I have the following jquery ajax request
$(document).ready(function(){
var friendrequest = $(".friendrequest").val();
$(".afriendreq").click(function(){
$(".afriendreq").hide();
$.ajax({ type: "POST", url:"functions/ajaxfriends.php", data:"friendrequest" ,success:function(result){
$(".cfriendreq").show();
}});
});
});
And it gets the input from here
while ($row = mysql_fetch_array($search)) {
d
?>
<div id="search_container">
开发者_开发百科 <div id="search_image"><img src="<?php echo $row['picture'] ?>"></img></div>
<input type="hidden" value="<?php echo $row['id'] ?>" id="friendrequest" ></input>
<div id="search_name"><a href="profile.php?id=<?php echo $row['bigid'] ?>.'"> <?php echo $row['first_name'] . " " . $row['last_name']; ?></a> </div>
<div id="search_friend"><a class="afriendreq">Send Friend Request</a><a class="cfriendreq" style="display:none;">Cancel Friend Request</div>
</div><?php
echo "<br />";
}
?>
Unfortunately it's not working though. Can anyone find the problem because it's bugging me? Also the functions/ajaxfriends.php is
<?php
include 'functions.php';
if (isset($_POST['friendrequest'])){
$friendrequest= $_POST['friendrequest'];
$friendrequest= filter ($_POST['friendrequest']);
$sql_connectfriend = " INSERT INTO friends (`useridone` ,`useridtwo` ,`request`) VALUES ('$_SESSION[user_id]', '$friendrequest', '1' ";
$search = mysql_query($sql_connectfriend, $link) or die("Insertion Failed:" . mysql_error());
}
?>
Can anyone see why? Thanks in advance.
Try:
$.ajax({
url: 'functions/ajaxfriends.php',
type: 'POST',
data: { friendrequest: friendrequest },
success: function(result) {
$('.cfriendreq').show();
}
});
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