Question

如果enablePrettyUrl设置为false，则以下代码将按预期执行：

<?php
$script = <<< JS
   $('#zip_code').change(function(){
       var zipId=$(this).val();
       $.get('index.php?r=locations/get-city-province',{zipId:zipId},function(data){
       var data=$.parseJSON(data);
       alert(data.city+" liegt in "+data.province+"! Die Id ist "+zipId);
       $('#customers-city').attr('value',data.city);
       $('#customers-province').attr('value',data.province);
   });
});

JS;
$this->registerJS($script);
?>

以下代码不会，如果enablePrettyUrl设置为true：

 $.get('locations/get-city-province',{zipId:zipId},function(data){

以下是UrlManager的规则：

'urlManager' => [
    'class' => 'yii\web\UrlManager',
    'enablePrettyUrl' => true,
    'showScriptName' => true,
    'enableStrictParsing' => true,
    'rules' => [
        'locations' => 'locations/index',
        'locations_create' => 'locations/create',
        'locations_delete' => 'locations/delete',
        'locations_update' => 'locations/update',
        'locations_SaveAsNew' => 'locations/save-as-new',
        'locations_pdf' => 'locations/pdf',
        'locations_view' => 'locations/view',
        // ...
    ],
],

任何想法，如何以正确的方式编码$.get？这是Controller的方法：

public function actionGetCityProvince($zipId) {
    $location = Locations::findOne($zipId);
    echo Json::encode($location);
}

Answer 1

使用Url::to()或Url::toRoute()

<?php
$url = yii\helpers\Url::to(['locations/get-city-province']);
$script = <<< JS
    $('#zip_code').change(function(){
        var zipId = $(this).val();
        $.get('$url', {zipId:zipId}, function(data){
            var data = $.parseJSON(data);
            alert(data.city+" liegt in "+data.province+"! Die Id ist "+zipId);
            $('#customers-city').attr('value',data.city);
            $('#customers-province').attr('value',data.province);
        });
   });
JS;
$this->registerJS($script);
?>

如何使用enablePrettyUrl =＆gt; true定义$ .get url

1 个答案: