Updating products in a mysqli database using PHP - php

Question

Trying to update 1 row at a time using php.
I want to enable users to update products they have already added to a database, I have a simple form with the relevant fields:
<fieldset><legend><span> Update a product in the database! </span></legend>
<form id ="productsform" method="post" onsubmit="return false;" enctype="multipart/form-data">
<label> Product name: <input type="text" id="name" name="name"/> </label>
<label> Product quantity: <input type="number" id="quantity" name="quantity"/> </label>
<label> Product description: <input type="text" id="description" name="description"/> </label>
<label> Product price: <input type="text" id="price" name="price"/> </label>
</br>
<input id="submit" name="submit" type="button" class="reg" value="Update Product">
<div id="update"></div>
</form>
I am using ajax which is working correctly according the console, but im struggling with the php side of updating the rows:
<?php
include("dbase/config_database.php");
$id = $_POST["id"];
$name = $_POST['name'];
$quantity = $_POST['quantity'];
$description = $_POST['description'];
$price = $_POST['price'];
$query = "UPDATE products SET name = '{$name}', quantity = '{$quantity}', description = '{$description}', price = '{$price}'
WHERE id = {$id}";
mysqli_query($mysqli, $query);
?>
Here is the initial file I use to add the products to the database:
<?php
include("dbase/config_database.php");
//Stores all information passed through AJAX into the query
$name = $_POST['name'];
$quantity = $_POST['quantity'];
$description = $_POST['description'];
$price = $_POST['price'];
//Adds information to database
$query = "INSERT INTO products (name, quantity, description, price) VALUES ('$name','$quantity','$description','$price')";
//Runs the query
$result = $mysqli->query($query) OR die("Failed query $query");
echo $mysqli->error."<p>";
echo "Product Added";
$querynew = ("SELECT id as 'collectid' from products WHERE name = '$name'and quantity = '$quantity'and description ='$description'and price = '$price'");
$resultnew = $mysqli->query($querynew) OR die("Failed query $querynew");
while($info = mysqli_fetch_array( $resultnew)){
$productid = $info['collectid'];
}
$image = $_FILES['file1']['name'];
$type = $_FILES['file1']['type'];
$size = $_FILES['file1']['size'];
$tmp_name = $_FILES['file1']['tmp_name'];
$imgpath = "images/".$productid.".jpg";
// Run the move_uploaded_file() function here
$moveResult = move_uploaded_file($tmp_name, $imgpath);
// Evaluate the value returned from the function if needed
$querytwo = ("SELECT * FROM products WHERE name = '$name' and quantity = '$quantity' and description = '$description' and price = '$price'");
$resulttwo = $mysqli ->query($querytwo) OR die ("Failed query $querynew");
$info = array();
while($row = mysqli_fetch_assoc($resulttwo)){
$product = array("id" => $row ['id'],
"name" => $row ['name'],
"quantity" => $row ['quantity'],
"description" => $row ['description'],
"price" => $row ['price'],
);
array_push($info,$product);
}
$json_output = json_encode($info);
echo $json_output;
?>
Any help is much appreciated! I have messed around with the update php because im sure the problem is in there but cant find it.

Answer 1

You are not getting the $_POST["id"] from the form,because there is no input element with name id.
when you get all the data from the table then put the id in the form as a hidden field
like
<input type="hidden" name="id" value="<?=$row['id']?>">
then after submitting the form you'll get the id value in $_POST['id']
Always try to debug your code thoroughly try to print the query then you can easily know what is happening actually.All the best