How create json encode by one object with mutiple array in php? - php

Question

I have two table names are channel and department.
$result = mysql_query("SELECT * FROM channels");
while($json = mysql_fetch_assoc($result)){
$fetch = mysql_query("SELECT * FROM departments where channelid='".$json['id']."'");
$json2 = array();
while ($row = mysql_fetch_array($fetch)){
$json2[] = array('departmentname' => $row["departmentname"],'departmentid' => $row["did"]
);
}
$json['department'] = $json2;
echo json_encode($json);
}
Ouput:
{
"id": "1",
"channelname": "con",
"channelurl": "http:\/\/xxx.net\/YY\/",
"department": [
{
"departmentname": "xxx Travel",
"departmentid": "1"
},
{
"departmentname": "xxxx Virtual Assist",
"departmentid": "2"
},
"departmentname": "xxx Premier",
"departmentid": "3"
},
{
"departmentname": "xxxx Events",
"departmentid": "4"
}
]
}{
"id": "2",
"channelname": "Slim",
"channelurl": "http:\/\/xxxxx.net\/slim\/",
"department": [
{
"departmentname": "Virtualvideo",
"departmentid": "5"
}
]
}
Expected Result:
[
{
"id": "1",
"channelname": "xxxxx",
"channelurl": "http://XXXXX.net/yyy/",
"department": [
{
"departmentname": "XXX Travel",
"departmentid": "1"
},
{
"departmentname": "XXXX Virtual Assist",
"departmentid": "2"
},
{
"departmentname": "XXXX Premier",
"departmentid": "3"
},
{
"departmentname": "XXXX Events",
"departmentid": "4"
}
]
},
{
"id": "2",
"channelname": "Slim",
"channelurl": "http://XXXXXX.net/slim/",
"department": [
{
"departmentname": "Virtual video",
"departmentid": "5"
}
]
}
]

Answer 1

try this:
$json_channel = array();
$json_final = array();
$result = mysql_query("SELECT * FROM channels");
while($json = mysql_fetch_assoc($result))
{
$fetch = mysql_query("SELECT * FROM departments where channelid='".$json['id']."'");
$json_dept = array();
while ($row = mysql_fetch_array($fetch))
{
$json_dept[] = array('departmentname' => $row["departmentname"],'departmentid' => $row["did"]);
}
$json_channel = array('id' => $json["ch_id"], 'channelname' =>$json["channelname"], 'department' => $json_dept);
$json_final[] = $json_channel;
}
echo json_encode($json_final);

Answer 2

$output = array();
$result = mysql_query("SELECT * FROM channels");
while($json = mysql_fetch_assoc($result)){
$fetch = mysql_query("SELECT * FROM departments where channelid='".$json['id']."'");
$json2 = array();
while ($row = mysql_fetch_array($fetch)){
$json2[] = array('departmentname' => $row["departmentname"],'departmentid' => $row["did"]
);
}
$json['department'] = $json2;
$output[] = $json;
}
echo json_encode($output);