php Laravel 4/5 搜索表单如
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Laravel 4/5 search form like
提问by erm_durr
I have a search form and functions, all done. Just to ask, is it possible to do it in Eloquent a query like this:
我有一个搜索表单和功能，全部完成。只是想问一下，是否可以在 Eloquent 中执行这样的查询：
SELECT * FROM players WHERE name LIKE '%".$name."%'
To display some possible matching names. My current controller function:
显示一些可能的匹配名称。我目前的控制器功能：
public function search()
{
$name = Input::get('character');
$searchResult = Player::where('name', '=', $name)->paginate(1);
return View::make('search.search')
->with('name', $name)
->with('searchResult', $searchResult);
}
And my view:
而我的观点：
<form id="custom-search-form" class="form-search form-horizontal pull-right" action="{{ URL::action('CharactersController@search') }}" method="get">
<div class="input-append spancustom">
<input type="text" class="search-query" name="character" placeholder="Character/guild name">
<button type="submit" class="btn"><i class="icon-search"></i></button>
</div>
</form>
Thanks in advance.
提前致谢。
回答by rmobis
Hmmm, yes, just set likeas your comparison operator, and send the string with %'s. Something like this:
嗯，是的，只需设置like为您的比较运算符，并发送带有%'s的字符串。像这样的东西：
Player::where('name', 'LIKE', "%$name%")->get();
回答by Yaroslav
If you need to frequently use LIKE, you can simplify the problem a bit. A custom method like () can be created in the model that inherits the Eloquent:
如果你需要经常使用LIKE，你可以把问题简化一点。可以在继承 Eloquent 的模型中创建像 () 这样的自定义方法：
public function scopeLike($query, $field, $value){
return $query->where($field, 'LIKE', "%$value%");
}
So then you can use this method in such way:
那么你可以这样使用这个方法：
User::like('name', 'Tomas')->get();
回答by mnv
With quote of string:
用字符串引用：
$value = DB::connection()->getPdo()->quote('%' . strtolower($value) . '%');
$query->whereRaw('LOWER(your_table.your_column) LIKE ' . $value);
回答by Fokwa Best
public function get_student($match){
return Student::where('name', 'like', $match .'%');
}
Try the above
试试上面的
