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POSave button: how to update another table with results?
 

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    <p>I have a table of recipes and there is a save button next to each recipe. When this is clicked, I would like the recipe name to be retrieved and saved to the user's profile (add to database).</p> <p>However, I am encountering some problems. I have a save button and this is linked to 'save_recipe.php'. Currently when I select save, I am being directed to a blank page. I have a feeling it is something to do with the way I have created the save button, as no echo statements are being displayed.</p> <p>The code for the button is:</p> <pre><code>&lt;html&gt; &lt;body&gt; &lt;form action="../user profile/save_recipe.php" method="post"&gt; &lt;input type="submit" name="submit" class="btn" value="Save Recipe"&gt;&lt;/td&gt; &lt;/form&gt; &lt;/body&gt; &lt;/html&gt; </code></pre> <p>The code for save_recipe is:</p> <pre><code>&lt;!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"&gt; &lt;html xmlns="http://www.w3.org/1999/xhtml"&gt; &lt;head&gt; &lt;meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /&gt; &lt;title&gt;BMI Result&lt;/title&gt; &lt;/head&gt; &lt;link rel="stylesheet" href="css.css" type="text/css" /&gt; &lt;body&gt; &lt;?php error_reporting(E_ALL &amp;~ E_NOTICE); // Start the session session_start(); // If the session vars aren't set, try to set them with a cookie if (!isset($_SESSION['user_id'])) { if (isset($_COOKIE['user_id']) &amp;&amp; isset($_COOKIE['username'])) { $_SESSION['user_id'] = $_COOKIE['user_id']; $_SESSION['username'] = $_COOKIE['username']; } } echo gfgfg; // Make sure the browser is transmitting in UTF-8 header('Content-type: text/html; charset=utf-8'); // Clear the error message $error_msg = ""; //if (isset($_SESSION['user_id'])) // echo $_SESSION['user_id']; $dbc = mysqli_connect('localhost', 'root', 'root', 'help_me_be_healthy') or die("Error " . mysqli_error($dbc)); mysqli_set_charset($dbc, "utf8"); echo gfdgfdg; if(isset($_POST['submit'])) { $recipe_name = mysqli_result(mysqli_query("SELECT `recipe_1_name`, `recipe_2_name` FROM `carbohydrates`"); //"SELECT `recipe_1_name` OR `recipe_2_name` FROM `carbohydrates` WHERE `user_id` = '" . $_SESSION['user_id'] . "'"; $query = "UPDATE `users` SET `recipes_saved` = $recipe_name WHERE `user_id` = '" . $_SESSION['user_id'] . "'"; $data= mysqli_query($dbc,$query); echo '$recipe_name'; ); } ?&gt; </code></pre> <p>Any advice would be much appreciated, I'm fairly new to this!</p>
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    1. USSarah Al
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    1. COFirstly, this will throw an error `echo gfgfg;` so please don't use pseudo-code, if that's the case. Second, use `echo $recipe_name;` without quotes. 3rd, you're not echoing your [**query**](http://us2.php.net/mysqli_query) correctly. You need to use a `while` or `foreach` condition and fetch.
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      1. USFunk Forty Niner
    2. CO...4th there's no field in the form other than the `submit` button, 5th a cookie is pointlessly duplicated in `$_SESSION`, 6th the `SELECT` query seems pointless to...
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    3. CO... and the beat goes on ♫ ;-)
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      1. USFunk Forty Niner
 

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