Ajax and Mysql Inserting 2 same row for one click

Question

I have a button which has attribute onclick to start javascipt function and my javascript ajax script has xmlhttp object. It is communicating with dbaction.php

When I click button everything is okay about script and mysql but the problem is about rows. It is inserting TWO same row for each click until it rowCount>0 and when after first click echo prints "0" for rowCount(), then if I click again now echo prints "2" for rowCount()

Where is the problem?

enter image description here

<body>

<script type="text/javascript">

function loadName() {
    var xmlhttp;
    xmlhttp = new XMLHttpRequest();

    xmlhttp.onreadystatechange = function()
    {
        if (xmlhttp.readyState === 4 && xmlhttp.status === 200)
        {
            document.getElementById("sample_div").innerHTML = xmlhttp.responseText;

            $("#my_button").attr("value", "Sent");
            $("#my_button").attr("style", "color:red;");
        }
    };
    xmlhttp.open("GET", "dbaction.php?sender_id=5&receiver_id=6", true);
    xmlhttp.send();
}
  </script>

<input type="submit" id="buton" onclick="loadName();" value="Submit"/>
 <div id="sample_div" style="background-color: khaki; width:50%; margin: 0 auto;">     
 </div> 
 </body>

dbaction.php

<?php
//Database connection using PHP PDO(php database objects)
try {
$db = new PDO("mysql:dbname=member;host=localhost", "root", "",    array(PDO::MYSQL_ATTR_INIT_COMMAND => "SET NAMES utf8"));
 } 
catch ( PDOException $e ) {
echo $e->getMessage();
}

$sql= "SELECT * FROM friendship WHERE sender_id=".$_GET["sender_id"]." AND    receiver_id=".$_GET["sender_id"]."";
$rs=$db->prepare($sql);
$rs->execute();
$rs->fetch();
$count = $rs->rowCount();

if ($count>0){
 echo "You already sent request...";
 echo $count;
}
else {
$query = $db->query("INSERT INTO friendship (sender_id, receiver_id, is_approved) VALUES (".$_GET["sender_id"].",".$_GET["receiver_id"].",0)" );
$query->execute();

echo "Request sent";
echo $count;
}

check your console panel and see if only one request is sent , i think loadName() is calling twice — M Khalid Junaid
– M Khalid Junaid, Commented Jun 25, 2013 at 18:35
what do you mean about console panel? how can i monitorize it? — hakki
– hakki, Commented Jun 25, 2013 at 18:36
browser's console panel in firefox there is a addon firebug install it and check — M Khalid Junaid
– M Khalid Junaid, Commented Jun 25, 2013 at 18:39
if you are using Google Chrome then press f12 (It will open Developer tools) before clicking on the input and click on Network tab. — bystwn22
– bystwn22, Commented Jun 25, 2013 at 18:51

Rajeev Ranjan · Accepted Answer · 2013-06-25 18:39:08Z

1

check the difference :-

<?php
//Database connection using PHP PDO(php database objects)
try {
$db = new PDO("mysql:dbname=member;host=localhost", "root", "",          array(PDO::MYSQL_ATTR_INIT_COMMAND => "SET NAMES utf8"));
 } 
catch ( PDOException $e ) {
 echo $e->getMessage();
}

$sql= "SELECT * FROM friendship WHERE sender_id=".$_GET["sender_id"]." AND       receiver_id=".$_GET["sender_id"]."";
$rs=$db->prepare($sql);
$rs->execute();
$rs->fetch();
 $count = $rs->rowCount();

if ($count>0){
 echo "You already sent request...";
 echo $count;
 }
  else {
 $query = $db->query("INSERT INTO friendship (sender_id, receiver_id, is_approved) VALUES (".$_GET["sender_id"].",".$_GET["receiver_id"].",0)" );
 //$query->execute();

   echo "Request sent";
   echo $count;
}

in last else block you are using query() and execute() both,used for execute sql queries one query() directly and execute() for prepared statement

answered Jun 25, 2013 at 18:39

Rajeev Ranjan

4,1523 gold badges30 silver badges42 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

hakki Over a year ago

Ok, this is working properly. But without executing $query, how insert query inserting a row?

Rajeev Ranjan Over a year ago

@hakiko see the documentation here ellislab.com/codeigniter/user-guide/database/queries.html

bystwn22 · Accepted Answer · 2013-06-25 18:49:44Z

like @dianuj said in his comment, if you saw two requests at the same time in console, then change your js function like below

<script type="text/javascript">
  var in_request  = false;
  function loadName() {
    if ( in_request ) {
      return false;
    }
    in_request  = true;
    var xmlhttp;
    xmlhttp = new XMLHttpRequest();
    xmlhttp.onreadystatechange  = function() {
      if ( xmlhttp.readyState === 4 && xmlhttp.status === 200 ) {
        in_request  = false;
        document.getElementById( "sample_div" ).innerHTML = xmlhttp.responseText;
        $("#my_button").attr( "value", "Sent" );
        $("#my_button").attr( "style", "color:red;" );
      }
    };
    xmlhttp.open( "GET", "dbaction.php?sender_id=5&receiver_id=6", true );
    xmlhttp.send();
  }
</script>

Collectives™ on Stack Overflow

Ajax and Mysql Inserting 2 same row for one click

2 Answers 2

2 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related