Cannot add values in Database using PHP

Question

I am currently doing a project in adding values using database but I seem to have a problem. I am sure that my query is correct since I tried adding it manually in mysql. Only some of the fields seem to be able to get what I input. I get the error

"Error: INSERT INTO inventory (itemCode, dateReceived, typeOfFabric, details, unitOfMeasurement, amount, assignedOrderUse, section, row) VALUES ('', '', '', 'White', '', '5', '', 'C', 'C')"

    <?php

$host = "localhost";
$user = "root";
$pass = "";
$db = "gracydb";

if (isset($_POST['addInventory']))
{
    if(isset($_POST['itemCode'])){ $itemcode = $_POST['itemCode']; } 
    if(isset($_POST['dateReceived'])){ $inventoryDateReceived = $_POST['dateReceived']; } 
    if(isset($_POST['typeOfFabric'])){ $fabric = $_POST['typeOfFabric']; }
    if(isset($_POST['details'])){ $details = $_POST['details']; } 
    if(isset($_POST['unitOfMeasurement'])){ $measurement = $_POST['unitOfMeasurement']; }
    if(isset($_POST['amount'])){ $amount = $_POST['amount']; } 
    if(isset($_POST['assignedOrderUse'])){ $order = $_POST['assignedOrderUse']; } 
    if(isset($_POST['section'])){ $section = $_POST['section']; }
    if(isset($_POST['row'])){ $row = $_POST['row']; }

    $conn = mysql_connect($host, $user, $pass);
    $db_selected = mysql_select_db($db, $conn);

    $sql = "INSERT INTO inventory (itemCode, dateReceived, typeOfFabric, details, unitOfMeasurement, amount, assignedOrderUse, section, row)
    VALUES ('$itemcode', '$datereceived', '$fabric', '$details', '$measurement', '$amount', '$order', '$section', '$row')";

    if (mysql_query($sql)) {
        echo "New record created successfully";
    } else {
        echo "Error: " . $sql . "<br>" . mysql_error($conn);
    }

    mysql_close($conn);
    //header ('Location: .php');
}

?>

<form action = "<?php echo htmlspecialchars($_SERVER["PHP_SELF"]); ?>" method = "POST">
Item Code: <input type = "text" name = "itemcode"><br>
Date Received: <input type = "date" name = "inventoryDateReceived"><br>
Type of Fabric: <input type = "text" name = "fabric"><br>
Unit of Measurement: 
<select name = "measurement">
<option value = "Grams">Grams</option>
<option value = "Kilograms">Kilograms</option>
</select><br>
Amount: <input type = "number" name = "amount"><br>
Assigned Order/Use: <input type = "text" name = "order"><br>
Section: <input type = "text" name = "section"><br>
Row: <input type = "text" name = "row"><br>
<input type = "submit" value = "submit" name = "addInventory">
</form>

Check that you can insert into itemCode, dateReceived, typeOfFabric, unitOfMeasurement, assignedOrderUse empty values — Firas Rassas
– Firas Rassas, Commented Feb 22, 2016 at 7:35
So what is the actual error? You show us the SQL, but not the error itself. — Sergey Vidusov
– Sergey Vidusov, Commented Feb 22, 2016 at 7:56

devpro · Accepted Answer · 2016-02-22 07:39:51Z

1

These indexes not matched with your input form names:

$_POST['itemCode']
$_POST['dateReceived']
$_POST['typeOfFabric']

These should be:

$_POST['itemcode']
$_POST['inventoryDateReceived']
$_POST['fabric']

Check your form inputs:

<input type = "text" name = "itemcode">
<input type = "date" name = "inventoryDateReceived">
<input type = "text" name = "fabric">

answered Feb 22, 2016 at 7:39

devpro

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Comments

SebTM · Accepted Answer · 2016-02-22 07:46:45Z

I don't see any sense in this part of the code:

if(isset($_POST['itemCode'])){ $itemcode = $_POST['itemCode']; } 
if(isset($_POST['dateReceived'])){ $inventoryDateReceived = $_POST['dateReceived']; } 
if(isset($_POST['typeOfFabric'])){ $fabric = $_POST['typeOfFabric']; }
if(isset($_POST['details'])){ $details = $_POST['details']; } 
if(isset($_POST['unitOfMeasurement'])){ $measurement = $_POST['unitOfMeasurement']; }
if(isset($_POST['amount'])){ $amount = $_POST['amount']; } 
if(isset($_POST['assignedOrderUse'])){ $order = $_POST['assignedOrderUse']; } 
if(isset($_POST['section'])){ $section = $_POST['section']; }
if(isset($_POST['row'])){ $row = $_POST['row']; }

Your are just setting values (if isset) to new variables - but if they not exists you will still use undefined variables. Also there is no escaping to prevent sql-injections and validation of the given values!

I think you will get this error because of a missing variable.

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