asked on

php and mysqli

This code is working:

<?php
if(isset($_POST["submit"])){
$servername = "localhost";
$username = "root";
$password = "xxxxxxx";
$dbname = "ticktock";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}

$sql = "INSERT INTO itdocs (title, details, loc, Type)
VALUES ('".$_POST["doc_name"]."','".$_POST["doc_details"]."','".$_POST["doc_location"]."','".$_POST["doc_type"]."')";

if ($conn->query($sql) === TRUE) {
echo "<script type= 'text/javascript'>alert('New record created successfully');</script>";
} else {
echo "<script type= 'text/javascript'>alert('Error: " . $sql . "<br>" . $conn->error."');</script>";
}

$conn->close();
}
?>

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What I need to do is add in another section from a second form on the same page which will add data to the same database fields
The field names for demo sake will be name1, name2, name3, name4
Can someone give me a pointer on how to add the second query
I just need the structure

SOLUTION

Dave Baldwin

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doctorbill

ASKER

Ok - point taken
But what about my question - can I have a pointer please

SOLUTION

Chris Stanyon

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doctorbill

ASKER

This is what I have so far:
The first set of code between the first php tags works from a form on the page
The second one from a second form on the page does not

<?php
if(isset($_POST["submit"])){
$servername = "localhost";
$username = "root";
$password = "xxxx";
$dbname = "ticktock";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}

$sql = "INSERT INTO itdocs (title, details, loc, Type)
VALUES ('".$_POST["doc_name"]."','".$_POST["doc_details"]."','".$_POST["doc_location"]."','".$_POST["doc_type"]."')";

if ($conn->query($sql) === TRUE) {
echo "<script type= 'text/javascript'>alert('New record created successfully');</script>";
} else {
echo "<script type= 'text/javascript'>alert('Error: " . $sql . "<br>" . $conn->error."');</script>";
}

$conn->close();
}
?>

<?php
if(isset($_POST["it_doc_name"])){
$servername = "localhost";
$username = "root";
$password = "xxxx";
$dbname = "ticktock";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}

$sql = "INSERT INTO itdocs (title, details, loc, Type, file_view)
VALUES ('".$_POST["it_doc_name"]."','".$_POST["it_doc_details"]."','".$_POST["it_doc_location"]."','".$_POST["it_upload"]."','".$_POST["it_doc_type"]."')";

if ($conn->query($sql) === TRUE) {
echo "<script type= 'text/javascript'>alert('New record created successfully');</script>";
} else {
echo "<script type= 'text/javascript'>alert('Error: " . $sql . "<br>" . $conn->error."');</script>";
}

$conn->close();
}
?>

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SOLUTION

Chris Stanyon

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doctorbill

ASKER

The second set of php code does not add the details to the database

doctorbill

ASKER