ajax - Display database value in array PHP MYSQLi -


i"m attempting display data i've sent ajax php file, reason not displaying on page. way works enter search term input field, , ajax script post value php script, return database value requested back.

error_reporting(e_all);  ini_set('display_errors', '1');   if (isset($_post['name']) === true && empty($_post['name']) === false) {     //require '../db/connect.php';      $con = mysqli_connect("localhost","root","root","retail_management_db");      $name = mysqli_real_escape_string($con,trim($_post['name']));         $query = "select `names`.`location` `names` where`names`.`name` = {$name}";     $result = mysqli_query($con, $query);     if (mysqli_num_rows($result) > 0) {         while ($row = mysqli_fetch_array($result)) {             $loc = $row['location'];              echo $loc;            }//close while loop                                 } else {         echo $name . "name not found";     } }

html form:

<!doctype html>  <html>     <head>     <meta charset="utf-8">     <title>retail management application</title>     </head>      <body>                      name: <input type="text" id="name">          <input type="submit" id="name-submit" value="grab">             <div id="name-data"></div>         <script src="http://code.jquery.com/jquery-1.11.2.min.js"></script>         <script src="js/global.js"></script>     </body> </html>

you're appending mysql error result query, , you're trying query query result, try following:

$query = "select `names`.`location` `names` where`names`.`name` = '$name'"; $result = mysqli_query($con, $query); if (mysqli_num_rows($result) > 0) {

edit:

{$name} string , should quoted instead.

change '$name' in where clause.

using:

$result = mysqli_query($con, $query) or die(mysqli_error($con));

will provide reason why query failed.


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