将唯一的提交值保存到数据库表中,并使用 CodeIgniter 返回每个值的 id

问题描述 投票:0回答:2

我的表成分由(ingredient_id,name)组成,类别由(category_id,name)组成,category_ingredients由(ingredient_id,category_id)组成。我创建了一个按类别添加许多成分的表单,我想检查 1 种或多种成分是否已经存在,那么我只需要获取成分的 id,然后其他不存在的成分将插入到我的数据库中。你们能帮我吗?

这是我的代码: 查看:

    <?php echo form_open('dashboard/uploadIngredients', 'class="form-horizontal" enctype="multipart/form-data"'); ?>
        <div class="form-group">
            <div class="col-sm-10">

                <select required class="form-control" name="ingredient_category">

                    <option value="" selected disabled>Select Ingredient Category</option>
                <option value="All">All</option>
                <?php foreach($this->products_model->getCategory() as $row): ?>
                    <option value="<?php echo $row->category_id ?>"><?php echo $row->category_name; ?></option>
                <?php endforeach; ?>
                </select>

            </div>
        </div>
        <div class="form-group">
            <div class="col-sm-10">
                <textarea class="form-control" name="ingredients" rows="5" placeholder="Ingredients (EX. onion, oil, pasta)" required></textarea> 
            </div>
        </div>

        <div class='form-group'>
            <div class="col-sm-10">
                <button class="btn btn-lg btn-positive" type="submit"><i class="glyphicon glyphicon-ok"></i> Save Ingredient</button>
            </div>
        </div>
    <?php echo form_close(); ?>

控制器:

 public function uploadIngredients()
{

    foreach(explode(',', $this->input->post('ingredients')) as $key => $value)
    {
        $saveData[] = array('ingredient_id' => null,
                            'name'  => trim($value)
        );  
    }

    $ingredient_id = $this->products_model->saveIngredients($saveData); 
    foreach (explode(',', $this->input->post('ingredient_category')) as $key => $value)
    {
     foreach ( $ingredient_id as $key => $str ){
        $joinData[] = array(
                            'ingredient_id'     => $str,
                            'category_id'       => intval($value)
        );
}
        //var_dump($joinData); die();
        $this->products_model->saveCategoryIngredients($joinData);

        redirect('dashboard/add_ingredients');
        }


}

型号:

 public function saveIngredients($ingredient_id)
{
    foreach($ingredient_id as $row => $value) {
        $query=$this->db->where('ingredient_id', $value->ingredient_id);
            $this->db->insert('ingredient', $value);
            $insert_id[] = $this->db->insert_id();  
    }


    return $insert_id;
}


public function saveCategoryIngredients($data)
{

     foreach($data as $row => $value)
     {
        $this->db->insert('category_ingredient', $value);
        $insert_id[] = $this->db->insert_id();  
     }          
     return $insert_id;}
}
php codeigniter sql-insert exists submission
2个回答
0
投票

您只需向模型添加一个函数,如下所示:

<?php

public function getIngredientByName($name) {
    return $this->db
    ->from('ingredient I')
    ->where('I.name', $name)
    ->get()->row(); //Will return the row of ingredient if ingredient exists, else null
}

在你的控制器中:

<?php
foreach(explode(',', $this->input->post('ingredients')) as $key => $value)
{
    if (!$this->products_model->getIngredientByName($value)) {
        $saveData[] = array(
            'ingredient_id' => null,
            'name'  => trim($value)
        );  
    }
}

0
投票

感谢 Gwendal 回答这个问题,我正在稍微修改这个答案以防止重复插入,例如:- 如果用户插入 SuGar CaNe 但我们的数据库中有 Sugar Cane 那么使用答案的代码我们将有 2 个 sugar cane 为了避免这些类型的插入,我们可以将此代码用于模型

<?php

public function getIngredientByName($name) {
    return $this->db
    ->from('ingredient I')
    -> where("( REPLACE( LOWER(I.name), ' ', '') LIKE '".strtolower(preg_replace('/\s+/', '', $name)) ."%')")
    ->get()->row(); //Will return the row of ingredient if ingredient exists, else null
}

控制器与

相同
<?php
foreach(explode(',', $this->input->post('ingredients')) as $key => $value)
{
    if (!$this->products_model->getIngredientByName($value)) {
        $saveData[] = array(
            'ingredient_id' => null,
            'name'  => trim($value)
        );  
    }
}
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