PHP - Form input inside table array

Question

I'm having an issue with a form inside my table. I created a table that shows a list of recent status's that the people the user follows posts. I created a form inside the table so the community can "like" or "unlike".

Every time I try to submit the form, the post is being posted in the URL and stops at my button name "likestatus="

example of the URL returned:

account.php?likesid=5&likerid=117&likestatus=

The input of the form works, as each button shows the corresponding "likesid" and "likerid" in the URL

I just don't understand why it's failing.

Here is the code for my "$_POST['likestatus']"

<?php
if(isset($_POST['likestatus']))
{


//inputs from the form
$likesid =mysqli_real_escape_string($conn, $_POST['likesid']);
$likerid =mysqli_real_escape_string($conn, $_POST['likerid']);

//query to check if already likes status
$sql = "SELECT likesid, likerid FROM user_likes WHERE (likesid='$likesid') AND (likerid='$likerid')";
$result = mysqli_query($conn, $sql);

$alreadylike = mysqli_num_rows($result);
if ($alreadylike > 0) {
    echo "You already liked this status";
}
else {      

    $hostname = "localhost";
    $username = "root";
    $password = "changed for post";
    $databaseName = "changed for post";     
    $connect = mysqli_connect($hostname, $username, $password, $databaseName);      

    //inputs from the form
    $likesid =mysqli_real_escape_string($connect, $_POST['likesid']);
    $likerid =mysqli_real_escape_string($connect, $_POST['likerid']);


    //query to insert into table
    $query = "INSERT INTO `user_likes` SET `likesid`='".$likesid."', `likerid`='".$likerid."'";

    $result = mysqli_query($connect, $query);

    if($result)
    {           
        echo 'updated';
    }
    else
    {
        echo 'not updated';
    }   
    mysqli_close($connect);
    }
}
?>

The form code:

<td style="width: 100px;">  
<p align="center">
<form action="account.php" method="post">   
    <input type="hidden" name="likesid" value="<?php echo $statusid; ?>">
    <input type="hidden" name="likerid" value="<?php echo $_SESSION['id']; ?>">
    <button style="color: #FFFFFF; width: 50%; margin-left: 35px; background-color: transparent; cursor: pointer; text-shadow: 1px 1px 4px #2F3A85" type="submit" name="likestatus">
    <label class="myLabel"><span class="w3-xlarge" style="font-family: 'Righteous'; border: 1px; border-radius: 5px; border-color: #FFFFFF;">like</span></label>
    </button>
</form></p>     
</td>

The page does not return any errors that could point me in any direction, it simply just reloads the page with post info in the URL.

I am assuming the complications start with the fact that the form is inside a table array and maybe the forms submit value is remaining the same throughout all tables?

What's not working ? Your data are correctly saved in the DB, ain't they ? Also your code is vulnerable to user input manipulation; anyone could change the likerid in the form to force other people to like something. — NaeiKinDus
– NaeiKinDus, Commented May 6, 2017 at 19:24
The data is not being saved in the DB at this point, thanks for the tip on the manipulation, I'll make that easy fix — sicksixstrings
– sicksixstrings, Commented May 6, 2017 at 19:35

Edward · Accepted Answer · 2017-05-06 19:05:19Z

0

Because your submit button is a button, not an input.

Change this:

<button style="color: #FFFFFF; width: 50%; margin-left: 35px; background-color: transparent; cursor: pointer; text-shadow: 1px 1px 4px #2F3A85" type="submit" name="likestatus">

to this:

<input style="color: #FFFFFF; width: 50%; margin-left: 35px; background-color: transparent; cursor: pointer; text-shadow: 1px 1px 4px #2F3A85" type="submit" name="likestatus">

answered May 6, 2017 at 19:05

Edward

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3 Comments

Agam Banga Over a year ago

The button has already type=submit. Check this w3schools.com/tags/att_button_type.asp

Edward Over a year ago

@AgamBanga Maybe?: stackoverflow.com/questions/1800709/…

sicksixstrings Over a year ago

I tried changing the button to input before posting this question, but it gave me the exact same results. Any other ideas?

sicksixstrings · Accepted Answer · 2017-05-06 20:34:45Z

0

Apparently I was trying to run a form inside of a form, which explains why the post was just hanging out in the URL and not submitting into my database.

I deleted the form tags on the outside of the form that I was trying to submit and it all works as desired.

As embarrassing as it is, I figured I'd post the answer in hopes to save someone else the time I spent...

answered May 6, 2017 at 20:34

sicksixstrings

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