using Jquery variable from mysql

Question

I was searching for the answer but I couldn't found one for the specific problem, someone says something about ajax or json but I have no idea how to use this things, so I decided to ask a question.

Actualy I'm saving the values from a skill calculator in mysql using php/mysql and jquery for the calculating.

What I need at moment is take a number from mysql and use it as variable, maybe somthing like this:

ranksalvo[0] == 
    <?php
        mysql_select_db("skill");
        $query = mysql_query("SELECT * FROM skillrank WHERE ID = '71' ");
        $rows = mysql_fetch_array($query);
        $rank1 = $rows['rank1'];
    ?>;

if(ranksalvo[0] == 1){      
    $('.archerrank2').slideDown(1000);
}

As you can see its just a very bad example, but I think it shows exactly what I want also why I couldn't find a good enough answer when I was searching it.

Thank you!

<form action="conexao/conexao.php" method="post">
    <script src="jquery.js"></script> 
    <script type="text/javascript">

        var ranksalvo = [0,0,0,0,0,0];
        var skillsalvo = [0,0,0,0,0,0];

        var ranksalvo[0] =  
        <?php
            mysql_select_db("skill");
            $query = mysql_query("SELECT * FROM skillrank WHERE ID = '71' ");
            $rows = mysql_fetch_array($query);
            $rank1 = $rows['rank1'];
            echo "$rank1";
        ?>;

        if(ranksalvo[0] == 1){

            $('#rank1s').slideDown(1000);
            $('#rank2s').slideDown(1000);
            $('#rank3s').slideDown(1000);
            $('#rank4s').slideDown(1000);
            $('#multishot').slideDown(1000);
            $('#fulldraw').slideDown(1000);
            $('#swiftstep').slideDown(1000);
            $('#kneelingshot').slideDown(1000);         
            $('.archerrank2').slideDown(1000);
        }

    </script>

to assign values you use only one equal sign ranksalvo[0] = ........ '==' is to compare... — Leo Javier, Aug 27 '15 at 15:32
Ops, sorry that double ==, but it was just an example, I haven't tried this code. Its just an example and I can't belive its right. ^^ — Cleber Benedet, Aug 27 '15 at 15:35
"... exactly what I want ..."!! What is the scenario - initial page build or fetching content for a page that's already served? — Roamer-1888, Aug 27 '15 at 16:01
Well, the page is already built, I'm changing it to receive value from mysql. — Cleber Benedet, Aug 27 '15 at 16:05
In that case, it sounds like you correctly identified the need for AJAX, which means intercommunication of a displayed web page with a sever (typically the same server that served the page). — Roamer-1888, Aug 27 '15 at 16:11
I understand, is it possible to take a number from mysql and use it as variable in a jquery code with ajax? — Cleber Benedet, Aug 27 '15 at 16:17

score 0 · Answer 1 · answered Aug 27 '15 at 15:34

0

would be something like:

var ranksalvo[0] = 
    <?php
        mysql_select_db("skill");
        $query = mysql_query("SELECT * FROM skillrank WHERE ID = '71' ");
        $rows = mysql_fetch_array($query);
        $rank1 = $rows['rank1'];
    ?>;

if(ranksalvo[0] == 1){      
    $('.archerrank2').slideDown(1000);
}

answered Aug 27 '15 at 15:34

Leo Javier

1,383
12
19

Gonna try straight away! – Cleber Benedet Aug 27 '15 at 15:39
it should work.... just make sure you catch the error if your database returns a null or a NaN – Leo Javier Aug 27 '15 at 15:44
you would need to run it on your server since you are retrieveing data from a database to initialize the variable. yo can use if you don't want to query a database... just to test... – Leo Javier Aug 27 '15 at 15:54
Unfortunately the php code is crashing the whole code, if I erase only the php party the code backs to work normally. – Cleber Benedet Aug 27 '15 at 16:11
try with instead of the database call – Leo Javier Aug 27 '15 at 16:14
It won't work, cuz the value needs to be the value from mysql. – Cleber Benedet Aug 27 '15 at 16:17
can you debug and make sure you are actually getting the expected value from the database? in this case you should be getting the value 1 (as a number... not as a string) – Leo Javier Aug 27 '15 at 16:37
i'm getting this error: Uncaught SyntaxError: Unexpected token [ – Cleber Benedet Aug 27 '15 at 16:49
please share the code to take a look – Leo Javier Aug 27 '15 at 16:51
I'm doing it local, gonna need to create a new db somewhere and post to show it. I can like compact everything and upload it somewhere... – Cleber Benedet Aug 27 '15 at 17:01
just paste the piece of code that retrieve the value from the database... – Leo Javier Aug 27 '15 at 17:25
Done, check up there where I have edited the code. – Cleber Benedet Aug 27 '15 at 17:29
you are not getting anything from the database you are passing echo "$rank1" as a string it should be echo $rank1; without the quotes – Leo Javier Aug 27 '15 at 17:37
The if statement of your javascript will never be true because ranksalvo[0] value would be "$rank1" – Leo Javier Aug 27 '15 at 17:39
Gonna try without quotes – Cleber Benedet Aug 27 '15 at 17:45
Still the same problem, the php code crash all the code, since my code is really giant its funny to see what happens when it crash. ^^ – Cleber Benedet Aug 27 '15 at 17:58
you need to make sure you are getting a value from your db man... it looks like you are not getting any value.... and if thats the case, the condition will never be true – Leo Javier Aug 27 '15 at 19:26
I found the problem, for some reason this code doesn't work, if I just put echo $rank1; its enough to get syntax error again, then I made it different. I created an input where the value is the php code and after I made the Jquery get the value from this input, and finally it worked. – Cleber Benedet Aug 27 '15 at 20:38

using Jquery variable from mysql

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