php - getting data from database as a json response -

here trying data database , want display json response user can fetch each field.

here how user can perform query

http://localhost/safari/index.php?getbranch=true 

this should give branch details tables.

here php code it

<?php     if(isset($_get['getbranch']))     {         $getbranch = $_get['getbranch'];         if($getbranch == 'true')         {             getbranch($getbranch);         }     }     function getbranch($getbranch)     {         $con = mysqli_connect('127.0.0.1', 'root', '', 'safari');         if (mysqli_connect_errno())         {             echo "failed connect mysql: " . mysqli_connect_error();             return;         }            $today = date("ymd");                     $result = mysqli_query($con,"select division, branch,branchofficename,branchofficecode,status tbl_branchoffice");         while ($row = @mysqli_fetch_array($result))         {             $result1 = json_encode($row);                      }         echo $result1;     } 

what's wrong wit this?

json response:

[{"0":"3","sno":"3","1":"2","division":"2","2":"2","branch":"2","3":"saffari travels","branchofficename":"saffari travels","4":"gfgbhghfhf","branchofficecode":"gfgbhghfhf","5":"active","status":"active"}, 

{"0":"4","sno":"4","1":"2","division":"2","2":"chennai","branch":"chennai","3":"chennai","branchofficename":"chennai","4":"br01","branchofficecode":"br01","5":"active","status":"active"},{"0":"5","sno":"5","1":"3","division":"3","2":"delhi","branch":"delhi","3":"delhi","branchofficename":"delhi","4":"br02","branchofficecode":"br02","5":"notactive","status":"notactive"},{"0":"6","sno":"6","1":"2","division":"2","2":"bangalore","branch":"bangalore","3":"bangalore","branchofficename":"bangalore","4":"br03","branchofficecode":"br03","5":"active","status":"active"},{"0":"7","sno":"7","1":"3","division":"3","2":"pune","branch":"pune","3":"pune","branchofficename":"pune","4":"br04","branchofficecode":"br04","5":"notactive","status":"notactive"}]

change while loop

$result1 = array(); while ($row = @mysqli_fetch_array($result)) {            array_push($result1 , $row);             } 

by doing so, have collected result in $result1

now can encode it

echo  $result1 = json_encode( $result1);   

i prefer use array, ignor json_encode line code,

foreach($result1 $resultset){  //resultset contains 1 single row of table     foreach($resultset $column => $columnvalue){   //assuming table column name 'city'         if($column == 'city' && $columnvalue == 'pune' ){          //displaying array of table satisfies condition              var_dump($resultset );          }     } }