使用 laravel 和 jquery 分页无法按预期工作
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【中文标题】使用 laravel 和 jquery 分页无法按预期工作【英文标题】：Paginate doesn't work as expected uisng laravel and jq 【发布时间】：2020-05-04 22:21:48 【问题描述】：我正在处理过滤数据。因为有<input>，如果有人开始打字，我正在调用jq 函数向我的控制器发送请求。它工作正常。即使我也得到过滤数据。但是，如果我单击第 2 页，则会影响设计。所以当我知道它是改变我的url。
举个例子—— 过滤数据后（这工作正常）
code.test/?page=1
但是如果你点击第2页，它会重定向到
code.test/filter?page=2
这是main.blade-的代码-
<div class="container">
<div class="row">
<div class="col-md-9 col-sm-12">
<div class="form-group">
<input type="text" class="form-control" id="search" placeholder="Enter email" name="email">
</div>
</div>
<div class="col-md-3 col-sm-12">
<div class="form-group">
<select id="department">
<option value="0">All Departments</option>
@foreach($department as $d)
<option value="$d->id">$d->name</option>
@endforeach
</select>
</div>
</div>
</div>
<div id="filter">
<div class="row">
<?php
$count = count($data);//dd($data[0]->fname);
?>
@if($count > 0)
@foreach($data as $d)
<div class="col-md-12">
$d->fname, $d->lname<br>
$d->profile<br>
<b>$d->departments->name</b>
</div>
@endforeach
@else
No data found
@endif
</div>
</div>
$data->appends($data)->links()
</div>
这是我的 jq 函数 -
function filter()
var str = $("#search").val();
var dep = $('#department option:selected').val();
// /alert(dep);
$.ajaxSetup(
headers:
'X-CSRF-TOKEN': $('meta[name="csrf-token"]').attr('content')
);
$.ajax(
type: "GET",
url: '/filter',
data:
str: str,
dep: dep,
,
success: function(data)
console.log(data);
$('#filter').html(data);
,
);
$(document).ready(function()
$("#search").on('input', function()
filter();
);
$("#department").change(function ()
filter();
);
);
现在在controller，我在过滤后返回view-
public function filter(Request $request)
$str = $request->str;
$dep = $request->dep;//dd($dep);
$s = new Staff;
$d = new Department;
//If input and dropdown values are available
if($str != null && $dep != 0)
$data = $s::with('departments')
->where('department', $dep)
->where(function($q) use ($str)
$q->where('fname', 'like', '%'.$str.'%')
->orWhere('lname', 'like', '%'.$str.'%');
)
->paginate(10)
->appends(['dep'=> $dep, 'str'=> $str]);
$data_count = count($data);
return view('search', compact('data', 'data_count'));
else if($str != null && $dep == 0) //If input value is set and dropdown value set to all departments
$data = $s::with('departments')
->where(function($q) use ($str)
$q->where('fname', 'like', '%'.$str.'%')
->orWhere('lname', 'like', '%'.$str.'%');
)
->paginate(10)
->appends(['dep'=> $dep, 'str'=> $str]);
$data_count = count($data);
return view('search', compact('data', 'data_count'));
else if($str == null && $dep != 0) //If dropdown value is not null and input is null
$data = $s::with('departments')
->where('department', $dep)
->paginate(10)
->appends(['dep'=> $dep, 'str'=> $str]);
$data_count = count($data);
return view('search', compact('data', 'data_count'));
else if($str == null && $dep == 0) //If dropdown value is null and input is null
$data = $s::with('departments')->paginate(10)->appends(['dep'=> $dep, 'str'=> $str]);
$data_count = count($data);
return view('filter', compact('data', 'data_count'));
请帮帮我。
提前谢谢你。
【问题讨论】：
【参考方案1】：现在回答有点晚了，但这可能会对你有所帮助 -
-
在您的
main.blade 中，您可以将该数据传递给其他view，然后将@include 新的blade 文件传递给main.blade，而不是直接加载数据。像这样 -
<div class="ui-block">
<div class="ui-block-content">
<div class="row">
<div class="col col-xl-9 col-lg-9 col-md-9 col-sm-12 col-12">
<div class="form-group">
<input type="text" class="form-control" id="search" placeholder="Start typing keywords.." name="search">
</div>
</div>
<div class="col col-xl-3 col-lg-3 col-md-3 col-sm-12 col-12">
<div class="form-group">
<select id="department" class="form-control">
<option value="0">All Departments</option>
@foreach($department as $d)
<option value="$d->id">$d->name</option>
@endforeach
</select>
</div>
</div>
</div>
</div>
</div>
<div id="filter">
@include('search')
</div>
以search.blade 为例，创建新刀片并将<div id="filter"> /* ---- This lines ---*/ 中的任何内容粘贴到search.blade。
在你jsmain.blade的函数更改url-
$.ajax(
type: "GET",
url: '/',//change this to main.blade's url
data:
str: str,
dep: dep,
,
success: function(data)
console.log(data);
$('#filter').html(data);
,
);
-
在
controller 中进行更改。只需检查请求是否来自ajax，然后处理该数据。
public function main(Request $request)
$data = Staff::with('departments')->orderBy('created_at', 'DESC')->paginate(10);
$department = Department::all();
//If request is from ajax, then processing to filter data
if($request->ajax())
$str = $request->str;
$dep = $request->dep;//dd($dep);
$s = new Staff;
$d = new Department;
if($str != null && $dep != 0) //If input and dropdown values are available
$data = $s::with('departments')
->where('department', $dep)
->where(function($q) use ($str)
$q->where('fname', 'like', '%'.$str.'%')
->orWhere('lname', 'like', '%'.$str.'%')
->orWhere('profile', 'like', '%'.$str.'%');
)
->orderBy('created_at', 'DESC')
->paginate(10);
else if($str != null && $dep == 0) //If input value is set and dropdown value set to all departments
$data = $s::with('departments')
->where(function($q) use ($str)
$q->where('fname', 'like', '%'.$str.'%')
->orWhere('lname', 'like', '%'.$str.'%')
->orWhere('profile', 'like', '%'.$str.'%');
)
->orderBy('created_at', 'DESC')
->paginate(10);
$data_count = count($data);
return view('search', compact('data', 'data_count'));
else if($str == null && $dep != 0) //If dropdown value is not null and input is null
$data = $s::with('departments')
->where('department', $dep)
->orderBy('created_at', 'DESC')
->paginate(10);
else if($str == null && $dep == 0) //If dropdown value is null and input is null
$data = $s::with('departments')->orderBy('created_at', 'DESC')->paginate(10);
// returning data to view
return view('search', ['data' => $data])->render();
//returning data if request is not from ajax
return view('main', compact('data', 'department'));
希望这对你有用。谢谢。
【讨论】：
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