Question

I have a form that has 3 types of field:

1. (2) text fields that submit to “tblCocktail”
2. (4) select fields that are populated with values from “tblIngredient”
   and submit to “tblRecipe”
3. (4) select fields with pre-set options that submit to “tblRecipe”

Form Code (There are 4x of each “selectingred” & “quantity” drop-downs):

<form method="POST" action="addcocktail.php" >
                    Cocktail Name: <input type="text" name="cocktailname" /> 
                    How To: <input type="text" name="howto" /> 
                    <br> 
                    <select id="selectingred1" name="selectingred1">
                      <?php
                      $sql = "SELECT ingredientID, name FROM tblIngredient ".
                      "ORDER BY name";

                      $rs = mysql_query($sql);

                      while($row = mysql_fetch_array($rs))
                      {
                        echo "<option value=\"".$row['ingredientID']."\">".$row['name']."</option>\n  ";
                      }
                      ?>
                    </select>
                    <select id="quantity1" name="quantity1">
                      <option></option>
                      <option>1</option>
                      <option>2</option>
                      <option>3</option>
                      <option>4</option>
                    </select>
                    <br>
<input type="submit" value="add" />
                </form>

addcocktail.php:

 <?php include("databasecon.php"); ?>

<?php
mysql_select_db("mwheywood", $con);

//insert cocktail details
$sql="INSERT INTO tblCocktail (name, howto)
VALUES
('$_POST[cocktailname]','$_POST[howto]')";

$sql2="INSERT INTO tblRecipe (ingredientID, quantity)
VALUES
('$_POST[selectingred1]','$_POST[quantity1]'),
('$_POST[selectingred2]','$_POST[quantity2]'),
('$_POST[selectingred3]','$_POST[quantity3]'),
('$_POST[selectingred4]','$_POST[quantity4]')";



if (!mysql_query($sql,$con))
  {
  die('Error: you fail at life' . mysql_error());
  }
echo "cocktail added";

if (!mysql_query($sql2,$con))
  {
  die('Error: you fail at life' . mysql_error());
  }
echo "ingredients added";

mysql_close($con);

?>

This is currently only adding the “selectingred4” and “quantity4” values into “tblRecipe”. It’s ignoring the two inserts for the text boxes, and the first 3 select box entries.

My other problem was also that Im grabbing the “ingredientID” and “name” from the php in my form, but when I submit the form it’s not submitting “ingredientID” into “tblRecipe” either.

-any help would be appreciated -Matt

Answer 1

You keep overwriting your SQL query and only the last one actually exists to be executed. Here’s a fixed version of your code:

<?php include("databasecon.php"); ?>

<?php
mysql_select_db("mwheywood", $con);

//insert cocktail details
$sql="INSERT INTO tblCocktail (name, howto)
VALUES
('$_POST[cocktailname]','$_POST[howto]')";

if (!mysql_query($sql,$con))
{
die('Error: you fail at life' . mysql_error());
}

//insert recipe details
$sql="INSERT INTO tblRecipe (ingredientID, quantity)
VALUES
('$_POST[selectingred1]','$_POST[quantity1]')";

$sql="INSERT INTO tblRecipe (ingredientID, quantity)
VALUES
('$_POST[selectingred2]','$_POST[quantity2]')";

$sql="INSERT INTO tblRecipe (ingredientID, quantity)
VALUES
('$_POST[selectingred3]','$_POST[quantity3]')";

$sql="INSERT INTO tblRecipe (ingredientID, quantity)
VALUES
('$_POST[selectingred4]','$_POST[quantity4]')";
if (!mysql_query($sql,$con))
  {
  die('Error: you fail at life' . mysql_error());
  }
echo "cocktail added";

mysql_close($con);

?>

Furthermore, you can combine the last four queries into one query:

<?php include("databasecon.php"); ?>

<?php
mysql_select_db("mwheywood", $con);

//insert cocktail details
$sql="INSERT INTO tblCocktail (name, howto)
VALUES
('$_POST[cocktailname]','$_POST[howto]')";

if (!mysql_query($sql,$con))
{
die('Error: you fail at life' . mysql_error());
}

//insert recipe details
$sql="INSERT INTO tblRecipe (ingredientID, quantity)
VALUES
('$_POST[selectingred1]','$_POST[quantity1]'),
('$_POST[selectingred2]','$_POST[quantity2]'),
('$_POST[selectingred3]','$_POST[quantity3]'),
('$_POST[selectingred4]','$_POST[quantity4]')";
if (!mysql_query($sql,$con))
  {
  die('Error: you fail at life' . mysql_error());
  }
echo "cocktail added";

mysql_close($con);

?>