[英]How to properly format PDO::FETCH_NUM into JSON results
我正在返回這樣的JSON文檔:
ug_gpid: [[4], [2], [3]]
這不是預期的格式,我想要這樣:
ug_gpid: [4, 2, 3]
這是我的代碼:
$user = json_decode($resp, true);
$sql = "SELECT ug_gpid FROM user_group WHERE ug_usid =:id AND ug_status = 1";
try {
$db = getConnection();
$stmt = $db->prepare($sql);
foreach ($user as $item => $value) {
$newUser[$item] = $value;
$stmt->bindParam("id", $value["us_id"]);
$stmt->execute();
$ugroup = $stmt->fetchAll(PDO::FETCH_NUM);
$newUser[$item]['ug_gpid'] = $ugroup;
}
$db = null;
$response = $app->response();
$response->header('Content-Type', 'application/json');
echo json_encode($newUser, JSON_NUMERIC_CHECK);
} catch(PDOException $e) {
$error = array("queryGroupError"=> array("text"=>$e->getMessage()));
echo json_encode($error);
}
我將結果存儲在$ugroup
。 如何獲得所需的輸出?
問題是PDOStatement::fetchAll
返回一個數組數組。 您要構建單個值的平面數組。
我會改用這個...
$stmt = $db->prepare($sql);
$stmt->bindParam(':id', $us_id);
foreach ($user as $item => $value) {
$newUser[$item] = $value;
$us_id = $value['us_id']; // this is for the bind
$stmt->execute();
while ($gpid = $stmt->fetchColumn()) {
$newUser[$item]['ug_gpid'][] = $gpid;
}
}
它無法按預期運行的原因已經得到解答; 但是,使用單個查詢會更有效:
$sql = sprintf(
"SELECT ug_usid, GROUP_CONCAT(ug_gpid)
FROM user_group
WHERE ug_usid IN (%s) AND ug_status = 1
GROUP BY ug_usid",
str_pad('?', count($user) * 2 - 1, ',?');
);
// ...
// run query and build search result map
$stmt = $db->prepare($sql);
$stmt->execute(array_column($user, 'us_id'));
$groups = $stmt->fetchAll(PDO::FETCH_KEY_PAIR);
// extend user information with search results
array_walk($user, function(&$u) use ($groups) {
if (isset($groups[$user['us_id']])) {
$u['ug_gpid'] = explode(',', $groups[$user['us_id']]);
} else {
$u['ug_gpid'] = [];
}
});
echo json_encode($user, JSON_NUMERIC_CHECK);
請注意,它假定您使用MySQL作為數據庫引擎,因為它使用GROUP_CONCAT
,它是SQL的MySQL擴展。
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