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Nikola242016-06-15 21:36:02
PHP
Nikola24, 2016-06-15 21:36:02

$_Post does not accept / return a value. Why?

Hello everyone! For some reason, post either does not accept, or does not correctly give the value. The result is null, although it gives the correct value in alert. There is also some error with the output:

mysql_fetch_row

Warning: mysql_fetch_row() expects parameter 1 to be resource, boolean given in Y:\home\localhost\www\article.php on line 7

Send and receive code
<script type="text/javascript" charset="utf-8">
    var art_id;
    function reply_click(art_id)
    {
      alert(art_id);
      jQuery(document).ready(function(){
      jQuery.ajax({
        url:"http://localhost/article.php",
        type:"POST",
        dataType:"json",
        data: "art_id" + art_id,
        success: function(html){
          var ART = jQuery('#ART');
          console.log(html);
          jQuery.each(html, function(key, value){
            console.log(value);
            ART.append(value.article_text);
          });
        }, error: function(e){
          alert(e);
        }
      });
    });
    }
  </script>

Base call code:
<?php 
  $id = $_POST['art_id'];
  print_r(json_encode($id));
  $sqlconn = mysqli_connect('localhost','root','','carhelper') or die(mysqli_error());
  $query = "SELECT article_text FROM `table_article` where id=$id ";
  $res =  mysql_query($query);
  $row = mysql_fetch_row($res);
  print_r(json_encode($row[0]));
?>

Please tell me what can be done about it!

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1 answer(s)
A
Alexander Sisyukin, 2016-06-15
@Nikola24

var art_id; /// it's not clear at all what you could put there when you just declared it
in json, accepted it, then decode json-decode and do not encode json-encode,
otherwise you send json to the selection

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