ANdrww Posted December 10, 2013 Share Posted December 10, 2013 I am trying to create a script (from a tutorial I try to adapt to my own needs) and I am running into a few problems. Here I go... I have an /includes folder which contains the following: database.php and functions.php with the following content: database.php <?php // Database connectivity stuff $host = "localhost"; // Hostname for the database. Usually localhost $username = "root"; // Username used to connect to the database $password = "root"; // Password for the username used to connect to the database $database = "blog"; // The database used // Connect to the database using mysqli_connect $connection = mysqli_connect($host, $username, $password, $database); // Check the connection for errors if (mysqli_connect_errno($connection)) { // Stop the whole page from loading if errors occur die("<br />Could not connect to the database. Please check the settings and try again.") . mysqli_connect_error() . mysqli_connect_errno(); } ?> functions.php <?php // Functions file for the system function show_posts($user_id) { $posts = array(); $sql = "SELECT body, stamp from posts where user_id = '$user_id' order by stamp desc"; $result = mysqli_query($connection, $sql); while ($data = mysqli_fetch_assoc($result)) { $posts = array( 'stamp' => $data->stamp, 'user_id' => $user_id, 'body' => $data->body ); } return $posts; } function show_users() { $users = array(); $sql = "SELECT id, username FROM users WHERE status = 'active' ORDER BY username"; $result = mysqli_query($connection, $sql); while ($data = mysqli_fetch_array($result)) { $users[$data->id] = $data->username; } return $users; } function following($user_id) { $users = array(); $sql = "SELECT DISTINCT user_id FROM following WHERE follower_id = $user_id"; $result = mysqli_query($connection, $sql); while ($data = mysqli_fetch_assoc($result)) { array_push($users, $data->user_id); } return $users; } ?> And here's the code that I run to display the users in users.php <?php $users = show_users(); foreach ($users as $key => $value) { echo $key . " " . $value; } ?> That throws me the following errors: Notice: Undefined variable: connection in /Applications/MAMP/htdocs/blog/includes/functions.php on line 22Warning: mysqli_query() expects parameter 1 to be mysqli, null given in /Applications/MAMP/htdocs/blog/includes/functions.php on line 22Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in /Applications/MAMP/htdocs/blog/includes/functions.php on line 24 I've tried to pass the $connection value to the show_users(); function with no results... any help would be appreciated. I am able to display the users from the database if I introduce the following code in users.php, but I want to try and keep things as clear as possible, without cluttering the files unnecessary. <?php $users = array(); $query = "SELECT id, username FROM users WHERE status = 'active' ORDER BY username"; $result = mysqli_query($connection, $query); if ($result) { while ($member = mysqli_fetch_assoc($result)) { echo strip_tags($member['username']) . "<br />"; } } else { echo "There are no posts to display. Why not add a new one?"; } mysqli_free_result($result); mysqli_close($connection); ?> PS: The users.php file also has require('includes/database.php'); and require('includes/functions.php'). I guess I'm trying to recreate the last bit of code into a working function... and I can't seem to do it. Any help is appreciated. I hope it makes an sense... -- Andrei Quote Link to comment Share on other sites More sharing options...
Ch0cu3r Posted December 10, 2013 Share Posted December 10, 2013 (edited) You need to read up on variable scope as this the issue with your code. The problem is the $connection variable is defined outside of your functions. Functions have their own variable scope, meaning variables defined outside of it are not accessible. The solution is to either define $connection as global or pass it to your functions as an argument. Edited December 10, 2013 by Ch0cu3r Quote Link to comment Share on other sites More sharing options...
KevinM1 Posted December 10, 2013 Share Posted December 10, 2013 You need to read up on variable scope as this the issue with your code. The problem is the $connection variable is defined outside of your functions. Functions have their own variable scope, meaning variables defined outside of it are not accessible. The solution is to either define $connection as global or pass it to your functions as an argument. Please don't suggest using globals. We strive to teach best practices here, and using 'global' is anything but. Quote Link to comment Share on other sites More sharing options...
ANdrww Posted December 10, 2013 Author Share Posted December 10, 2013 Thanks for the reply Ch0cu3r. I've tried passing the $connection variable to my function as follows: function show_users($conection) { $users = array(); $sql = "SELECT id, username FROM users WHERE status = 'active' ORDER BY username"; $result = mysqli_query($connection, $sql); while ($data = mysqli_fetch_array($result)) { $users[$data->id] = $data->username; } return $users; } That throws the folloing error: Warning: Missing argument 1 for show_users(), called in /Applications/MAMP/htdocs/blog/users.php on line 18 and defined in /Applications/MAMP/htdocs/blog/includes/functions.php on line 19Notice: Undefined variable: connection in /Applications/MAMP/htdocs/blog/includes/functions.php on line 22Warning: mysqli_query() expects parameter 1 to be mysqli, null given in /Applications/MAMP/htdocs/blog/includes/functions.php on line 22Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in /Applications/MAMP/htdocs/blog/includes/functions.php on line 24 I should pass the $connection variable when calling the function in users.php ? Like this? <?php $users = show_users($connection); foreach ($users as $key => $value) { echo $key . " " . $value; } ?> Thanks Kevin for pointing that out. Quote Link to comment Share on other sites More sharing options...
Ch0cu3r Posted December 10, 2013 Share Posted December 10, 2013 Yes you need to pass the connection when calling the show_users function. You have spelt connection incorrectly here function show_users($conection) { two n's not one Make sure you are including database.php before you call the function too. Quote Link to comment Share on other sites More sharing options...
kicken Posted December 10, 2013 Share Posted December 10, 2013 I should pass the $connection variable when calling the function in users.php ? Like this?Yes, you should. You need to define it as a parameter when declaring the function (as you did with function show_users($connection){) and you need to actually pass in the value at the time you call the function (as you showed). The errors you posted are caused by you forgetting to actually pass in the value when calling the function. Quote Link to comment Share on other sites More sharing options...
ANdrww Posted December 10, 2013 Author Share Posted December 10, 2013 Yes, you should. You need to define it as a parameter when declaring the function (as you did with function show_users($connection){) and you need to actually pass in the value at the time you call the function (as you showed). The errors you posted are caused by you forgetting to actually pass in the value when calling the function. Thanks a lot.. still not quite there though :\ Functions.php function show_users($conection) { $users = array(); $sql = "SELECT id, username FROM users WHERE status = 'active' ORDER BY username"; $result = mysqli_query($connection, $sql); while ($data = mysqli_fetch_array($result)) { $users[$data->id] = $data->username; } return $users; } Users.php <?php $users = show_users($connection); foreach ($users as $key => $value) { echo $key . " " . $value; } ?> Error Notice: Undefined variable: connection in /Applications/MAMP/htdocs/blog/includes/functions.php on line 22 Warning: mysqli_query() expects parameter 1 to be mysqli, null given in /Applications/MAMP/htdocs/blog/includes/functions.php on line 22 Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in /Applications/MAMP/htdocs/blog/includes/functions.php on line 24 It still doesn't get the connection. What am I missing? I've included the database.php file in users.php, I have the variable $connection = mysqli_connect($host, $username, $password, $database); Frustrating Quote Link to comment Share on other sites More sharing options...
KevinM1 Posted December 10, 2013 Share Posted December 10, 2013 Where is $connection coming from? It needs to exist before you can use it. Quote Link to comment Share on other sites More sharing options...
Solution ANdrww Posted December 10, 2013 Author Solution Share Posted December 10, 2013 Sorry, sorry sorry. I've just read that I misspelled $connection. It's now working as expected, and I'm currently trying to figure out the other functions I've wrote as well. Please don't close the topic (though solved) as I might (will for sure) ask other questions as well. Thanks a lot. Turns out I need to be more careful when writing the names. Quote Link to comment Share on other sites More sharing options...
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