspace:
    <select id="Diskspace">
        <option value="0 AND 1">$0 - 1GB</option>
        <option value="1 AND 5">$1 - 5GB</option>
        <option value="5 AND 10">$5 - 10GB</option>     
    </select></br></br> 
    Price:
    <select id="Price">
        <option value="0 AND 5">$0 - $5</option>
        <option value="1 AND 5">$5 - $10</option>
        <option value="10 AND 20">$10 - $20</option>
        <option value="20 AND 40">$20 - $40</option>
        <option value="40 and 500">>$40</option>            
    </select></br></br> 
    <input type="submit" id="submit" value="enter">
</form> 
<div id="output"></div>

JS

$(document).ready(function(){

    $("div#output").hide();
    $("#submit").click(function(){
        $.post('join.php', {
            diskspace: $("#diskspace").val(),
            price: $("#price").val() 
        },
        function(data){
            $("div#output").html(data); 
            $("div#output").show();     
        }
        );
        return false;
    });     
});
</script>

join.php

<?php
if (isset($_POST['diskspace'])){
mysql_connect("localhost","root","") or die('Could not connect');
mysql_select_db("webhost") or die ('Could not find DB');

$diskspace = $_POST['diskspace'];
$price =$_POST['price'];

$query =  mysql_query("
    SELECT * FROM data WHERE Diskspace BETWEEN $diskspace
    AND Price BETWEEN $price 
");

$count = mysql_num_rows($query);

if ($count == 0){
        $output = "No such results, sorry.";
    }else{
        while($row = mysql_fetch_array($query)){
            $diskspace = $row['Diskspace'];
            $price = $row['Price'];
            $host = $row['Provider'];
            $output .= '<div>'.$host.'  '.$diskspace.'  '.$price.'</div>'; 
        }
}
echo $output;
}
?>

Answer 1

Your HTML element has an id of Diskspace and you're looking for an element of diskspace .

Take the following for example:

<select id="Diskspace">
    <option value="hello">hello</option>
</select>

<script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script type="text/javascript">
$(function(){
    console.log( $('#diskspace').val() );
});
</script>

The result is undefined, and in new versions of jQuery, if the element cannot be found it will not pass the variable through ajax/post.

The same applies to your Price element.

Answer 2

Your select id is not same in your jquery. Change it like this:

 <form id="form" method="POST">
        Diskspace:
        <select id="diskspace">
            <option value="0 AND 1">$0 - 1GB</option>
            <option value="1 AND 5">$1 - 5GB</option>
            <option value="5 AND 10">$5 - 10GB</option>     
        </select></br></br> 
        Price:
        <select id="price">
            <option value="0 AND 5">$0 - $5</option>
            <option value="1 AND 5">$5 - $10</option>
            <option value="10 AND 20">$10 - $20</option>
            <option value="20 AND 40">$20 - $40</option>
            <option value="40 and 500">>$40</option>            
        </select></br></br> 
        <input type="submit" id="submit" value="enter">
    </form> 
    <div id="output"></div>

Answer 3

replace:

diskspace: $("#diskspace").val(),

with

diskspace: $("#Diskspace").val(),

Answer 4

You didn't give your <select> s names:

<select id="Diskspace" name="Diskspace">
                       ^^^^^^^^^^^^^^^^--- missing

No name, no form submission for that field. id does NOT count - it's used purely for DOM operations, and is not relevant for form submissions.

Beyond that, you're vulnerable to SQL injection attacks .