How to build json for two mysql queries? - php

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

Answer 1

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));

Answer 2

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?"

Answer 3

$SQL = mysql_query("SELECT * FROM `receipts` WHERE DATE(date) = '2011-08-03'");
if(mysql_num_rows($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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