I have a form in which a user uses to comment on a post. Now the hidden input of this form is correct, it gets the correct 'streamidcontent' yet when I send it through ajax and into the database it always changes to the last created status id '4076' and adds it to that post at the very top of the feed. So I'm wondering, what I'm doing wrong.
streamdata_comments
1 comment_id int(11) No None AUTO_INCREMENT
2 comment_poster int(11) No None
3 comment_streamitem int(11) No None
4 comment_datetime datetime No None
FORM
<form id="mycommentform" method="POST" class="form_statusinput">
<input type="hidden" name="streamidcontent" id="streamidcontent" value="'.$streamitem_data['streamitem_id'].'">
<input type="input" name"content" id="content" placeholder="Say something" autocomplete="off">
<input type="submit" id="button" value="Feed">
</form>
COMMENT_ADD.PHP
<?php
session_start();
require"include/load.php";
error_reporting(E_ERROR | E_WARNING | E_PARSE | E_NOTICE);
if(isset($_POST['streamidcontent'])&isset($_POST['content'])){
$content = htmlspecialchars($_POST['content']);
$streamid = htmlspecialchars($_POST['streamidcontent']);
$content = preg_replace('/(?<!S)((http(s?):\/\/)|(www.))+([\w.1-9\&=#?\-~%;\/]+)/','<a href="http$3://$4$5">http$3://$4$5</a>', $content);
$insert = "INSERT INTO streamdata_comments(comment_poster, comment_streamitem, comment_datetime, comment_content) VALUES (".$_SESSION['id'].",'$streamid',UTC_TIMESTAMP(),'$content')";
$add_post = mysqli_query($mysqli,$insert) or die(mysqli_error($mysqli));
}
AJAX
<script>
$(document).ready(function(){
$("form#mycommentform").submit(function(event) {
event.preventDefault();
var streamidcontent = $("#streamidcontent").val();
var content = $(this).children('#content').val();
$.ajax({
type: "POST",
url: "comment_add.php",
cache: false,
dataType: "json",
data: { streamidcontent: streamidcontent, content: content},
success: function(html){
$("#containerid").html("<div class='stream_comment_holder' id='comment_holder_"+html['comment_streamitem']+"'><div id='comment_list_"+html['comment_streamitem']+"'><div class='stream_comment' id='comment_"+html['comment_id']+"'>div class='stream_comment_holder' id= style='display:;'><div class='stream_comment'><table width='100%'><tbody><tr><td valign='top' width='30px'><img class='stream_profileimage' style='border:none;padding:0px;display:inline;' border=\"0\" src=\"imgs/cropped"+html['id']+".jpg\" onerror='this.src=\"img/no_profile_img.jpeg\"' width=\"40\" height=\"40\" ></td><td valign='top' align='left'><a href='profile.php?username="+html['username']+"'>"+html['first']+" </a>"+html['comment_content']+"</td></tr></tbody></table></div></div></div></div>");
}
});
return false
});
});
</script>
AND MY OLD AJAX WHICH INSERTED FINE. But I need to add my like, dislike and delete button to this. So changed it to the above AJAX which doesn't work properly.
function addcomment(streamid,content,containerid,posterid,postername,postid){
var obj = document.getElementById(containerid);
$.post("../comment_add.php", { streamid: streamid,content:content} );
obj.innerHTML = obj.innerHTML + "<div class='stream_comment'><table width='100%'><tbody><tr><td valign='top' width='30px'><img style='border:none;padding:0px;height:30px;width:30px;border-radius:0px;' src='imgs/cropped"+posterid+".jpg' onerror='this.src="img/no_profile_img.jpeg";'></td><td valign='top' align='left'><a href='profile.php?username="+posterid+"'>"+postername+" </a>"+content+"</td></tr></tbody></table></div>";
}
Have you ever thought that the error might actually be your feed, and not the actual insert? Or it could be the AJAX?
To verify the insert is working or not, try:
echo $streamid; // Shows you're getting the right value from AJAX.
$add_post = mysqli_query($mysqli,$insert) or die(mysqli_error($mysqli));
if($add_post) {
$NewID = mysqli_insert_id(mysqli);
$query = 'SELECT * FROM streamdata_comments WHERE comment_id=' . $NewID;
if ($result = mysqli_query($mysqli, $query)) {
while ($row = $result->fetch_object()){
var_dump($row); // Check the actual input is as you expected.
}
mysqli_free_result($result);
}
}
If using AJAX, you'll need to monitor results through one of the network tool (press F12 in Chrome, NET panel; or use Fiddler for Win or Charles for Mac)
There are some strange things happening in your code:
$content = $_POST['content'];
$content = strip_tags($_POST['content']);
You are filling the $content variable twice with different values. The same goes for the id:
$streamid = $_POST['streamidcontent'];
$streamid = strip_tags($_POST['streamidcontent']);
I seriously doubt you should use strip_tags anyway before inserting it into the database. Just use htmlspecialchars() when you display it. Or use htmlpurifier when displaying it. Also you have a SQL Injection vulnerability in your application. Read more about preventing this here: How can I prevent SQL injection in PHP?.
With the code you have shared the new record will be assigned a new id. How did you check you got the same id everytime?.
Finally you can get the last inserted id in mysqli by doing:
echo $mysqli->insert_id;
after the query().
Also have you checked the rendered HTML to see what the id is in the first place?