Undefined index PHP/MySQL when trying to query database - php

Question

This one is working fine #1 but when I try to relate other tables on the second one, it gives an error
Notice: Undefined index: recipe.id in C:\xampp\htdocs\hyukies\public\samples.php on line 30
Notice: Undefined index: recipe.ingredientid in C:\xampp\htdocs\hyukies\public\samples.php on line 31
RecipeID:
IngredientID:
Notice: Undefined index: recipe.id in C:\xampp\htdocs\hyukies\public\samples.php on line 30
Notice: Undefined index: recipe.ingredientid in C:\xampp\htdocs\hyukies\public\samples.php on line 31
RecipeID:
IngredientID:
Notice: Undefined index: recipe.id in C:\xampp\htdocs\hyukies\public\samples.php on line 30
Notice: Undefined index: recipe.ingredientid in C:\xampp\htdocs\hyukies\public\samples.php on line 31
RecipeID:
IngredientID:
<?php
$sql = mysqli_query($connection, "SELECT id, location FROM location");
$userinfo = array();
while ($row_user = mysqli_fetch_assoc($sql))
$userinfo[] = $row_user;
foreach ($userinfo as $user) {
echo "ID: {$user['id']}<br />"
. "Location: {$user['location']}<br /><br />";
}
?>
This one is not working error problem please help......
<?php
$sql_2 = mysqli_query($connection, "SELECT recipe.id, recipe.ingredientid, ingredients.id, ingredients.quantity, ingredients.`name` FROM recipe INNER JOIN ingredients ON ingredients.id = recipe.ingredientid");
$userinfo_2 = array();
while ($row_user_2 = mysqli_fetch_assoc($sql_2))
$userinfo_2[] = $row_user_2;
foreach ($userinfo_2 as $user_2) {
echo "RecipeID: {$user_2['recipe.id']}<br />"
. "IngredientID: {$user_2['recipe.ingredientid']}<br /><br />";
}
?>

Answer 1

Remove the back ticks from 'name' in your query
$sql_2 = mysqli_query($connection, "SELECT recipe.id, recipe.ingredientid, ingredients.id, ingredients.quantity, ingredients.`name` FROM recipe INNER JOIN ingredients ON ingredients.id = recipe.ingredientid");
Should be
$sql_2 = mysqli_query($connection, "SELECT recipe.id, recipe.ingredientid, ingredients.id, ingredients.quantity, ingredients.name FROM recipe INNER JOIN ingredients ON ingredients.id = recipe.ingredientid");
Also, remove 'recipe' from $user_2['recipe.id']. It should be $user2['id'], but it won't be clear which id you are getting. To selected them independently you can use 'AS' in your query to give them a unique identifier.
SELECT recipe.id AS id1, ingedients.id AS id2

Answer 2

The keys in the associative array just contain the column names, not the table names. So it should be:
echo "RecipeID: {$user_2['id']}<br />"
. "IngredientID: {$user_2['ingredientid']}<br /><br />";

Answer 3

The table names are not included in the result set, so the columns are only named id, ingredientid etc. You really should use as for naming them, like recipe.id as recipeid. There is no warning about there being to id fields and PHP will map them together.
You don't really need both id fields anyway, since you also have the ingredientid, but most probably you will get ingredients.id in the id field also since it's latter.
So update your query to something like
SELECT recipe.id AS recipeid,
recipe.ingredientid,
ingredients.quantity,
ingredients.name
FROM recipe
INNER JOIN ingredients ON ingredients.id = recipe.ingredientid
and then use recipeid to get the recipe.id. This will always be better, since it's clear what id you are getting.