jQuery AJAX call to a database query

Question

I have an AJAX function as so:

function admin_check_fn(type)
{

    //$("#notice_div").show();

    //var allform = $('form#all').serialize();

    $.ajax({
        type: "POST",
        //async: false,
        url: "<?php bloginfo('template_url'); ?>/profile/adminquery_check.php",
        data: { type: type },
        //data: 'code='+code+'&userid='+userid,
        dataType: "json",
        //dataType: "html",

        success: function(result){
            var allresult = result.res
            $('#result').html(  allresult  );

            alert(allresult);
            //$("#notice_div").hide();  
        }
    })
}

And server-side:

$queryy="SELECT * FROM wp_users";
$name = array();    
$resultt=mysql_query($queryy) or die(mysql_error()); ?>

<?php while($rowss=mysql_fetch_array($resultt)){  

    $name = $rowss['display_name']; 

}

echo json_encode( array( 

    "res" =>  array($rowss['display_name']),
    "fvdfvv" => "sdfsd"

    ) 
);

Basically for some reason it is not displaying all of the returned values from the query to the users table in the database. It works when I query another table with just one entry in it, so im thinking it could be something to do with the fact there is an array it is not parsing correctly?

Just wondered if anyone else has came accross this problem?

Thanks

You're trying to inject the json object you get from the server into the markup without any transformation, and this is not going to work. Anyway did you check the json that you get back, and is the success callback correctly called? — mamoo
– mamoo, Commented Mar 2, 2012 at 13:54
I would strongly encourage the use of vaiable names that do not look like type-o's. — halfpastfour.am
– halfpastfour.am, Commented Mar 2, 2012 at 14:01

charlietfl · Accepted Answer · 2012-03-02 13:57:46Z

3

Your ajax success is treating the returned data as html but it is json.

var allresult = result.res
 /* assumes allResult is html and can be inserted in DOM*/
 $('#result').html(  allresult  );

You need to parse the json to create the html, or return html from server

Also php loop:

$name = $rowss['display_name'];

Should be more like:

$name[] = $rowss['display_name'];

answered Mar 2, 2012 at 13:57

charlietfl

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Smamatti · Accepted Answer · 2012-03-02 13:56:00Z

2

You always overwrite the name:

<?php while($rowss=mysql_fetch_array($resultt)) {  
    $name = $rowss['display_name']; 
}

But this part is not within your loop:

"res" =>  array($rowss['display_name']),

Therefore you only get one result (the last one).

answered Mar 2, 2012 at 13:56

Smamatti

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1 Comment

JamesG Over a year ago

Yes thats right, I can grab a value by putting in part of the array i want i.e.[1] but how do i go about getting all of it?

halfpastfour.am · Accepted Answer · 2012-03-02 13:59:42Z

1

Smamatti has a good point.

To fix this:

<?php
$query = "SELECT * FROM wp_users";
$names = array();    
$result = mysql_query($query) or die(mysql_error());

while( $rows = mysql_fetch_array( $result ) ){
    $names[] = $rows['display_name']; 
}

echo json_encode( array( 
    "res" =>  $names,
    "fvdfvv" => "sdfsd"
    ) 
);

answered Mar 2, 2012 at 13:59

halfpastfour.am

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