Level 104
If you wanted to persist the search filter across GET requests you might be better to append the search to the query string. Then the pagination links will automatically have the filter as well as the page number
1 like
Jul 27, 2018
9
Level 5
Problem in pagination after ajax return in laravel
Guys I am working with a project milkfarm
I am having a form with search box and text box.
In the textbox event on change i wrote a jquery to search the value
Everything is file it searches and appends with the
But after search is done if I move to page 2 then it is refreshed the
search option disappears and all the data are displayed
Say for example I will seach time=”am” then it fetches 7 records
because there are only 7 record matches. But I have given
pagination(5). So only 5 records are displayed in the page 1 to see the
balance 2 records if I click page 2 the the page is refreshed so search
done is cancelled
How to rectify it
This is my view.blade
<tbody >
@foreach($exs as $ex)
<tr>
<td>{{ $ex->date }}</td>
<td>{{ $ex->time }}</td>
<td>{{ $ex->no_of_litre }}</td>
<td>{{ $ex->note }}</td>
</tr>
@endforeach
</tbody>
This is my javascript
<script type="text/javascript">
$('#search_time').on('keyup',function(){
$.ajax({
dataType: 'json',
type : 'get',
url : '{{URL::to('search_excess_milk_details')}}',
data:{
'search_time': $('#search_time').val()
},
success:function(data)
{
var res='';
$.each (data, function (key, value) {
res +=
'<tr>'+
'<td>'+value.date+'</td>'+
'<td>'+value.time+'</td>'+
'<td>'+value.no_of_litre+'</td>'+
'<td>'+value.note+'</td>'+
'</tr>';
});
$('tbody').html(res);
}
});
})
</script>
This is my controller search
public function search_excess_milk_details( Request $request )
{
$xcess_milks = Excess_milk::where(function ($query) use ($request)
{
if (!empty($request->search_time)) {
$query->Where('time', 'LIKE', '%' . $request->search_time . '%');
}
})->orderBy('date','asc')->get;
return Response($xcess_milks);
}
Kindly some one help please..!!!
Level 5
Level 5
@tykus kindly reply for this also??
please
Level 2
Try your code like
controller
public function search() {
$input = Input::get('search');
$posts = Post::where('title','LIKE',$input)->paginate(5);
if (Request::ajax()) {
return View::make('posts.search')->withPosts($posts);
}
return View::make('posts.index')->withPosts($posts);
}
1 like
Level 5
@pardeepkumar thankz for your reply
can u slightly explain what will happen in this code??
1 like
Level 2
First i am get search input value by using$input = Input::get('search'); then
by help of Post model with where condition search particular value and then paginate it.
then by ajax request with if condition to get desire output .
1 like
Level 24
public function search_excess_milk_details( Request $request )
{
$searchTime = $request->filled('search_time') ? $request->query('search_time') : null;
$query = Excess_milk::query();
$queryStr = [];
if (!empty($searchTime))
{
$query->Where('time', 'LIKE', '%' . $searchTime . '%');
$queryStr['search_time'] = $searchTime;
}
$xcess_milks = $query->orderBy('date', 'asc')->paginate(5);
// append the query string, eg, http://localhost/something?search_time=searchTime
$xcess_milks->appends($queryStr);
return Response($xcess_milks);
}
and view
@if ($xcess_milks->hasPages())
{{ $xcess_milks->links() }}
@endif
this just an example, you need modify by yourself
Level 75
can slightly explain with coding please.
If you don't understand the query string why don't you take a break from laravel and take a couple of quality tutorials before getting too deep into laravel.
It should go without saying that those parameters have to be passed somewhere.
1 like
Level 5
@jlrdw @newbie360 @pardeepkumar
thank you guys
Let me try..
Kindly if possible refer my thread and give a solution for that
How to use form tag in ajax return in laravel?
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