I'm using colorbox for my PM system. I added a trash-btn so users can delete their messages whenever they like. When the user opened his PM (colorbox) and clicks the trash-btn to delete a message, it does not delete it. While whenever I browse directly to a message and hit the button, it does work. So it only doenst work when colorbox has been opened.
Since I'm a new to this and dont know much of javascript, I would like it if anyone could help me out abit. Here's my EDITED code
read_message.php form (this opens in the colorbox)
echo '
<div class="inboxMessage">
<div class="inboxMessageImg NoNewMsg"></div>
<div class="inboxMessageHeader">
<a id="ajax" class="inboxMessageLink" onclick="showMessage('.$row['message_id'].')">'.$row['message_title'].'</a>
<p class="inboxMessageStatus Read">'.$inboxMessageStatus_Read.'</p>
</div>
<div class="inboxMessageDescription">'.$inboxMessageDescription.'</div>
<div class="inboxMessageActions">
<form method="post" action="message/delete_message.php">
<input type="hidden" value="'.$row['message_title'].'" name="message_title">
<input type="hidden" value="'.$row['message_id'].'" name="message_id">
<input type="submit" class="deleteMessageIcon" value="" name="deleteMessage">
</form>
</div>
</div>';
This is delete_message.php page
<?php
include '../../includes/db_connect.php';
sec_session_start();
//Delete bericht uit db
if (isset($_POST['deleteMessage'])) {
$msgID = $POST['message_id'];
$msgTitle = $POST['message_title'];
$deleteMessage = mysqli_query($mysqli,"DELETE FROM messages WHERE message_title = '$msgTitle' AND message_id = '$msgID'") OR die(mysql_error($mysqli));
if($deleteMessage) {
echo "Deleted";
}else{
echo "Error - Try again";
}
}
?>
If you need more information, please let me know so.
Thanks in advance!
PS - I know anyone is able to change the message id and title in the form so he's able to delete messages. First of all, the user needs to know the whole title of anyone elses message. Besides that, I'll create a random number which will be the message_id so its not easy to get/find the message id - IM WORKING ON IT. First, the delete function needs to work properly.
I think I saw your error:
if the query is returning and doing nothing is because the condition is Wrong. If you try the query in PHPMyAdmin the Query is OK, the only thing I can think is in this part:
$msgID = $POST['message_id'];
$msgTitle = $POST['message_title'];
it should be:
$msgID = $_POST['message_id'];
$msgTitle = $_POST['message_title'];
if it doesn't work make:
print_r("DELETE FROM messages WHERE message_title = '$msgTitle' AND message_id = '$msgID'");
and let us know the output
Related
i have a form for uploading files / images. I have no problem uploading and displaying the images , but i want to forward the images back to the same page that the imageas where selected, and use the POST data to place the image chosen in the 1st place , so the user can view and then if wanting to change the image for another.
The Problem i am having though is posting the flie back.
I have tried using it as a veriable and just using the variable to display an image but when i send that back to the image selection page, it does not display anything and if the user does not select a new image , the image data is not sent either..
i know im short cutting my explanation but i hope sum1 can help
here is my code (simplyfied)
THE INPUT FORM
<input type='file' class='fileinput' id='sellimage0' name='sellimage0' onchange='addimg0(this , img0, mainimage);' accept='image/*' /></input>
THE RECEIVER
if (!empty($_FILES['sellimage0']['name'])){
$image_path0 = $_FILES['sellimage0']['name'];
$image_path0 = filter_var($image_path0, FILTER_SANITIZE_STRING);
$image_path0 = strip_tags($image_path0);
$i0url=$image_path0;
} else {$i0url='';}
THE DISPLAY
<?php if (!empty($i0url)) { echo "
<div class='image0' id='image0' title='1st Image' >
<img id='img0' src='{$i0url}' class='imageclass'></img>
<input type='hidden' id='img0' name='sellimage0' value='{$_FILES['sellimage1']['name']}' ></input>
</div>
"; } ?>
THE POST BACK & RECEIVE
if (!empty($_FILES['sellimage0']['name'])){
$image_path0 = $_FILES['sellimage0']['name'];
$image_path0 = filter_var($image_path0, FILTER_SANITIZE_STRING);
$image_path0 = strip_tags($image_path0);
$i0url=$image_path0;
} else {$i0url='';}
<img id='img0' <?php if (!empty($i0url)) { echo " src='{$i0url}' "; } else { echo " src='' style='opacity:0;' "; } ?> onclick=' document.getElementById("sellimage0").click();' class='imageclass' ></img>
<div class='cancel' onClick='cancelimage(sellimage0 , img0);'> </div>
<div class='magnify' > </div>
<input type='file' class='fileinput' id='sellimage0' name='sellimage0' onchange='addimg0(this , img0, mainimage);' accept='image/*' /></input>
I hope this is making sense to you all.
i wish to choose an image, then post that choice to a view page
then from the view page , user has option to go back and edit
but how do i send the form data back to be received correctly so i can send the file back and forth between edit and review...
The way i have been trying just removes the selected images when i go back to the mainpage again.. iv tried many methods, but im lost on how to do this correctly.. Iv looked on the net to but finding my specific problem seems to be difficult.
Thank you.
I have worked on this for a few days now.
This snippet finds the correct row(s) from the database. It works fine. However, when the Button is clicked, I only want for the comment it is next to targeted, not all the comments on that page. It's hard to explain.. Let me show you image.
What I am doing currently is selecting the database rows and outputting them for debugging, instead of removing them (I don't want to go around deleting sections of my database with broken code)
$db is mysqli connection to the database
Snippet of the while loop:
TL/TR
I want to do the following with this in the end:
Remove the comment from the database.
Give the feedback to an admin.
This may sound like a noob question, but please help me.
Many thanks!
Just can't wrap my head around this, made like 6 pages single-handed and now I am stuck in a thing this "easy" for a d**n week and i still can't get it right.
EDIT
Latest snippet (Still doing it, What the hell!?!)
$query = $db->query("SELECT * FROM comments WHERE post_id='$id'");
while($row = $query->fetch_object()){
echo "<h5>".$row->name."</h5>","<br>";
$strip_comment = strip_tags($row->comment);
$delComment = $row->comment_id;
$strip_comment_shlashes = stripslashes($strip_comment);
echo "<blockquote>".$strip_comment_shlashes,"<br><br></blockquote>";
//button stuff
if($is_admin){
$query1 = $db->prepare("SELECT comment_id FROM comments WHERE comment_id = '$delComment'");
$query1->execute();
$query1->bind_result($commId);
while($row2 = $query1->fetch()):
?>
<form action="<?php echo $_SERVER['PHP_SELF']."?id=$id"?>" method="post">
<input type="submit" class="closeButton" name="deleteComment" value="<?php echo $commId; ?>" />
</form>
<?php
if(isset($_POST['deleteComment'])&& $is_admin){
if($is_admin && $commId){
echo "Comment ID <b>$commId</b> Removed";
}
}
endwhile;
}
Edit: NON-OBJECT error..
$delComment_2 = $_POST['deleteComment'];
$query2 = $db->prepare("SELECT * FROM comments WHERE comment_id='$delComment_2'");
$query2->execute();
Edit 2: (31/10)
Please, Could someone fix this snippet and post it? I don't usually ask for working snippets, but this one is driving me crazy. I just am too noob to understand how this goes. Thanks.
Change the input to this:
<input type="submit" class="closeButton" name="deleteComment" value="$comment_id_here" />
And then just use the comment id to delete a specific one from the database. (check $_POST['deleteComment'])
Of course, there are numerous other ways to do this - but the point is the same: you need to pass the comment id (not the post id) from your page to the query string. Whatever way you use to accomplish that is up to you, I just gave you an example.
I have put up some code to make an ajax, php, mysql chat in real time...
this is my code:
-jQuery/Ajax
$(function(){
setInterval(function() {
$('#chMsgCont').load('./ajax/msg.php');
var div = $('#chMsgCont');
var o = div.offset().top;
div.scrollTop( o + 12302012 );
}
,1000);
});
"./ajax/msg.php"
<?php
include("../system/config.site.php");
$query = mysql_query("SELECT * FROM chat_msg ORDER BY timestamp ASC");
while($p = mysql_fetch_assoc($query)) {
$auth = mysql_fetch_assoc(mysql_query("SELECT * FROM chat_users WHERE id = '".$p['auth_id']."'"));
?>
<div class="chatMsg">
<p id="chatPMsg">
<span class="chatTime"><?php echo date("H:i", $p['timestamp']); ?></span>
<b><?php echo $auth['name']." ".$auth['surname']; ?></b><br />
<?php echo stripslashes($p['msg']); ?>
</p>
<p id="chatImg">
<img src="./images/thumb<?php echo $p['auth_id']; ?>.png" />
</p>
<div style="clear:both;"><!– –> </div>
</div>
<?php
}
?>
I haven't found any method for not loading every second the "msg.php" file... I thought: is it possible to load initially the messages and then with setIntervall check for new messages to append to #chMsgCont? I think it is possible, but i can't figure out how to code it :/, can anyone help me?
Yes it is possible to load all messages upon initial page load; then subsequently update the chat window with new messages (have you ever wondered how Gmail auto populates new mail into your Inbox without a page refresh? Lots of AJAX requests).
You're on the right track. Just make sure you have a column in your MySQL database that holds a "yes/no" value on whether its a new message or not. Then everytime a new message is fetched using ajax/msg.php, in addition to fetching and outputting the content, perform a SQL update to update the fetched data as well.
Once you have that working you can further optimize your chat application by creating a simplier MySQL SELECT statement that simply checks the count of "new" rows (SELECT COUNT(new_message) FROM messages). Then check the value returned and whether it's greater than 0 before attempting to append content. I would also check for empty return data in your JavaScript to prevent from unnecessarily manipulating DOM elements.
How do I Limit some information displayed from the database and add a link eg "More" to enable read all information in a drop down using PHP. such as what is on facebook (Read more...). I am dealing with a lot of content and I dont want it all displayed at once.
Here is part of the code
echo "<p>".$row['Firstname']." ".$row['Lastname']."</p>";
echo "<p>".$row["Course"]." | ".$row["RegID"]."</p>";
echo "<p>".$row["Email"]."</p>";
echo "<p>"."Tel:".$row["Telephone"]."</p>";
echo "<p>".$row["info"]."</p>";
The code is running well only that I want to limit the information
echo "<p>".$row["info"]."</p>";
so that not all is displayed
Thanks
Use Jquery-ui click on "view source" and you'll see it's very simple really, just set the row that you want as the header (what's clicked to show the rest) and store the rest in a div below.
Split info into two strings, one intro, and the rest. Display only the intro to begin with. Insert a link that displays the rest when clicked.
$intro = substr($row['info'], 0, 200);
$rest = substr($row['info'], 200);
echo sprintf(
<<<HTML
<p>
<span class="intro">%s</span><span class="rest" class="display: none">%s</span>
Show more
</p>
HTML
, htmlentities($intro)
, htmlentities($rest)
);
displayRest is a Javascript-function that, given a link, finds the previous span with class rest, shows it and removes the link. I leave it as an exercise to implement this in a way that fits your project. You can go with native Javascript, or use a library such as jQuery, YUI, MooTools, Prototype etc.
if(isset($_POST['more']))
{
$query="select col1,col2,col3, ... ,colN from tableName ";
}
else
{
$query="select col1,col2,col3 from tableName ";
}
//HTML
<form method="post">
<input type="submit" name="more" value="More" />
</form>
//PHP
$records=mysql_query($query);
while($row=mysql_fetch_assoc($records))
{
//Display
}
The limit must be fixed on the SQL request.
// If you want to transmit limitation with a GET PARAMETER.
// You can also $_POST ur data.
$limitation = $_GET['limit'];
//..... And in your SQL REQUEST
$sql = "SELECT * FROM your_table LIMIT 0 , $limitation";
//And in the link....
echo 'Show only 10 Results'
?>
You can optimize that and add security precaution to prevent errors when $limitation receive empty or non numeric parameters.
<?php if(isset($_GET['limit']) && !empty($_GET['limit']) && !preg_match(EXPRESSION, $_GET['limit'])){
//YOU CAN DO THE LIMITATION WHITOUT SQL ERRORS
}
else{
//ERROR DIRECTIVE
}
?>
I'm new to forms and post data ... so I don't know how solve this problem!
I've a php page (page1) with a simple form:
<form method="post" action="/page2.php">
<input type="search" value="E-Mail Address" size="30" name="email" />
<input type="submit" value="Find E-Mail" />
</form>
How you can notice ... this form post the 'email' value to the page2. In the page2 there is a small script that lookup in a database to check if the email address exist.
$email = $_POST['email'];
$resut = mysql_query("SELECT * FROM table WHERE email = $email");
.
.
.
/* do something */
.
.
.
if($result){
//post back yes
}
else{
//post back no
}
I don't know how make the post back in php! And how can I do to the post back data are read from a javascript method that shows an alert reporting the result of the search?
This is only an example of what I'm trying to do, because my page2 make some other actions before the post back.
When I click on the submit button, I'm trying to animate a spinning indicator ... this is the reason that I need to post back to a javascript method! Because the javascript function should stop the animation and pop up the alert with the result of the search!
Very thanks in advance!
I suggest you read up on AJAX.
Here's a PHP example on W3Schools that details an AJAX hit.
Hi i think you can handle it in two ways.
First one is to submit the form, save the data in your session, check the email, redirect
back to your form and display the results and data from session.
Like
session_start();
// store email in session to show it on form after validation
$_SESSION['email'] = $_POST['email'];
// put your result in your session
if ($results) {
$_SESSION['result'] = 'fine';
header(Location: 'yourform.php'); // redirect to your form
}
Now put some php code in your form:
<?php
session_start();
// check if result is fine, if yes do something..
if ($_SESSION['result'] == 'fine) {
echo 'Email is fine..';
} else {
echo 'Wrong Email..';
}
?>
More infos : Sessions & Forms
And in put the email value back in the form field
<input type="search"
value="<?php echo $_SESSION['email']; ?>"
size="30"
name="email" />
Please excuse my english, it is horrible i know ;)
And the other one the ajax thing some answers before mine !
As a sidenote, you definitly should escape your data before using it in an SQL request, to avoid SQL injection
As you are using mysql_* functions, this would be done with one of those :
mysql_escape_string
or mysql_real_escape_string
You would not be able to post in this situation as it is from the server to the client. For more information about POST have a look at this article.
To answer your question you would want to do something like this when you have done your query:
if(mysql_num_rows($result)){ //implies not 0
$data = mysql_fetch_array($result);
print_r($data);
}
else{
//no results found
echo "no results were found";
}
The print_r function is simply printing all the results that the query would have returned, you will probably want to format this using some html. $data is just an array which you can print a single element from like this:
echo $data['email'];
I hope this helps!
<?php
echo " alert('Record Inserted ');"
OR
echo " document.getElementByID('tagname').innerHtml=$result;"
?>
OR
include 'Your Html file name'