i have two tables, in yearselection
two columns are there and in testtable2
three columns are there, based on first table id's using in second table., i want to display json responce as below using php, these two tables.
yearselection:
id year
6 2014-2015
2 2010-2011
3 2011-2012
4 2012-2013
5 2013-2014
1 2009-2010
7 2015-2016
testtable2:
id name yearselection
1 test1 2
2 test2 1
3 test3 1
4 test4 1
5 test5 2
6 test6 3
i want display like this in json format:
{
"2009-2010": [
{
"id": "2",
"name": "test2"
},
{
"id": "3",
"name": "test3"
},
{
"id": "4",
"name": "test4"
}
],
"2010-2011": [
{
"id": "1",
"name": "test1"
},
{
"id": "5",
"name": "test5"
}
],
"2011-2012": [
{
"id": "6",
"name": "test6"
}
]
}
mycode
public function actionArchives()
{
//echo $keyword=$_POST['keyword'];
$query= Yii::app()->db->createCommand("select * from yearselection ORDER BY id ASC")->queryAll();
$arr = array();
if(count($query) > 0) {
foreach ($query as $queryElement) {
$query2= Yii::app()->db->createCommand("select * from testtable2 where yearselection='".$queryElement['id']."' ORDER BY id ASC")->queryAll();
$arr[] = $queryElement;
}
}
# JSON-encode the response
$json_response = json_encode($arr);
// # Return the response
echo $json_response;
//exit;
}
you have write your query first like this
public function actionArchives()
{
//echo $keyword=$_POST['keyword'];
$query= Yii::app()->db->createCommand("SELECT y.year,y.id as year_id,t.id as test_id,t.name
FROM yearselection as y JOIN testtable2 as t
ON y.id=t.yearselection")->queryAll();
$arr=array();
if(count($query) > 0)
{
$arr = $query;
// Now you run loop to change indexes of array,i.e.
$count=count($arr);
for($i=0;$i<$count;$i++)
{
$year=$arr[$i]['year'];
$arr[$year]=$arr[$i];
unset($arr[$i]);
}
// Now your array has index as your year column,
}
# JSON-encode the response
$json_response = json_encode($arr);
// # Return the response
echo $json_response;
//exit;
}
Now once you write the above query you will get your data with columns year,year_id,test_id,name
Now you have take whole data in an array as you have done above in variable $arr[]
(Not touching the $query
varible to preserve query data).
Now you can do simply json_encode($arr);
Note:-
Please don't hit DB in a loop as if your loop length is suppose 100 then it will hit DB 100 times. So you can apply JOINS in those cases so that it will take data from 2 tables based on a single parameter.
Hope this solves your query.
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