I'm trying to edit a post using this snippiet of from my query.php page:
echo '<button type="submit" onclick="openModal(this)" id="btn-edit"
name="edit" value='.$postID.'></button>';
function openModal(id)
{
var editpost = id.value;
$.ajax({
url:"query.php",
method:"POST",
data:{ editpost : editpost },
success:function(data)
{ //
},
error: function () {//
}
});
$('#edit').modal('show');
}
The function to get the data from the post selected is on the same php page. This is the code:
if(isset($_POST['editpost']))
{
session_start();
$editpostid = $_POST['editpost'];
if($editpostid != "")
{
$sql = "SELECT * FROM post WHERE PostId = '" . $editpostid . "'";
}
if($sql != "")
{
$qry = mysqli_query($connection, $sql);
if (mysqli_num_rows($qry) > 0)
{
foreach($qry as $row)
{
$_SESSION['editpostdesc'] = $row['PostDesc'];
$_SESSION['editpostfile'] = $row['PostFile'];
$_SESSION['editpostid'] = $row['PostId'];
}
}
}
}
But the modal is on the other page (main page). I want to get the data of the post and display it on the modal, so I tried using $_SESSION
. Yes, I got the data and was able to display it on the modal, but the problem is when I try to edit another post, the first value assigned to the session cannot be replaced. Is there is any other way that I can pass the values without using session? I'm really running out of ideas, I'm just starting my first web project.