Editing a database in PHP with mysqli - php

Question

Working my way though Creating, Reading, Updating & Deleting (CRUD) information from a database I have done C,R & D but for some reason I can not update.
What am trying to do:
Edit a categorizes title
My Results:
when clicking edit in the table of the cat title I want to change I can get the cat title to echo into a form, where it can be changed then when i try and change the cat title, click update the form goes away as I wanted but the cat title stays the same.
as well am not getting any query errors
What should I look for when debugging code that has no errors?
Can someone see my problem?
if (isset($_GET['edit'])) {
$cat_id = $_GET['edit'];
include "includes/update.php";
}
?>
<form action="categories.php" method="post">
<div class="form-group">
<label for="cat_title">Edit Category</label>
<?php
if (isset($_GET['edit'])) {
$cat_id = $_GET['edit'];
$query = "SELECT * FROM categories WHERE cat_id = $cat_id ";
$select_categories_id = mysqli_query($connection,$query);
while ($row = mysqli_fetch_assoc($select_categories_id)) {
$cat_id = $row['cat_id'];
$cat_title = $row['cat_title'];
?>
<input value="<?php if(isset($cat_title)){echo $cat_title;} ?>" class="form-control" type="text" name="cat_title">
<?php
}
}
if(isset($_POST['edit_category'])){
$edit_cat_title = $_POST['cat_title'];
$query = "UPDATE FROM categories SET cat_title = '{$edit_cat_title}' WHERE cat_id = {$cat_id} ";
$edit_query = mysqli_query($connection,$query);
if (!$edit_query) {
die(mysqli_error($edit_query));
}
}
?>
</div>
<input class="btn btn-primary" type="submit" name="edit_category" value="Edit Category">
</form>

Answer 1

In the following line you have a small error:
$query = "UPDATE FROM categories SET cat_title = '{$edit_cat_title}' WHERE cat_id = {$cat_id} ";
You need to remove the "FROM" in there, will look like this:
$query = "UPDATE categories SET cat_title = '{$edit_cat_title}' WHERE cat_id = {$cat_id} ";
and it should work as expected.

Answer 2

you're calling the update query with
$query = "UPDATE FROM categories SET cat_title = '{$edit_cat_title}' WHERE cat_id = {$cat_id} ";
where $cat_id probably is not set,
once to get this value you're using an if condition
if (isset($_GET['edit'])) {
$cat_id = $_GET['edit'];
where the $_GET[ 'edit' ] can be empty, and the next error is to overwrite the $cat_id variable in a loop: $cat_id = $row['cat_id'];

Answer 3

if (isset($_GET['edit'])) {
$cat_id = $_GET['edit'];
include "includes/update.php";
}
?>
<form action="categories.php" method="post">
<div class="form-group">
<label for="cat_title">Edit Category</label>
<?php
if (isset($_GET['edit'])) {
$cat_id = $_GET['edit'];
}
if(isset($_POST['edit_category'])){
$edit_cat_title = $_POST['cat_title'];
$query = "UPDATE FROM categories SET cat_title = '{$edit_cat_title}' WHERE cat_id = {$cat_id} ";
$edit_query = mysqli_query($connection,$query);
if (!$edit_query) {
die(mysqli_error($edit_query));
}
}
Delete that FROM, it should be UPDATE categories SET ...
You realise that $edit_cat_title is from a POST value, but $cat_id is from a GET value, but your form (as far as I can see) has an action value of action = "categories.php" which will contain NO GET VARIABLES.
Add the property enctype to your <form> such as:
<form ... enctype='multipart/form-data' ... >
To be honest this last point is good practise but I'd be surprised if that was why your POST data was not being populated.
Possible solutions:
1) Use $_REQUEST['cat_id'] and insert the cat_id as a POSTed field in your form, so it can use variables given by either $_POST or $_GET [or $_COOKIE].
2) change your action to goto : action ="categories.php?edit=XXX" to submit the form to an address with a valid GET value as required.
3) Use sessions to hold data from previous pages (such as cat_id). This is my prefered option.
Extra
From question comments it becomes clear that the part if isset($_POST['edit_category']) is never true, so this means that your form is incorrect -- either you have not got a form field named name='edit_cateogry' or your form field is never filled in, or never sent with the form (perhaps the input is placed after the </form> form closing tag?
Anyhow, your issue is that your $_POST value you are looking for is never set.