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AuthorTopic: jQuery.post()  (Read 1733 times)

Offline aegkaluk

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« on: 19, 12, 2012, 08:10:28 »
http://api.jquery.com/jQuery.post/

demo.php
Code: [Select]
<script type="text/javascript">

    $(document).ready(function() { 
$("#dialog").dialog({
   autoOpen: false,
   modal: true,
   buttons : {
"Confirm" : function() {
var wk_id = $("#callConfirm").attr("title");
//--alert("You have confirmed!");     
updateAck(wk_id);
$("#dialog").dialog("close");
},
"Cancel" : function() {
  $(this).dialog("close");
}
  }
});

$("#callConfirm").on("click", function(e) {
e.preventDefault();
$("#callConfirm").attr("disabled", "disabled");
$("#dialog").dialog("open");
});



    });

function updateAck(wk_id) {
$.post("updateAck.php", { id: id},
function(data){
   console.log(data.SQL);   
   console.log("update status accept : "+data.STATUS);   
       if(data.STATUS){
alert("Accept Success!");
   }else{
alert("Can't accept this time! Please try again!");
   }
   
},"json");
}

});
</script>

<button id="callConfirm" title="<?php echo$_GET['id'];?>">Accept</button>
<div id="dialog" class="dialog" title="Confirmation Required">
  Confirm this item ?
</div>


update.php
Code: [Select]
<?php
function connectSQL($servername,$databasename,$user_encode,$password_encode){
    
$user base64_decode($user_encode); 
    
$pass base64_decode($password_encode);
    
$connection_string ="DRIVER={SQL Server};SERVER=$servername;DATABASE=$databasename;AutoTranslate=no"
    
$conSQL odbc_connect($connection_string,$user$pass) or die ("can't connect server"); 
    return 
$conSQL;
}
$connection=  connectSQL('server_name''db_name'base64_encode('user'), base64_encode('password'));

$id$_POST['id'];

$sql="Update <your_table> Set field_name='<value>' Where <field_name>='{$value}'";
$objQuery=odbc_exec($connection,$sql);
if(
$objQuery){
$STATUS=true;
}else{
$STATUS=false;
}

echo 
json_encode(array("STATUS"=>$STATUS,"SQL"=>$sql));

?>

« Last Edit: 19, 12, 2012, 08:13:35 by aegkaluk »