I have implemented my project in Yii. I displaying search results in view part.
I wrote a search query for fetching the results from one table which is Recipe. In this table name, course_id, cuisinename,type, colorie_count respectively, course_id,cuisinename, type are columns respectively.
In my controller i write code like this:
$result="SELECT * FROM recipe WHERE name LIKE '%$name%' AND `cuisinename` LIKE '$cuisine1%' AND course_id LIKE '%$course1%' AND `type` LIKE '%$type1%' AND `calorie_count` LIKE '%$calorie1%' ORDER BY recipe_id DESC LIMIT 15";
values are getting. if i give condition based to display the search result text. not displaying all name. but those are displaying based on names.
I added below my view part condition and code:
$query= Course::model()->find("course_id=$as1");
$course2=$query->course_name;
$query1= Cuisine::model()->find("id=$as4");
$cuisine2=$query1->cuisinename;
$query3= RecipeType::model()->find("id=$as3");
$type2=$query3->recipeType_name;
<?php echo '<p align="center"> Showing results for : '.'Name'.$getval.', '.'Course-'.$course2.', '.'Cuisine-'.$cuisine2.', '.'Type-'.$type2;', ';'</p>';
echo ', '.'Calories-'.$calorie;
?>
You need to create relations between tables look there. For Recipe model it should be
public function relations()
{
return array(
'cuisine'=>array(self::BELONGS_TO, 'Cuisine', 'cuisine_id'),
'type'=>array(self::BELONGS_TO, 'RecipeType', 'type_id'),
);
}
Then you can get values as $model->cuisine->name. If you don't understand creating relations, generate models (it tables must be correct foreign keys) with gii.