Reputation:
i am trying to create an ajax favorite button in php similar to instagram and twitter.
my code seems to be fine and i am doing everything correctly and tried to get it working this way:
php code:
$user_id = $_SESSION['active_user_id'];
extract($_POST);
extract($_GET);
if(isset($_GET['message']))
{
$id=$_GET['message'];
$q=$db->prepare("SELECT msgid,date,text
FROM messages
WHERE to_id=? and msgid=?");
$q->bindValue(1,$user_id);
$q->bindValue(2,$id);
$q->execute();
$row2=$q->fetch();
$d=$row2['date'];
$fav_questionq=$db->prepare("SELECT *
FROM messages
LEFT JOIN users
ON messages.to_id=users.id
WHERE users.id=? AND messages.msgid=?
");
$fav_questionq->bindValue(1,$user_id);
$fav_questionq->bindValue(2,$id);
$fav_questionq->execute();
$frow=$fav_questionq->fetch();
$fquestion= $frow['text'];
$result = $db->prepare("SELECT * FROM fav_messages
WHERE username=? AND message=?");
$result-bindValue(1,$user_id);
$result-bindValue(2,$id);
$result->execute();
if($result->rowCount()== 1 )
{
$deletequery=$db->prepare("DELETE FROM fav_messages WHERE message=?");
$deletequery->bindValue(1,$id);
$deletequery->execute();
echo("<script>location.href = 'index.php?a=recieved';</script>");
}
else
{
$insertquery = $db->prepare("INSERT INTO fav_messages (username,message,fav_question,fav_date) values(?,?,?,?)");
$insertquery->bindValue(1,$user_id);
$insertquery->bindValue(2,$id);
$insertquery->bindValue(3,$fquestion);
$insertquery->bindValue(4,$d);
$insertquery-execute();
}
echo("<script>location.href = 'index.php?a=recieved';</script>");
}
javascript code:
<script>
function GetXmlHttpObject() {
var xmlHttp=null;
try
{
// Firefox, Opera 8.0+, Safari
xmlHttp=new XMLHttpRequest();
}
catch (e)
{
// Internet Explorer
try
{ xmlHttp=new ActiveXObject("Msxml2.XMLHTTP"); }
catch (e)
{ xmlHttp=new ActiveXObject("Microsoft.XMLHTTP"); }
}
return xmlHttp;
}
function ajaxfav(){
var xmlHttp=GetXmlHttpObject();
var url="favorite.php?message="+document.msgidform.fav_message.value;
xmlHttp.onreadystatechange=function(){
if(xmlHttp.readyState==4){
alert("Message is favorited");
}
}
xmlHttp.open("GET",url,true);
xmlHttp.send(null);
}
</script>
HTML form and link code:
<form name="msgidform" method="post">
<input type="hidden" name="fav_message" id="" <?php echo "value= '$msg_id'"; ?>></p>
</form>
<a class="msg-icon" href="" onclick="ajaxfav();"><img
src="images/linedfav.png" id='img'></img></a>
but i kept getting an error in the console that reads:
"Uncaught TypeError: Cannot read property 'value' of undefined at ajaxfav" but i've fixed it and now it immediately goes to the alert message, but nothing is inserted into the database, meaning that the data was not sent to the php file. can someone advise me on what i can do? network tab of the ajax call
please i would appreciate any help or suggestion. the php file is called but does not insert anything into the database.
Upvotes: 0
Views: 84
Reputation: 1943
You can use
document.msgidform.elements['fav_message'].value
in your ajaxfav() function to get name field value.
Upvotes: 1