5 replies to this topic

[Stasonix]

for example, I have query like this

$sql = "SELECT `name`,`surname`,`age` FROM `table` WHERE (`name`='".$name,"') AND (`age`=20)";

$query = mysql_query($sql);

$result = mysql_fetch_array($query);

so I need to echo only 2 records is name and age and to recieve it in JQuery(ajax-method), what I am gonna do next?

[John]

Iterate through the results echoing name and age.

[codehutch]

while ($result = mysql_fetch_array($query, MYSQL_ASSOC)) {

    echo $result["name"] . " - " . $result["age"];

}

I don't know what you mean about the "ajax thing" though. Maybe you can make it more clearer.

[Calgon]

Create a separate page which calculates everything using the code codehutch provided, and then use jquery's ajax functions to echo out the results dynamically, here's an example:

$.ajax({

  type: "GET",

  url: "page_query_calc_thing.php",

}).done(function( msg ) {

  $('#div-to-update').html(msg); 

});

[SoN9ne]

If you are using it VIA AJAX, do not echo the data. echo the data as JSON. This is a standard for handling data VIA AJAX and it allows you more control over the data.

[Orjan]

Agree with SoN9ne. You just read out your data and fill an array with it, and then, there are json_encode() function, just echo it's result to the ajax call.

---------- Post added at 12:18 AM ---------- Previous post was at 12:17 AM ----------

unnecesary post as I missed the SoN9nes link to the function, still I agree with him.