所有,從一個數組
一直在努力解決這個問題有一段時間沒有任何進展匹配的結果...任何反饋是更受歡迎。
我們使用基於用戶呼叫者ID和YoB的簡單認證 - 如果他有獨特的YoB,他可以訪問該服務,如果該呼叫者ID下有另一個用戶與YoB相同:他會收到一條消息,需要。
函數的目標是返回一個匹配用戶的記錄 - 此刻它總是返回最後一條記錄。
PHP代碼:
// $response -> XML record with firstName, lastName and dob (dd/mm/yyyy)
// $yob -> YoB user has entered for authentication
function parseResponseYOB($response, $yob)
{
$duplicates = 0; // If there are users with same YoB, display message that additional info is required
if(empty($response->CallerMembers->CallerMemberDetails->dob))
{
$iterateArr = $response->CallerMembers->CallerMemberDetails;
}else{
$iterateArr[] = $response->CallerMembers->CallerMemberDetails;
}
foreach($iterateArr as $result)
{
$parseResult['firstName'] = $result->firstName;
$parseResult['lastName'] = $result->lastName;
$parseResult['yob'] = substr($result->dob, -4);
if($parseResult['yob'] == $yob)
{
$duplicates++;
}else{
continue;
}
}
// Check for duplicate YoBs
if($duplicates > 1)
{
return "Multiple members with the same YOB";
}elseif($duplicates < 1){
return "No members with the specified YOB found";
}
return $parseResult; // No duplicates, return the record of the matching user
// PROBLEM: Always the last record is returned... not the one with matching YOB?
}
}
歡迎來到Stack Overflow! – 2012-08-01 19:44:19