问题描述
在我的项目中,我创建了一个搜索框.当我键入一些内容并单击获取信息"按钮时,我会使用ajax调用在控制台窗口中获取所有信息.现在我想在yii2 gridview中填充此数据.数据在每个运行时都会有所不同.我想将此数据提供给gridview的$ dataprovider吗?
In my project, I have created a search box. when I type something and click the button "get info" I get all information in console window using ajax call. now I want to populate this data in yii2 gridview. data will different at every runtime. I want to give this data to $dataprovider of gridview is it possible??
这是代码-CompaniesController.php
here is code-CompaniesController.php
public function actionCompanyinfo(){
$text_in_search = $_GET['text_in_search'];
$left_items_cat = ltrim($_GET['left_items_cat']);
if($left_items_cat == "Companies"){
$query = (new \yii\db\Query())
->select(['c.name', 'c.id'])
->from(['companies as c'])
->where('c.name LIKE :query')
->addParams([':query'=>'%'.$text_in_search.'%'])
->all();
$response['comapnies_matching'] = $query;
return \yii\helpers\Json::encode([
$response
]);
}
}
companies/index.php
companies/index.php
$form = ActiveForm::begin();
$typeahead = $form->field($model, 'name')->textInput(['maxlength' => true]);
$getinfobtn = Html::SubmitButton( 'Get info', [ 'class' => 'btn btn-success' , 'id' =>'getinfo']) ;
ActiveForm::end();
myjsfile.js
myjsfile.js
$("#getinfo").click(function(){
var text_in_search = $("#companies-name").val();
var left_items_cat = $('#left-items li.active').text();
var url = "index.php?r=companies/companyinfo";
$.ajax({
url: url,
dataType: 'json',
method: 'GET',
data: {text_in_search,left_items_cat},
success: function (data, textStatus, jqXHR) {
// $( "#country"+id ).html(data[0].countries);
console.log(data[0]);
// **want to show this data in yii2 grid view**
},
error: function (jqXHR, textStatus, errorThrown) {
console.log('An error occured!');
alert('Error in ajax request');
}
});
控制台窗口
comapnies_matching
:
Array(3)
0
:
{name: "ADC Therapeutics Sarl", id: "402"}
1
:
{name: "ADC Therapeutics Sarl", id: "407"}
2
:
{name: "ADC Therapeutics Sarl", id: "412"}
如何在gridview中显示/填充此数据?
how to show/ populate this data in gridview??
推荐答案
CompaniesController.php
CompaniesController.php
public function actionCompanyinfo(){
$text_in_search = $_GET['text_in_search'];
$left_items_cat = ltrim($_GET['left_items_cat']);
if($left_items_cat == "Companies"){
$dataProvider= (new \yii\db\Query())
->select(['c.name', 'c.id'])
->from(['companies as c'])
->where('c.name LIKE :query')
->addParams([':query'=>'%'.$text_in_search.'%'])
->all();
return $this->renderPartial('gridview', [
'dataProvider' => $dataProvider,
]);
}}
您需要在gridview.php视图文件中呈现gridview
You need to render the gridview in gridview.php view file
myjsfile.js
myjsfile.js
$("#getinfo").click(function(){
var text_in_search = $("#companies-name").val();
var left_items_cat = $('#left-items li.active').text();
var url = "index.php?r=companies/companyinfo";
$.ajax({
url: url,
method: 'GET',
data: {text_in_search,left_items_cat},
success: function (data, textStatus, jqXHR) {
//set the ajax response as your html content
$('#myDiv').html(data);
},
error: function (jqXHR, textStatus, errorThrown) {
console.log('An error occured!');
alert('Error in ajax request');
}
});
这篇关于如何使用Yii2从ajax调用填充gridview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!