Use of undefined constant id - assumed 'id'

html-table, php

Solution

Replace

<input type='hidden' name='id' value='".$db->bacaData(id,$id)."'>

with

<input type='hidden' name='id' value='".$db->bacaData('id',$id)."'>
                              ------------------------^^^^---

you are using `id` and php will count it as constant so wrap it inside single quote.

Even you have same error for other declaration too wrap first argument inside of `''`.

Correct Code

elseif(isset($_GET['aksi']) && $_GET['aksi']=='edit')
{
    $id=$_GET['id'];
    echo"<br>
        <form method=POST action='?aksi=update_data'>
            <table>
                <tr><td>Id</td><td><input type=text name='id' value='".$db->bacaData('id',$id)."'></td></tr>
                <tr><td>Nama</td><td><input type=text name='nama' value='".$db->bacaData('nama',$id)."'></td></tr>
                <tr><td>Email</td><td><input type=text name='email' value='".$db->bacaData('email',$id)."'></td></tr>
                <tr><td>Pesan</td><td><input type=text name='pesan' value='".$db->bacaData('pesan',$id)."'></td></tr>
                <tr><td></td><td><input type=submit value='simpan'></td></tr>
            </table>
        <input type='hidden' name='id' value='".$db->bacaData('id',$id)."'>
        </form>
    ";
}

Problem

Original source