Code Monkey home page Code Monkey logo

yii2-jqgrid-widget's People

Contributors

himiklab avatar rshamyan avatar

Stargazers

avatar avatar avatar avatar avatar avatar avatar avatar avatar avatar avatar avatar avatar avatar avatar avatar avatar avatar avatar avatar

Watchers

avatar avatar avatar avatar

Forkers

heartshare nuengcpe rinat-wc wersion rshamyan apaoww drsdre psolom osolodkin fernandotw bspavel gavin-k adargor buqento fredww

yii2-jqgrid-widget's Issues

Problem installing extension through composer

(Cross-post from http://www.yiiframework.com/forum/index.php/topic/61776-getting-himiklabs-gridjs-widget-for-yii2-working/ )

When trying to install yii2-jqgrid-widget through composer, by adding the following line to my composer.json - "himiklab/yii2-jqgrid-widget" : "*" - (with minimum stability "stable") and running composer update, I am greeted with a problem message:

Your requirements could not be resolved to an installable set of packages.

Problem 1

  • himiklab/yii2-jqgrid-widget 1.0.8 requires bower-asset/jqgrid 4.7.0 -> no matching package found.
  • himiklab/yii2-jqgrid-widget 1.0.7 requires bower-asset/jqgrid 4.7.0 -> no matching package found.
  • Installation request for himiklab/yii2-jqgrid-widget * -> satisfiable by himiklab/yii2-jqgrid-widget[1.0.7, 1.0.8].

Jqgrid and DepDrod

Здравствуйте, подскажите пожалуйста, как можно организовать передачу выбранного значения depdrop'а, чтобы можно было загрузить данные в jqgrid, зависящие от выбранных значений в depdrop'e?
Т.е., например, чтобы виджет jqgrid срабатывал после того, как установлены все значения в depdrop'e?

default

Мой вид

<?php
use himiklab\jqgrid\JqGridWidget;
use yii\helpers\Url;
use yii\widgets\ActiveForm;
use yii\helpers\ArrayHelper;
use kartik\depdrop\DepDrop;
use app\models\Groups;
?>
<div class="wrap">
    <div class="row">
        <?php
        $form = ActiveForm::begin();
        $query = \app\models\FormObuch::find()->all();
        $items = ArrayHelper::map($query,'idf','name_f');
        ?>
        <div class="col-sm-3">
            <?php
            echo $form->field($model, 'name_f')->dropDownList($items,
                [
                    'id'=>'cat-id',
                    'prompt' => 'Выбор формы...',
                ]);
            ?>
        </div>
        <div class="col-sm-3">
            <?php
            // Child # 1
            $model = new Groups();
            echo $form->field($model, 'name_g')->widget(DepDrop::classname(), [
                //'type'=>DepDrop::TYPE_SELECT2,
                'options'=>['id'=>'subcat-id'],
                'pluginOptions'=>[
                    'depends'=>['cat-id'],
                    'placeholder'=>'Выбор группы...',
                    'url'=>Url::to(['/post/subcat'])
                ]
            ]);
            ?>
        </div>
        <?php
            ActiveForm::end();
        ?>
    </div>
    <?= JqGridWidget::widget([
        'requestUrl' => Url::to('jqgrid'),
        'gridSettings' => [
            'colNames' => ['Фамилия', 'Имя', 'Отчество', 'Дата рождения', 'Номер зачетки'],
            'colModel' => [
                ['name' => 'famil', 'index' => 'famil', 'editable' => true],
                ['name' => 'name', 'index' => 'name', 'editable' => true],
                ['name' => 'otchestvo', 'index' => 'otchestvo', 'editable' => true],
                [
                    'name' => 'date_birth', 'index' => 'date_birth', 'editable' => true,
                    'formatter' => "date",
                    'formatoptions' => ['srcformat' => "ISO8601Long", 'newformat' => "m/d/Y"],
                ],
                ['name' => 'num_zuch', 'index' => 'num_zuch', 'editable' => true]
            ],
            'rowNum' => 15,
            'autowidth' => true,
            'height' => 'auto',
        ],
        'pagerSettings' => [
            'edit' => true,
            'add' => true,
            'del' => true,
            'search' => ['multipleSearch' => true],
        ],
        'enableFilterToolbar' => true,
        'filterToolbarSettings' => [
            'autosearch' => true,
            'stringResult' => true,
            'searchOnEnter' => true,
            'defaultSearch' => "cn"
        ],
    ]) ?>
</div>

Мой контроллер

<?php
namespace app\controllers;

use app\models\FormObuch;
use app\models\Result;
use Yii;
use app\models\EntryForm;
use himiklab\jqgrid\actions\JqGridActiveAction;
use yii\helpers\Json;
![default](https://user-images.githubusercontent.com/25239310/33243859-9b9c3b36-d2fe-11e7-9362-c5f1dcf075d1.png)

class PostController extends CommonController
{
    public $layout = 'basic';

    public function actions()
    {
        return [
            'jqgrid' => [
                'class' => JqGridActiveAction::className(),
                'model' => Result::className(),
                'scope' => function (\yii\db\ActiveQuery $query) {
//вывод данных с idg=5
                    $query->select(['ids', 'famil', 'name', 'otchestvo', 'date_birth','num_zuch'])->where(['idg'=>5]);
                },
            ],
        ];
    }

    public function actionGroups(){
        $model = new FormObuch();
        return $this->render('groups', ['model' => $model]);
    }

    public function actionSubcat() {
        $out = [];
        if (isset($_POST['depdrop_parents'])) {
            $parents = $_POST['depdrop_parents'];
            if ($parents != null) {
                $cat_id = $parents[0];
                $out = FormObuch::getGroups($cat_id);
                // the getSubCatList function will query the database based on the
                // cat_id and return an array like below:
                // [
                //    ['id'=>'<sub-cat-id-1>', 'name'=>'<sub-cat-name1>'],
                //    ['id'=>'<sub-cat_id_2>', 'name'=>'<sub-cat-name2>']
                // ]
                $poc = Json::encode(['output'=>$out]);
                echo Json::encode(['output'=>$out, 'selected'=>'']);
                return;
            }
        }
        echo Json::encode(['output'=>'', 'selected'=>'']);
    }
}

Subtable as table

Здравствуйте! Воспользовался вашим виджетом и у меня возник вопрос, возможно ли в нем реализовать подтаблицу со своим CRUD в таком виде?
image
Пытался сделать тоже самое с вашим виджетом, но что-то не идет.
image

View

 <?= JqGridWidget::widget([
        'requestUrl' => Url::to('jqgrid'),
        'gridSettings' => [
            'colNames' => ['Дата рождения', 'Имя', 'Фамилия', 'Номер зачетки'],
            'colModel' => [
                ['name' => 'datebirth', 'index' => 'datebirth', 'editable' => true],
                ['name' => 'firstname', 'index' => 'firstname', 'editable' => true],
                ['name' => 'lastname', 'index' => 'lastname', 'editable' => true],
                ['name' => 'num', 'index' => 'num', 'editable' => true],
            ],
            'rowNum' => 15,
            'autowidth' => true,
            'height' => 'auto',
            'subGrid' => true,
            'mtype' => 'GET',
            'subGridUrl' => Url::to('jqsubgrid'),
            'subGridModel' => [
                    [
                        'name' => ['Рейтинг', 'Присутствие'],
                        'width' => [200, 200],
                        'align' => ['left','left'],
                    ]
            ],
        ],
        'pagerSettings' => [
            'edit' => true,
            'add' => true,
            'del' => true,
            'search' => ['multipleSearch' => true],
        ],
        'enableFilterToolbar' => true,
        'filterToolbarSettings' => [
            'autosearch' => true,
            'stringResult' => true,
            'searchOnEnter' => true,
            'defaultSearch' => "cn"
        ],

    ]) ?>

Controller

 public function actions()
    {

            return [
                'jqgrid' => [
                    'class' => JqGridActiveAction::className(),
                    'model' => NewStudents::className(),
                    'scope' => function (\yii\db\ActiveQuery $query) {
                        $query->select(['datebirth', 'firstname', 'lastname', 'num']);
                    },
                ],
                'jqsubgrid' => [
                    'class' => JqGridActiveAction::className(),
                    'model' => Raiting::className(),
                    'scope' => function (\yii\db\ActiveQuery $query) {
                        $query->select(['raiting', 'presence']);
                    },
                ],
            ];
    }

    public function actionGroups(){
        return $this->render('groups');
    }

filter won't work on filter toolbar but work on find records

I find my filter toolbar won't work but if I use find records to filter. It can work property.

My solution is to adjust the code in controller like below. so that now it can work. but I wonder if any solution is better.

public function actions()
    {  
        return [  
            'jqgrid' => [  
                'class' => JqGridActiveAction::className(),  
                'model' => Country::className(),  
                'scope' => function ($query) {  
                    $query->select(['*']);  
                    if (isset($_REQUEST['country_id'])){  
                        $query->andFilterWhere(['like', 'country_id', $_REQUEST['country_id']]);  
                    } elseif(isset($_REQUEST['country_code'])){  
                        $query->andFilterWhere(['like', 'country_code', $_REQUEST['country_code']]);  
                    } elseif(isset($_REQUEST['country_name'])){  
                        $query->andFilterWhere(['like', 'country_name', $_REQUEST['country_name']]);  
                    } elseif(isset($_REQUEST['dial_code'])){  
                        $query->andFilterWhere(['like', 'dial_code', $_REQUEST['dial_code']]);  
                    } elseif(isset($_REQUEST['support'])){  
                        $query->andFilterWhere(['like', 'support', $_REQUEST['support']]);  
                    }  
                },  
            ],  
        ];  
    }

I can't use your Extension, help me pls

I'm newbie in yii2 ,I install extension with composer and I have himiklab folder in vendor,I tried to use this extension from example but not work! when i call jqgrid
I have error "Invalid Configuration – yii\base\InvalidConfigException
Object configuration must be an array containing a "class" element."

siteController

public function behaviors()
{
return [
'access' => [
'class' => AccessControl::className(),
'only' => ['logout'],
'rules' => [
[
'actions' => ['logout'],
'allow' => true,
'roles' => ['@'],
],
],
],
'verbs' => [
'class' => VerbFilter::className(),
'actions' => [
'logout' => ['post'],
],
],

          'jqgrid' => [
           'class' => JqGridActiveAction::className(),
            'model' => Page::className(),
            'columns' => ['title', 'author', 'language']
        ],
    ];
}

Bad Request (#400): Не удалось проверить переданные данные.

Здравствуйте, помогите решить следующую проблему.
Мой PostController.

namespace app\controllers;
use app\models\Result;
use Yii;
use app\models\EntryForm;
use himiklab\jqgrid\actions\JqGridActiveAction;

class PostController extends CommonController
{
    public $layout = 'basic';

    public function actions()
    {
        return [
            'jqgrid' => [
                'class' => JqGridActiveAction::className(),
                'model' => Result::className(),
                'scope' => function ($query) {
                    $query->select(['ids', 'fio', 'num_zuch','idg']);
                },
            ],
        ];
    }
    public function actionIndex(){
        $model = new EntryForm();
        if($model->load(Yii::$app->request->post()) && $model->validate()){
            return $this->render('autoriz', ['model' => $model]);
        }else{
            return $this->render('autoriz', ['model' => $model]);
        }
    }
    public function actionProgress(){
        return $this->render('progress');
    }
}

Представление progress

<?php
use himiklab\jqgrid\JqGridWidget;
use yii\helpers\Url;

?>

<?= JqGridWidget::widget([
    'requestUrl' => Url::to('jqgrid'),
    'gridSettings' => [
        'colNames' => ['IDS', 'FIO', 'NUMBER', 'IDG'],
        'colModel' => [
            ['name' => 'ids', 'index' => 'ids', 'editable' => true],
            ['name' => 'fio', 'index' => 'fio', 'editable' => true],
            ['name' => 'num_zuch', 'index' => 'num_zuch', 'editable' => true],
            ['name' => 'idg', 'index' => 'idg', 'editable' => true]
        ],
        'rowNum' => 15,
        'autowidth' => true,
        'height' => 'auto',
    ],
    'pagerSettings' => [
        'edit' => true,
        'add' => true,
        'del' => true,
        'search' => ['multipleSearch' => true]
    ],
    'enableFilterToolbar' => true,
])

Модель Result

namespace app\models;

use yii\db\ActiveRecord;

class Result extends ActiveRecord
{
    public static function tableName()
    {
        return 'students'; 
    }
}

В адресной строке если прописываю следующий url (http://mysite.ru/web/post/jqgrid?action=request), то на отдельной странице выводятся данные в json формате, но они не передаются в представление.

Объясните пожалуйста что я упустил?

400

404 not found error

In My Controller

namespace frontend\controllers;

use himiklab\jqgrid\actions\JqGridActiveAction;
use backend\models\MasterCodePiee;

class ResultEntryController extends \yii\web\Controller
{
    public function actions()
    {
        return [
            'jqgrid' => [
                'class' => JqGridActiveAction::className(),
                'model' => MasterCodePiee::className(),
                'scope' => function ($query) {
                    /** @var \yii\db\ActiveQuery $query */
                    $query->select(['S_CODE', 'S_TITLE'])->limit(100);
                },
            ],
        ];
    }
    public function actionIndex()
    {
       // $results = StudentResult::find()->limit(10)->all();
        return $this->render('index'/*, ['results'=>$results]*/);
    }

}

In My View

/* @var $this yii\web\View */
use himiklab\jqgrid\JqGridWidget;
use yii\helpers\Url;
use yii\helpers\BaseJson;
//echo BaseJson::encode($results);
?>

<?= JqGridWidget::widget([
    'requestUrl' => Url::toRoute('jqgrid'),
    'gridSettings' => [
        'colNames' => ['S_CODE', 'S_TITLE'],
        'colModel' => [
            //['name' => 'MATRIC_NO', 'index' => 'MATRIC_NO', 'editable' => true],
            ['name' => 'S_CODE', 'index' => 'S_CODE', 'editable' => true],
            ['name' => 'S_TITLE', 'index' => 'S_TITLE', 'editable' => true]
        ],
        'rowNum' => 15,
        'autowidth' => true,
        'height' => 'auto',
    ],
    'pagerSettings' => [
        'edit' => true,
        'add' => true,
        'del' => true,
        'search' => ['multipleSearch' => true]
    ],
    'enableFilterToolbar' => true,
]) ?>

But I got error 404 not found

image

Very appreciate if you can help.

Recommend Projects

  • React photo React

    A declarative, efficient, and flexible JavaScript library for building user interfaces.

  • Vue.js photo Vue.js

    🖖 Vue.js is a progressive, incrementally-adoptable JavaScript framework for building UI on the web.

  • Typescript photo Typescript

    TypeScript is a superset of JavaScript that compiles to clean JavaScript output.

  • TensorFlow photo TensorFlow

    An Open Source Machine Learning Framework for Everyone

  • Django photo Django

    The Web framework for perfectionists with deadlines.

  • D3 photo D3

    Bring data to life with SVG, Canvas and HTML. 📊📈🎉

Recommend Topics

  • javascript

    JavaScript (JS) is a lightweight interpreted programming language with first-class functions.

  • web

    Some thing interesting about web. New door for the world.

  • server

    A server is a program made to process requests and deliver data to clients.

  • Machine learning

    Machine learning is a way of modeling and interpreting data that allows a piece of software to respond intelligently.

  • Game

    Some thing interesting about game, make everyone happy.

Recommend Org

  • Facebook photo Facebook

    We are working to build community through open source technology. NB: members must have two-factor auth.

  • Microsoft photo Microsoft

    Open source projects and samples from Microsoft.

  • Google photo Google

    Google ❤️ Open Source for everyone.

  • D3 photo D3

    Data-Driven Documents codes.