Laravel使用"if"查询情况? [英] Laravel query with "if" conditions?

问题描述

我正在尝试使用Laravel 4创建一个高级搜索表单，这是查询:

I'm trying to make an advanced search form with Laravel 4, and this is the query:

$result = DB::table('users_ads')
        ->join('ads', 'users_ads.ad_id', '=', 'ads.id')         
        ->orderBy($column, $method)
        ->where('status', TRUE)
        ->where(function($query) use ($input)
        {
            $query->where('short_description', $input['search'])
                    ->where('category', $input['category'])
                    ->where('product', $input['product']);

        })
        ->join('users', 'users_ads.user_id', '=', 'users.id')
        ->select('ads.id', 'ads.img1', 'ads.short_description', 'ads.category', 'ads.product', 'ads.price', 'users.city')
        ->get();

    return $result;

问题是用户可能不会使用所有输入字段.所以我想在这部分包括一些条件:

The problem is that the user might not use all the input fields. So i want to include some if conditions in this part:

$query->where('short_description', $input['search'])
                    ->where('category', $input['category'])
                    ->where('product', $input['product']);

..因此，如果输入为空，则删除"where"条件.

.. so if the input is empty, to remove the "where" condition.

推荐答案

$filters = [
    'short_description' => 'search',
    'category' => 'category',
    'product' => 'product',
];

.....

->where(function($query) use ($input, $filters)
{
    foreach ( $filters as $column => $key )
    {
        $value = array_get($input, $key);

        if ( ! is_null($value)) $query->where($column, $value);
    }
});

较新版本的Laravel具有when方法，这使此操作变得更加容易:

Newer version of Laravel have a when method that makes this much easier:

->where(function ($query) use ($input, $filters) {
    foreach ($filters as $column => $key) {
        $query->when(array_get($input, $key), function ($query, $value) use ($column) {
            $query->where($column, $value);
        });
    }
});

这篇关于Laravel使用"if"查询情况?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！