Update MySQL database using HTML/PHP table (inline) - php

Question

I'm trying to update my database using the html/php table with save button to confirm the changes and save in database.
Here's my Table
How can I edit the 'Quantity' row like text field and save with the button to confirm.
this is my table
<table class=" table-responsive table table-striped table-bordered table-hover ">
<thead>
<tr>
<th>Item #</th>
<th>Description</th>
<th>Unit</th>
<th>Quantity</th>
</tr>
</thead>
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbName = "db";
$port = "3306";
$conn = new mysqli($servername, $username, $password, $dbName, $port);
if($conn->connect_error)
{
die("Connection Failed: ".$conn->connect_error);
}
$sql = "SELECT * FROM product";
$result1 = $conn->query($sql);
$result2 = $conn->query($sql);
?>
<tbody>
<?php
while ($row = mysqli_fetch_array($result1)) {
echo "
<tr>
<td>" . $row[0] . "</td>
<td>" . $row[1] . "</td>
<td>" . $row[2] . "</td>
<td>0</td>
</tr>";
}
?>
</tbody>
</table>
Please help. Thanks!

Answer 1

Put your table inside a form.
<form action="give_your_action_page" method="post">
<table class=" table-responsive table table-striped table-bordered table-hover ">
<thead>
<tr>
<th>Item #</th>
<th>Description</th>
<th>Unit</th>
<th>Quantity</th>
</tr>
</thead>
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbName = "db";
$port = "3306";
$conn = new mysqli($servername, $username, $password, $dbName, $port);
if($conn->connect_error)
{
die("Connection Failed: ".$conn->connect_error);
}
$sql = "SELECT * FROM product";
$result1 = $conn->query($sql);
$result2 = $conn->query($sql);
?>
<tbody>
<?php
while ($row = mysqli_fetch_array($result1)) {
echo "
<tr>
<td>" . $row[0] . "</td>
<td>" . $row[1] . "</td>
<td>" . $row[2] . "</td>
<td><input type='number name='".$row[1]."' value='".$row[3]."'/></td>
</tr>";
}
?>
</tbody>
</table>
Now at your action page you can access the form values using $_POST array using the value in "Description" field.
For eg : $_POST["banana"] will give the quantity of banana.
Update the database using these values.

Answer 2

You're going to change a field that is not in the database?
because
<tr>
<td>" . $row[0] . "</td>
<td>" . $row[1] . "</td>
<td>" . $row[2] . "</td>
<td>0</td> ----------------> ? (empty) ?
</tr>
You must first create a field called Quantityin your database, then try again if it was a difficult ask questions.

Answer 3

Ok your Code is a really bad , well most of it , and your Q was really hard to understand , but its ok , try to be more accurate next time =) , this answer that i came with is depends on my understanding to what you tried to ask .
I suggest that you topic from w3shool : Select data from database
In my answer or in my Code if we can say that , I used PDO , Whats PDO ?
PDO is an acronym for PHP Data Objects. PDO is a lean, consistent way
to access databases. This means developers can write portable code
much easier. PDO is not an abstraction layer like PearDB. PDO is a
more like a data access layer which uses a unified API (Application
Programming Interface).
Search for it on Google and read some info about how it works and try to learn it , its simple and its the same as ( MYSQLI ) .
This is my Code which i tested it many times and it works , this code can be written in many ways but i tried to make it as simple as i can .
<?php
// Start of the first part ( connect to the database usnig PDO
$servername = "localhost";
$username = "root";
$password = "";
$database = "your database name";
try {
$conn = new PDO("mysql:host=$servername;dbname=$database", $username, $password);
// set the PDO error mode to exception
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
}
catch(PDOException $e)
{
echo "Connection failed: " . $e->getMessage();
}
// END of the first part
// second part : bring the new quantity from the field
if(isset($_POST['update']))
{
$newquantity = $_POST['newQu'];
$itemName = $_POST['item'];
$UpdateOnQuantity = $conn->prepare( "UPDATE hr SET Quantity = ? WHERE Item = ?");
$UpdateOnQuantity->execute(array($newquantity,$itemName));
if($UpdateOnQuantity)
{
}
else
{
echo "error";
}
}
$FetchNewQuantity = $conn->prepare("SELECT * FROM hr");
$FetchNewQuantity->execute();
?>
<html>
<head>
</head>
<body>
<table class=" table-responsive table table-striped table-bordered table-hover ">
<thead>
<tr>
<th>Item #</th>
<th>Description</th>
<th>Unit</th>
<th>Quantity</th>
</tr>
</thead>
<tbody>
<?php
while($FetchedData = $FetchNewQuantity->FETCH(PDO::FETCH_ASSOC))
{
?>
<tr>
<td><?php echo $FetchedData['Item']; ?></td>
<td><?php echo $FetchedData['Description']; ?></td>
<td><?php echo $FetchedData['Unit']; ?><td>
<td><?php echo $FetchedData['Quantity']; ?></td>
<td>
<form action="test2.php" method="POST">
<input type="hidden" name="item" value="<?php echo $FetchedData['Item']; ?>">
<input type="text" name="newQu">
<input type="submit" name="update">
</form>
</td>
</tr>
<?php } ?>
</tbody>
</table>
</body>
</html>
This is my code hope it help you , i tested many times as i said so if it did not work with you let me know , and if you did not like PDO just replace PDO code with the normal Mysqli code .
Edit : i edited the code again , did not noticed that you linked your table . i added a while loop so it can show all the data in your table .