How to change the option text in php? - php

Question

I tried to build an admin page. The admin will fill a form to add new product in the database and display it in the shop website. The problem is when I tried to select a gender from the dropbox, the new product doesn't add in the product table in the database as you can see below: (I want to select gender such as Boys)
The Admin Page and database result:
The code I used:
$host = "";
$userMS = "";
$passwordMS = "";
$connection = mysql_connect($host,$userMS,$passwordMS) or die("Couldn't connect:".mysql_error());
$database = "projectDataBase";
$db = mysql_select_db($database,$connection) or die("Couldn't select database");
if (isset($_POST['sAddProduct']))
{
addNewProduct();
}else if(isset($_POST['delete']))
{
$Product_ID=$_POST['Product_ID'];
$mysqlquery="delete from Product where Product_ID= ".$Product_ID."";
mysql_query($mysqlquery);
echo "Deleted successfully";
echo("<FORM><INPUT Type='button' VALUE='Back' onClick='history.go(-1);return true;'></FORM>");
}else{
showForm();
}
// add new product
function addNewProduct()
{
$ProductName = $_POST['Product_Name'];
$ProductPrice = $_POST['Price'];
$Gender = $_POST['Gender_ID'];
//database query to add product
$insertStringProduct = "INSERT into Product(Product_Name, Price,Gender_ID)
VALUE('$ProductName', '$ProductPrice', '$Gender')";
$result = mysql_query($insertStringProduct);
echo ("<p1>Product added Successfully</p1>");
echo("<FORM><INPUT Type='button' VALUE='Back' onClick='history.go(-1);return true;'></FORM>");
}
//function for the form page
function showForm()
{
//First form for adding new product
$self = htmlentities($_SERVER['PHP_SELF']);
echo("<form action = '$self' method='POST'>
<fieldset>
<legend>Adding New Product</legend>
Product Name: <input name='Product_Name' type='text' size = '40'>
<br /><br />
Price: <input name='Price' type='text' size = '20'><br><br />
Gender:
<select name='Gender_Description'>
<option value = '%'> <-- select--></option>");
$dbQuary = " SELECT DISTINCT Gender_Description from Gender";
$result = mysql_query($dbQuary);
while($row = mysql_fetch_row($result)){
echo("<option value ='$row[0]'> $row[0]</option>");
}
echo("
</select>
<br/><br/>
<input type='submit' name='sAddProduct' value = 'Add'/>
<input type='reset' value='Clear' />
</fieldset>
</form>");
}
The result ( nothing added)
However, when I change the code to
Gender:
<select name='Gender_ID'>
<option value = '%'> <-- select--></option>");
$dbQuary = " SELECT DISTINCT Gender_ID from Gender";
$result = mysql_query($dbQuary);
It's working
Can anyone help me with this?

Answer 1

In addNewProduct you are expecting $_POST['Gender_ID'] to be set. So of course, <select name='Gender_Description'> would not work, because Gender_Description != Gender_ID. That's also why it does work when you change it.
I'm assuming what you want to achive is to display the gender description, and it still to work. For that, you need both the id and the description:
$dbQuary = " SELECT DISTINCT Gender_ID, Gender_Description from Gender";
$result = mysql_query($dbQuary);
while($row = mysql_fetch_row($result)){
echo("<option value ='$row[0]'> $row[1]</option>");
}
Security
Your code is extremely unsafe. You are using mysql_* which is deprecated since 2013, and you are not sanitizing the input in any way, so your code is open to SQL injection (which is possibly in all kinds of queries; insert, update, delete, etc, and allows for data leaks, DOS, and possibly code execution and deletion/changing of data). The preferred way to prevent this are prepared statements (either using mysqli_* or PDO). They are not difficult to use, and the resulting code is also nicer.

Answer 2

You are not concatenating values as it should
Change
echo("<option value ='$row[0]'> $row[0]</option>");
to
echo("<option value =". '$row[0]' . "> ". $row[0]. "</option>");
OR
echo("<option value ='{$row[0]}'> {$row[0]}</option>");
EDIT:
Change your While-loop
while($row = mysql_fetch_array($result,MYSQL_BOTH)) {
echo("<option value ='{$row['gender_id']}'> {$row['gender_description']}</option>");
}
This will generate a Select list showing the Gender Description and and values will be numeric(of database)