php - The success Function in Ajax Doesn't work When using Json -

php - The success Function in Ajax Doesn't work When using Json -

i have login form shows in dialog box in jquery need pass variable php jquery using ajax & json method problem success function in ajax doesn't work , error function

here's both php , jquery code:

jquery:

$(document).ready(function () { $(".logbut").click( function () { $(".login").dialog({ width: 400, height:320, modal: true, autoopen: true, resizable: false, show: { effect: "blind", duration: 3000 }, hide: { effect: "drop", duration: 1000 }, buttons: { 'login': function() { $.ajax({ url:"login.php", type: 'post', datatype:"json", success: function(data) { if(data.status == 'success'){ alert("thank subscribing!"); }else if(data.status == 'error'){ alert("error on query!"); } }, error: function(){ alert("the ajax doesnt work"); } }) } } }) }) });

login.php:

<?php $servername = "amin-pc"; $connectioninfo = array( "database"=>"market"); $conn = new pdo("sqlsrv:server=amin-pc;database=market"); $conn->setattribute(pdo::attr_errmode, pdo::errmode_exception); $response_array = array(); if( $conn === false ) { die( print_r( sqlsrv_errors(), true)); } $passorg = $_request['passorg']; $passorg = $conn->quote($passorg); $userorg = $_request['userorg']; $userorg = $conn->quote($userorg); $sql = "select count(*) customers customers_user=$userorg , customers_pass=$passorg"; $stmt = $conn->query($sql); $result = $stmt->fetch(); $count = $result[0]; if($count >0) { $response_array['status'] = 'success'; } else{ $response_array['status'] = 'error'; } header('content-type: application/json'); echo json_encode($response_array); ?>

html:

<div class="logbut">login</div> <div class="login" style="display:none;"> <form method="post" class="logform" action="login.php"> <input type="text" name="userorg" style=" width:80px;" class="userorg" /> <label>username</label> <input type="password" name="passorg" style=" width:80px;" class="passorg"/> <label>password</label> <input type="submit" style="display:none" value="check"/> </form> </div>

thanks

can seek specifying info parameter follows in ajax request:

$.ajax({ url:"login.php", type: 'post', datatype:"json", data:{passorg:value of passorg}, success:....

php jquery ajax json