Question

I have this undefined value it said that on javascript console what is wrong with my functions?

What i am trying to do here is that, when i click the button it will submit the hidden value to my database. after submitting the value it will fadeout.

the jQuery below

 // AJAX/JQUERY FORM
$(function() {
$(".videoThumbS").click(function() {

var test = $("#v_w_id_value").val();
var dataString = test;

if(test=='') {
    alert("Please Select Video");
} else {

$.ajax({
    type: "POST",
    url: "api.php",
    data: dataString,
    cache: false,
    success: function(html){
    $(".videoThumb4").fadeout();
    $(".videoThumb4").hide();
}
});
} return false;
});
});

the form below

<form method="post" action="">
<input type="hidden" name="v_w_id" value="UOIDKSJLKH" id="v_w_id_value" />
<input class="videoThumbS" onClick="ALKJKLJLKMNF" type="button" name="selectSel" value="Select" id="selectbut" />
</form>

why i always got that?
the api.php below

<?php
include ( 'class/conn.php' );
if( isset( $_POST['selectSel'] ) ) {
$v_watch_id = $_POST['v_w_id'];
$selected_videos = mysql_query( "INSERT INTO s_vids VALUE('', '$v_watch_id', NOW() )" ) or die ( mysql_error() );


/* VIEW RECORDS */      
$result = mysql_query("SELECT * FROM s_vids");            //query
$array = mysql_fetch_row($result);                          //fetch result    

echo json_encode($array);
}
?>

Answer 1

Your dataString is just the value of $("#v_w_id_value"), so there is no $_POST['selectSel'] and $_POST['v_w_id'].

The data of ajax request should be something like:

data: {selectSel: 1, v_w_id: dataString },

And in fact you don’t need to pass the selectSel to the sever side, only need to check v_w_id.