having trouble using ajax to display the data with PHP and POST

Question

I am trying to run a php script that pulls data from a DB with the click of a button. I want to use ajax so the page doesn't refresh. I've tested using normal post / submit with page refresh and it works but I am having trouble using ajax to display the data. This is my code:

  <form method="get" id="ghosts">
         <div id=rightc>
              <h7>Select Hosts:</h7>
     <br></br>

      <select id="group" name="group" onchange='this.form.submit()'>
           <?php
        $link=mysql_connect($mysqlserver, $username, $password) or die ("Error connecting to mysql server: ".mysql_error());
        mysql_select_db($dbname, $link) or die ("Error selecting specified database on mysql server: ".mysql_error());
        $gquery=" SELECT groupname FROM groups";
        $gresult=mysql_query($gquery) or die ("Query to get data from firsttable failed: ".mysql_error());
        while ($grow=mysql_fetch_array($gresult)) {
        $groupname=$grow["groupname"];
            echo "<option>$groupname</option>";
        }
        ?>
    </select>
     <input type="hidden" name="ghosts" value="1" />
     <noscript> <input type="submit" name="ghosts" id="group" value="Choose Group" /></noscript>

  </form>

Use GET method and getting value using POST `isset($_POST['ahosts'])` — Saty, Aug 26 '15 at 12:22
You should edit your question with you form code and (if possible), your 'search.php' code. — Peter, Aug 26 '15 at 12:24
@Peter the form code is in there and the PHP is not neccessary, though confirmation that it returns data would be good. — Popnoodles, Aug 26 '15 at 12:24
If you can, you should [stop using `mysql_*` functions](http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php). They are no longer maintained and are [officially deprecated](https://wiki.php.net/rfc/mysql_deprecation). Learn about [prepared](http://en.wikipedia.org/wiki/Prepared_statement) [statements](http://php.net/manual/en/pdo.prepared-statements.php) instead, and consider using PDO, [it's really not hard](http://jayblanchard.net/demystifying_php_pdo.html). — Jay Blanchard, Aug 26 '15 at 12:26
@Popnoodles you're right about the form code, but I think that the PHP code may help because we could see the way he is getting the form data — Peter, Aug 26 '15 at 12:28
To use AJAX effectively you should put your PHP with its requests to MySQL in a separate file. You would then make the AJAX request to the PHP file [and return the data](http://jayblanchard.net/basics_of_jquery_ajax.html) — Jay Blanchard, Aug 26 '15 at 12:28
Hi thanks for all looking, the code is all self contained so this is search.php. The form is simple its just a button to trigger the script. It is show above in the code. Maybe there is a better way of doing this ? really what i want is a button that says search then runs the php code to pull data from the db using ajax — spyda46, Aug 26 '15 at 12:29
That's the same file that it's pulling in? So `` and `` and ` — Popnoodles, Aug 26 '15 at 12:30

Popnoodles · Answer 1 · 2015-08-26T12:40:18.537

If your search.php file returned HTML, and assuming that the ajax works (it looks like it should), you're trying to put the contents into an element that doesn't exist.

$(".result").html(data);

It needs to exist on the page. Add this somewhere.

<div class="result"></div>

However, your PHP doesn't return any data because of this condition if(isset($_POST['ahosts'])) which requests a parameter that isn't in the form and uses the wrong request method.

You're using GET ($.get()), so in the PHP it needs to read

if(isset($_GET['ahosts']))

And add this inside your form or give the button a name

<input type="hidden" name="ahosts" value="1" />
<!-- or -->
<button type="submit" name="ahosts" />

score 0 · Answer 2 · edited Aug 26 '15 at 12:33

0

I think that you make a mistake here :

if(isset($_POST['ahosts']))

You check if isset() POST request but actually you send GET request ...

read the documentation https://api.jquery.com/jquery.get/

So if you change your code to:

 if(isset($_GET['ahosts']))

this will return true and now you will have a response

edited Aug 26 '15 at 12:33

Jay Blanchard

34,243
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77
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answered Aug 26 '15 at 12:31

E. Stancheva

66
2

Guys many Thanks ! -- Popnoodles thanks very much your suggestions have rsolved the issue for me -- Great ! :) – spyda46 Aug 26 '15 at 12:56
Hi Guys, just a follow up question. I have a similar form to one above but this one uses a drop down menu to choose a "group" i've used the same logic with ajax as above, the page is working in the sense i get the results expected but the problem is the page is not refreshing, i added the new code above : – spyda46 Aug 26 '15 at 14:01
Hi, the main reason for using ajax is to prevent refreshing the page – E. Stancheva Aug 26 '15 at 14:11

having trouble using ajax to display the data with PHP and POST

2 Answers2