How to update multiple database records using a 2d array in CodeIgniter

Question

I have a table menu w/c contains of(recipe_id,ingredient_id,category_id).I was trying to update my ingredients yet it only update 1 ingredient_id. Like this

Here is my code:

CONTROLLER:

public function save_edit_recipe()
{
    foreach ($this->input->post('ingredients') as $key => $value) {
        $menuData[] = array(
            'recipe_id'     => intval($this->input->post('recipe_id')),
            'ingredient_id' => intval($value),
            'category_id'   => intval($this->input->post('recipe_category'))
        );
    }
    // var_dump($menuData); die();
    $this->products_model->updatemenu($menuData);
}

menudata is:

Array (
    [0] => Array (
        [recipe_id] => 2
        [ingredient_id] => 1
        [category_id] => 3
    )
    [1] => Array (
        [recipe_id] => 2
        [ingredient_id] => 2
        [category_id] => 3
    )
)

MODEL:

public function updatemenu($data)
{
    foreach ($data as $row => $value) {
        $this->db->where('ingredient_id', $data['ingredient_id']);
        $query = $this->db->update('menu', $value);
    }
    return $result;
}

Abdulla Nilam · Accepted Answer · 2016-01-29 23:03:41Z

3

Remove all data in controller and add in to model.

In Controller

$this->products_model->updatemenu();

In Model

public function updatemenu()
{
    foreach($this->input->post('ingredients') as $key => $value)
    {
        $menuData = array(
            'recipe_id'         => intval($this->input->post('recipe_id')),
            'ingredient_id'     => intval($value),
            'category_id'       => intval($this->input->post('recipe_category'))
        );
        $this->db->where('ingredient_id', $value);
        $this->db->update('menu', $menuData);
    }
}

answered Jan 29, 2016 at 23:03

Abdulla Nilam

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4 Comments

Abdulla Nilam Over a year ago

Check with this. Im not 100% sure. Before check change your data field as it was early too

Tpojka Over a year ago

@shikira I flagged your comment. You should delete that type of data set publicly.

devpro Over a year ago

I think no need to add ingredient_id in update array bro. Or any other column will use for where clause

mickmackusa Jul 20 at 13:09

This answer seemingly fixes nothing. It is not the responsibility of the model to access submission data. Doing so makes unit testing more tedious. It is better practice to pass all required data into the model method. The model should be querying the database and virtually nothing else.

mickmackusa · Accepted Answer · 2025-07-20 13:49:28Z

Your screenshot seems to indicate that:

ingredient_id is not a unique column in the table, and
ingredient_id 1 occurs twice, and
ingredient_id 2 does not exist.

Therefore, trying to UPDATE ingredient_id 1 will affect 2 records and ingredient_id 2 will affect no records. There is evidence that you unintentionally corrupted your database ingredient_id values during earlier development.

Regardless, if you want a process that can insert or update a record, you should make ingredient_id a unique primary key and call CodeIgniter's replace() method.

public function updatemenu(array $rows): int
{
    $affectedRows = 0;
    foreach ($rows as $row) {
        if ($this->db->replace('menu', $row, ['ingredient_id' => $row['ingredient_id']])) {
            $affectedRows += $this->db->affected_rows();
        }
    }
    return $affectedRows;
}

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