AJAX/JAVASCRIPT voting mysql

Question

<SCRIPT language="Javascript"> 
    function vote() 
    {
        <?php
            $date = date("Y-m-d");
            //$vote = mysql_num_rows(mysql_query("SELECT * FROM funny_thumbs WHERE ip='$ip' AND date='$date' AND id='$p'"));
            if (mysql_num_rows(mysql_query("SELECT * FROM funny_thumbs WHERE ip='$ip' AND date='$date' AND id='$p'")) < 1)
            { 
                mysql_query("INSERT INTO funny_thumbs SET ip='$ip', type='1', date='$date', id='$p'"); 
            }
            if (mysql_num_rows(mysql_query("SELECT * FROM funny_thumbs WHERE ip='$ip' AND date='$date' AND id='$p'")) >= 1)
            {
                ?>
                    alert("Jus jau balsavote !");
                <?php
            } 
        ?>
    }
</SCRIPT>
<td>
    <a href="#">
        <img src="img/thumbs_up.png" alt="LIKE" onclick="vote(); return false;"/>
    </a> 
</td>

I already set $ip and $p. When i refresh page, it insert string without clicking on the link. Why ?

zatatatata · Accepted Answer · 2011-07-29 14:49:54Z

1

You're trying to call PHP on an onclick event. PHP gets executed when the page is loaded (before you even see the HTML) and therefore it inserts the string. Then it displays the website to the user with a vote function that alerts the weird character string and does nothing else.

answered Jul 29, 2011 at 14:49

zatatatata

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1 Comment

zatatatata Over a year ago

@user There's no way to fix the current code, since it's fundamentally flawed. You clearly don't have the basics of AJAX and an understanding of what it is. I'd suggest reading jQuery tutorials, to get the fundametals. sitepoint.com/ajax-jquery seems to be a tutorial that's basic enough.

Localghost · Accepted Answer · 2011-07-29 14:57:20Z

There seems to be a bit of misunderstanding of the fundamentals of how PHP and Javascript work.

PHP is a server sided scripting language. That means, that it is only executed on the webserver, before the page is delivered to the client. When the page is delivered to the client, it consists of markup (HTML), client-sided scripts (Javascript), as well as other things such as flash objects etc...

Here is process:

Client requests page from server
Server processes request, and generates the HTML\Javascript\etc... through the use of server sided languages such as PHP\ASP etc...
Server delivers the payload generated in #2 to back to the client
The client's web browser interprets the payload delivered in #3 and displays it in some graphical way

Once the process finishes #2, you can no longer speak with the server and execute server sided code (such as PHP). In order to execute more PHP you must either reload the page completely, or utilize AJAX.

Lord Loh. · Accepted Answer · 2011-07-29 14:58:47Z

0

Looks like php is behaving the way you programmed.

The javascript corresponding to the like button shall be

function vote(){
alert("Jūs jau balsavote !");
}

if the vote has been casted and

function vote(){
}

if it has not been. You need to use AJAX or reload the page when the like button is pressed.

answered Jul 29, 2011 at 14:58

Lord Loh.

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