CODEWITHSUNDEEP
phpmysql
Lawrence Smith
Lawrence Smith

Reputation: 337

echo $_GET['id'] not displaying anything

I have a page in which a user chooses a recipe which has a unique ID, they are then sent to another page where the information for that recipe is displayed but it is not displaying anything.

$recipes = mysql_query("
        SELECT DISTINCT `name`, `step1`, `step2`, `step3`, `image`, `reference`
        FROM `recipe` r
        INNER JOIN `recipe_ingredients` ri
        ON r.id = ri.recipe_id
        WHERE ri.ingredient_id IN (".$ingredient1.",".$ingredient2.",".$ingredient3.")
    ");

This is the query on the main page, this allows the user to choose a set of ingredients and then retrieve recipes from them.

echo '<form action="recipe.php?id=<?php echo $recipe_id;?>" method="POST">';
        echo '<input id="search" name="Look" type="Submit" value="Look" >';
        echo '</form>';

Under each recipe there is a button a user presses, this will send the user to the specific recipe.

$sql = mysql_query("SELECT `name`, `image`
FROM `recipe` 
WHERE `id` = ".$_GET['id']." LIMIT 1");

This is the query I'm running on the results page (recipe.php)

Further down the page I ran this.

<?php
echo $_GET['id'];
?>

It didn't display an ID and in the url I was displayed with this:

"recipe.php?id="

There is data in the table so what am I doing wrong?

Upvotes: 0

Views: 2644

Answers (3)

Yoan Arnaudov
Yoan Arnaudov

Reputation: 4144

First of all before making any queries to the database make sure that all $variables that you use in the query are escaped, because you can get hacked very easily. Escape like this:

$_GET['variable'] = mysql_real_escape_string($_GET['variable']);

Integers can be escaped like this

$_GET['variable'] = (int) $_GET['variable'];

You are doing it wrong, you can add recipe_id as hidden input field and later read it after the form submittion. Here is example form.

<form action="/recipe.php" method="POST">
    <input type="hidden" name="recipe_id" value="<?= $recipe_id ?>" />
    <input id="search" name="Look" type="Submit" value="Look" />
</form>

And after the form is submitted you can do something like this:

$recipe_id = (int) $_POST['recipe_id'];
$sql = mysql_query("SELECT `name`, `image` FROM `recipe` WHERE `id` = $recipe_id LIMIT 1");

But I think you main problem is that $recipe_id in the form is just an empty variable, check that first :).

Upvotes: 1

brbcoding
brbcoding

Reputation: 13586

// change this
echo '<form action="recipe.php?id=<?php echo $recipe_id;?>" method="POST">';
// to this
echo '<form action="recipe.php?id=' . $recipe_id .'" method="POST">';

Change the line where you echo the form to something like this... You were trying to echo twice before (echo within the echo)... This works though.

Upvotes: -1

Samuel Cook
Samuel Cook

Reputation: 16828

The following echo syntax is incorrect:

echo '<form action="recipe.php?id=<?php echo $recipe_id;?>" method="POST">';

should be:

echo '<form action="recipe.php?id='.$recipe_id.'" method="POST">';

I don't see where you are setting $recipe_id, which appears to be the root of your problem.

It is unclear where you are trying to retrieve the elusive $recipe_id from.

If your trying to pull it in from $recipes then you should include it in your sql statement:

SELECT `id`,....

then pull it in after your $recipes->fetch_assoc():

$recipe_id = $recipes['id'];

Upvotes: 2

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