jquery - How to solve SyntaxError: JSON.parse: unexpected character at line 1 column 1 of the JSON data in ajax and php -

how solve error: syntaxerror: json.parse: unexpected character @ line 1 column 1 of json data

i sending data , ajax , php.

here ajax code:

flag = 111; var dt = $(this).serializearray(); dt.push({name: 'flag', value: flag });                               $.ajax({ url: 'emp.php', type: "post", async:true, data:  dt ,  datatype: 'html', contenttype: 'application/x-www-form-urlencoded; charset=utf-8', success: function(data){                     var x = json.parse(data); //this line shows error!! alert(x); $('#name').val(x.ename); $('#designation').val(x.designation); $('#department').val(x.department); $('#sd').val(x.secdivision); },                       error: function(jqxhr, textstatus, errorthrown) { console.log(textstatus, errorthrown); } }); 

here php:

$empid = (isset($_post['employeeid'])) ? $_post['employeeid'] : 'not'; $flag  = (isset($_post['flag'])) ? $_post['flag'] : 0; if($flag == 111){     $stid = oci_parse($conn, " begin                           :result :=  pkg_payroll.get_emp_by_id('<employee_id>$empid/employee_id>');                          end;" );       oci_bind_by_name($stid, ':result',$ru, 5000);        $output = oci_execute($stid);     $ru = new simplexmlelement($ru);         $json = json_encode($ru, json_numeric_check);     $jsonarray = json_decode($json ,true);       $jsn = $jsonarray['employee'];      $array = array('employee' =>   $jsn['empid'],                        'ename' => $jsn['ename'],                        'designation' => $jsn['designation'],                         'department'=>  $jsn['department'],                        'secdivision'=>  $jsn['secdivision']);                           echo json_encode($array); } 

updates: here sample of response data got in console after echo json_encode($array);

<br /> <font size='1'><table class='xdebug-error xe-notice' dir='ltr' border='1' cellspacing='0' cellpadding ='1'> <tr><th align='left' bgcolor='#f57900' colspan="5"><span style='background-color: #cc0000; color: #fce94f ; font-size: x-large;'>( ! )</span> notice: undefined index: employee in c:\wamp\www\payroll\emp.php  on line <i>24</i></th></tr> <tr><th align='left' bgcolor='#e9b96e' colspan='5'>call stack</th></tr> <tr><th align='center' bgcolor='#eeeeec'>#</th><th align='left' bgcolor='#eeeeec'>time</th><th align ='left' bgcolor='#eeeeec'>memory</th><th align='left' bgcolor='#eeeeec'>function</th><th align='left'  bgcolor='#eeeeec'>location</th></tr> <tr><td bgcolor='#eeeeec' align='center'>1</td><td bgcolor='#eeeeec' align='center'>0.0002</td><td bgcolor ='#eeeeec' align='right'>247040</td><td bgcolor='#eeeeec'>{main}(  )</td><td title='c:\wamp\www\payroll \emp.php' bgcolor='#eeeeec'>..\emp.php<b>:</b>0</td></tr> </table></font> {"employee":"fmcsc00015","ename":"tom","designation":"teacher","department":"english","secdivision":"academic" } 

parsererror syntaxerror: json.parse: unexpected character @ line 1 column 1 of json data

i confuse main reason of error, because did same type of coding json earlier.i checked php working fine.please me.thanks.

you returning json server , parsing html datatype in client side . so, in code change datatype

dattype: 'html' 

to

datatype: 'json', 

hope helps .