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 $insert_id;}
}
您只需向模型添加一个函数,如下所示:
<?php
public function getIngredientByName($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)
);
}
}
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)
);
}
}