How to build json for two mysql queries?

Question

this is the code i use to generate single json object

$SQL = mysql_query("SELECT * FROM `receipts` WHERE DATE(date) = '2011-08-03'");
    if(mysql_num_rows($SQL ) > 0){
        $i=0;
        $responce->success = true;
        while($SQL_RESULT = mysql_fetch_object($SQL)){
            $responce->data[$i]['reciept_no'] = $SQL_RESULT->reciept_no;
            $responce->data[$i]['time'] = $SQL_RESULT->date;
            $responce->data[$i]['user'] = $SQL_RESULT->user;
            $i++;
        }
    }
    else{
        $responce->success = false; 
        $responce->data = '';
        $responce->reason = "No Activity...";
    }

    echo json_encode($responce);

result is like

{"success":true,"data":[{"reciept_no":"2411","time":"09:33:56 AM","user":"test"},
{"reciept_no":"2412","time":"11:29:01 AM","user":" test "}]}

so there is another query which similar to this and generate exact same kind of output but from a another mysql table

i want to do is combine two results and send to javascript then decode it in javascript

like wrap first result with like table1 second result with table 2 or something

how to do that?

Sorry for the bad English

Regards

stewe · Accepted Answer · 2011-08-08 21:48:42Z

2

You could run both queries, one outputs to $response1 and the other to $response2, then you can use:

echo json_encode(array('table1'=>$response1,'table2'=>$response2));

answered Aug 8, 2011 at 21:48

stewe

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Comments

Marc B · Accepted Answer · 2011-08-08 21:19:47Z

0

JSON is just a text representation of a data structure. If you want to store two separate results in a single structure, then do

data['response #1'] = 'blah blah blah';
data['response #2'] = 'other other other';

You could store the two queries' data into a single sub-array, but then you'd need some extra data to be able to differentiate betweeen the two data sources. "did this record come from query #1? or from query #2?"

answered Aug 8, 2011 at 21:19

Marc B

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1 Comment

Gihan Lasita Over a year ago

posted data came from query1 but the query two also produce same kind of result. what im trying to do is load two extjs stores with one request

Jonas Schmid · Accepted Answer · 2011-08-09 05:27:00Z

$SQL = mysql_query("SELECT * FROM `receipts` WHERE DATE(date) = '2011-08-03'");
if(mysql_num_rows($SQL ) > 0){
    $i=0;
    while($SQL_RESULT = mysql_fetch_object($SQL)){
        $responce->data[$i]['reciept_no'] = $SQL_RESULT->reciept_no;
        $responce->data[$i]['time'] = $SQL_RESULT->date;
        $responce->data[$i]['user'] = $SQL_RESULT->user;
        $i++;
    }
}

$SQL = mysql_query("SELECT * FROM `receipts2` WHERE DATE(date) = '2011-08-03'");
if(mysql_num_rows($SQL ) > 0){
    $i=0;
    $responce->success = true;
    while($SQL_RESULT = mysql_fetch_object($SQL)){
        $responce->data[$i]['reciept_no'] = $SQL_RESULT->reciept_no;
        $responce->data[$i]['time'] = $SQL_RESULT->date;
        $responce->data[$i]['user'] = $SQL_RESULT->user;
        $i++;
    }
}
else{
    $responce->success = false; 
    $responce->data = '';
    $responce->reason = "No Activity...";
}

echo json_encode($responce);

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