CodeIgniter Forums

Update to 4.6.1

Posted by: serialkiller - 05-07-2025, 01:33 AM - Replies (12)

I'm updating some applications and one of the files that has changed is Config/Paths.php.
Compared to the previous one, it has a different value for the $systemDirectory variable, before it was:

PHP Code:
public string $systemDirectory = __DIR__ . '/../../vendor/codeigniter4/framework/system'; 

now instead I find:

PHP Code:
public string $systemDirectory = __DIR__ . '/../../system'; 

But this will never work, the system folder is not located there.

Am I doing something wrong?

systemDirectory conundrum

Posted by: codeus - 05-06-2025, 06:48 AM - Replies (2)

Hi all,
Just did a test install of CI v4.6.1 with composer. All worked fine and dandy.

I noticed that in app/Config/Paths the $systemDirectory is:

PHP Code:
    public string $systemDirectory = __DIR__ . '/../../vendor/codeigniter4/framework/system'; 

I'm sure that in v4.5 (and earlier) $systemDirectory was

PHP Code:
public string $systemDirectory = __DIR__ . '/../../system'; 

and that for every install I would have to edit this to _DIR__ . '/../../vendor/codeigniter4/framework/system' in Paths.php

Was this a bug? I'm curious because I've just seen a v4.4.0 installation (not mine) with system directory at the same level as app.

Don't know why system directory would be at the same level as app. Was this for a non-composer install??

TIA
Mike

Can't create new database with forge

Posted by: occitan - 05-05-2025, 09:01 PM - Replies (3)

I have tried this:

PHP Code:
$forge = \Config\Database::forge();
$forge->createDatabase('test_1002'); 

and this:

PHP Code:
$dbName = 'test_1002';
$sql = "CREATE DATABASE {$dbName}";
$query = $this->db->query($sql); 

and I get:

PHP Code:
CRITICAL - 2025-05-06 03:35:45 --> [Caused by] mysqli_sql_exception: Access denied for user 'development'@'%' to database 'test_1002' 

The server is Liquid Web Alma Linux, Maria DB, Plesk... the db user has full access to all functions and all databases.
I can create a db over ssh but it doesn't show up to Plesk, because Plesk only tracks it's own databases.
Does anyone know how to get past this, and create Plesk visible databases with CI4?

5 Things I Wish I Knew Before Starting a Startup as a Software Engineer

Posted by: php_rocs - 05-05-2025, 05:29 AM - Replies (1)

	StartUp advice (DevTo): https://dev.to/code42cate/5-things-i-wis...ineer-208i

Why Every Programmer Needs a Non Computer Hobby

Posted by: php_rocs - 05-05-2025, 05:26 AM - Replies (2)

	Good advice ( DevTo): https://dev.to/mahdijazini/why-every-pro...-hobby-fgf

Using Pager with an Array

Posted by: nneves - 05-04-2025, 01:55 PM - No Replies

	Hi I have some results from a Model that needs to be cached after being parsed to clean user comments. Is there's a way to use Pager with an Array? Thanks

CodeIgniter 4.6.1 has been released

Posted by: InsiteFX - 05-03-2025, 03:01 AM - Replies (2)

	CodeIgniter 4.6.1 has been released Download Upgrading from 4.6.0 to 4.6.1 Changes Bugs Fixed I' am sorry to say that while I was cleaning up the SPAM this morning I Accidently deleted the thread. Which I' am reporting to MyBB because the thread was not even checked a post was. Again sorry for my mistake. I have emailed Paul to let him know about it.

Feature to not append database prefix in QueryBuilder

Posted by: objecttothis - 05-02-2025, 02:37 AM - Replies (5)

Currently I'm not aware of the means to prevent query builder from appending the database prefix, so which you need to run a query where you do not want the prefix appended, you either can't use QueryBuilder or you have to use acrobatics to make it work. Please consider adding a parameter to QueryBuilder turning off the appending of the database prefix.

Take this code as an example of what I mean

PHP Code:
function dropForeignKeyConstraints(array $foreignKeys, string $table): void {
    $db = Database::connect();

    $prefixName = $db->getPrefix();
    $db->setPrefix('');
    $database_name = $db->database;

    foreach ($foreignKeys as $fk)
    {
        $builder = $db->table('INFORMATION_SCHEMA.TABLE_CONSTRAINTS');
        $builder->select('CONSTRAINT_NAME');
        $builder->where('TABLE_SCHEMA', $database_name);
        $builder->where('TABLE_NAME', $table);
        $builder->where('CONSTRAINT_TYPE', 'FOREIGN KEY');
        $builder->where('CONSTRAINT_NAME', $fk);
        $query = $builder->get();

        if($query->getNumRows() > 0)
        {
            $db->query("ALTER TABLE `$table` DROP FOREIGN KEY `$fk`");
        }
    }

    $db->setPrefix($prefixName);
} 

To prevent the query builder from appending the dbPrefix to INFORMATION_SCHEMA.TABLE_CONSTRAINTS, which would break the query, I need to capture the prefix, set it to an empty string, then put it back after I'm done with the QueryBuilder so that code below resumes the prefix appending. While this isn't needed all the time, it's better if I don't pepper the code with these extra three lines. I could write $prefixName = removePrefix() and restorePrefix($prefixName) helper functions, but that really only saves me one line of code. It would be better if I could call something like $builder = $db->table('INFORMATION_SCHEMA.TABLE_CONSTRAINTS', false); where the false parameter in the table() function returned the query without the prefix appended.

CI4 not loading after fresh install — blank page on localhost Post:

Posted by: sultanhassann6 - 04-30-2025, 01:13 AM - No Replies

I’m trying to set up CodeIgniter 4 on my local dev machine (Ubuntu 22.04, PHP 8.1) for a side project — a fresher job board I’m working on: https://247uaefreshers.com.
I followed the official CI4 install guide using Composer, no issues during setup. But when I navigate to

Code:
localhost:8080

, I just get a blank white page — no error, no logs, nothing in the browser console either.
I’ve tried:
• Running

Code:
php spark serve

(starts fine)
• Double-checked

Code:
.env

settings
• Permissions on writable/ seem okay
• Tried adding basic echo in

Code:
app/Controllers/Home.php

— still blank
Anyone seen this behavior before? Wondering if I missed something basic. Appreciate any tips ?

Model Find function Changed?

Posted by: dulongc - 04-29-2025, 02:41 PM - Replies (1)

In the past when I used the find() function on a model while passing the primary key, I would get a single result as the documentation states. It seems now in v. 4.6.0 it acts like it is going to return multiple results. This requires me to call $returnedArray[0]['name'] instead of just $returnedArray[0]['name']

My Relevant Code:

PHP Code:
$transID = $this->request->getPost('trans_id',FILTER_SANITIZE_NUMBER_INT);
$transModel = new \App\Models\TransactionModel();
$prevTran = $transModel->find($transID);

//Make sure this payment intent isn't already paid
if($prevTran['status'] == 'paid') {
                    
    //This transaction is already paid
    echo json_encode(array('status'=>'alread_paid','message'=> 'This transaction is already marked as paid.'));
    exit();
                    
} 

This results in error:

Quote:message 'Undefined array key "status"'

Upon inspecting the return from the database I can see the issue is that it's returning the row within a multidimensional array.

Expected Return:

Code:
$transModel = [

    'id'         => 1,

    'name'    => 'Registration',

    'status'    => 'paid',

]

Actual Return

Code:
$transModel = [

    [0]    =>    [

        'id'         => 1,

        'name'    => 'Registration',

        'status'    => 'paid',

    ],

]

I can do a work around by just overwriting the response. But am I wrong in believing this should already be the default response.

PHP Code:
$transID = $this->request->getPost('trans_id',FILTER_SANITIZE_NUMBER_INT);
$transModel = new \App\Models\TransactionModel();
$prevTran = $transModel->find($transID);

//Overwrite the return here
$prevTran = $prevTran[0];

//Make sure this payment intent isn't already paid
if($prevTran['status'] == 'paid') {
                    
    //This transaction is already paid
    echo json_encode(array('status'=>'alread_paid','message'=> 'This transaction is already marked as paid.'));
    exit();
                    
} 

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