This question already has an answer here:
I am trying to fetch the data from database for a an input field product code and i need to use its value to update the rest of the column values in the database but instead it is creating a different record and in the value field of the input box 'named' code, it shows and undefined variable error, please help.
HTML code:
<div class="small-8 columns">
<input type="text" id="right-label" placeholder="Product_code"
value="<?php echo "$pcode"?>" name="code">
</div>
PHP Script:
<?php
$servername="localhost";
$username="root";
$password="";
$dbname="bolt";
try{
$conn = new
PDO("mysql:host=$servername;dbname=$dbname",$username,$password);
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
if(isset($_POST["submit"])){
$pcode = ($_POST["code"]);
$pname = ($_POST["Pname"]);
$pdesc = ($_POST["desc"]);
$pimg = $_FILES["Img_name"]["temp_name"];
$imgExt = strtolower(pathinfo($pimg,PATHINFO_EXTENSION));
$valid_extensions = array('jpeg','jpg','png','gif','pdf');
$pqty = ($_POST["Pqty"]);
$pprice = ($_POST["Pprice"]);
$sql="UPDATE products SET product_name=$pname,product_desc=$pdesc,
product_img_name=$pimg,qty=$pqty,price
=$pprice) WHERE product_code=$pcode";
$stmt = $conn->exec($sql);
$stmt->execute();
echo $stmt->rowCount() . "new records added succesfully";
}
}
catch(PDOException $e){
echo $sql . "<br>" . $e->getMessage();
}
$conn = null;
?>
</div>
$sql is declared within the if condition, if if(isset($_POST["submit"])){ is false, you will get this error because $sql is not within the scope. Declare it on above condition and initialize it.