我想用PHP編寫一個包含數據庫連接和異常處理的程序。如果我插入一個不正確的用戶名,那麼它應該顯示其相應的錯誤信息,如果我插入不正確的數據庫,它應該顯示其相應的錯誤信息。預期異常消息不在PHP中顯示
但在下列程序中,如果我插入了不正確的數據庫或不正確的用戶名,它只會顯示消息「無法連接到數據庫」。
<?php
$hostname = "localhost";
$username = "root1";
$password = "";
$database = "php_thenewboston";
$conn = mysqli_connect($hostname,$username,$password);
$conn_db = mysqli_select_db($conn,$database);
class ServerException extends Exception{}
class DatabaseException extends Exception{}
try{
if(!$conn){
throw new ServerException('Could not connect to server.');
}elseif(!$conn_db){
throw new DatabaseException('Could not connect to database.');
}else{
echo "Connected.";
}
}catch(ServerException $ex){
echo "Error :".$ex->getMessage();
}catch(DatabaseException $ex){
echo "Error :".$ex->getMessage();
}
?>
我是PHP的初學者。 請在下面評論任何查詢。
編輯
由於要求通過@Hatef 下面是當用戶名不正確$conn
的var_dump
,密碼正確和數據庫名稱是正確的
object(mysqli)#1 (19) {
["affected_rows"]=>
int(-1)
["client_info"]=>
string(79) "mysqlnd 5.0.12-dev - 20150407 - $Id: 241ae00989d1995ffcbbf63d579943635faf9972 $"
["client_version"]=>
int(50012)
["connect_errno"]=>
int(0)
["connect_error"]=>
NULL
["errno"]=>
int(1044)
["error"]=>
string(68) "Access denied for user ''@'localhost' to database 'php_thenewboston'"
["error_list"]=>
array(1) {
[0]=>
array(3) {
["errno"]=>
int(1044)
["sqlstate"]=>
string(5) "42000"
["error"]=>
string(68) "Access denied for user ''@'localhost' to database 'php_thenewboston'"
}
}
["field_count"]=>
int(0)
["host_info"]=>
string(20) "localhost via TCP/IP"
["info"]=>
NULL
["insert_id"]=>
int(0)
["server_info"]=>
string(21) "5.5.5-10.1.16-MariaDB"
["server_version"]=>
int(50505)
["stat"]=>
string(132) "Uptime: 1072 Threads: 1 Questions: 16 Slow queries: 0 Opens: 18 Flush tables: 1 Open tables: 11 Queries per second avg: 0.014"
["sqlstate"]=>
string(5) "00000"
["protocol_version"]=>
int(10)
["thread_id"]=>
int(9)
["warning_count"]=>
int(0)
}
下面是$conn
時的var_dump用戶名正確,密碼正確,數據庫名稱不正確。
object(mysqli)#1 (19) {
["affected_rows"]=>
int(-1)
["client_info"]=>
string(79) "mysqlnd 5.0.12-dev - 20150407 - $Id: 241ae00989d1995ffcbbf63d579943635faf9972 $"
["client_version"]=>
int(50012)
["connect_errno"]=>
int(0)
["connect_error"]=>
NULL
["errno"]=>
int(1049)
["error"]=>
string(36) "Unknown database 'php_thenewboston1'"
["error_list"]=>
array(1) {
[0]=>
array(3) {
["errno"]=>
int(1049)
["sqlstate"]=>
string(5) "42000"
["error"]=>
string(36) "Unknown database 'php_thenewboston1'"
}
}
["field_count"]=>
int(0)
["host_info"]=>
string(20) "localhost via TCP/IP"
["info"]=>
NULL
["insert_id"]=>
int(0)
["server_info"]=>
string(21) "5.5.5-10.1.16-MariaDB"
["server_version"]=>
int(50505)
["stat"]=>
string(132) "Uptime: 1417 Threads: 1 Questions: 18 Slow queries: 0 Opens: 18 Flush tables: 1 Open tables: 11 Queries per second avg: 0.012"
["sqlstate"]=>
string(5) "00000"
["protocol_version"]=>
int(10)
["thread_id"]=>
int(10)
["warning_count"]=>
int(0)
}
在你嘗試調用mysqli_select_db()和它的結果之前檢查'mysqli_select_db()'**的結果... –
將你的mysqli_connect和mysqli_select_db放入你的try塊中,它會節省一些頭痛的。 – aynber
我使用'error_reporting(E_ALL); ini_set('display_errors',1);'在頁面頂部,所以PHP會返回非常有用的消息。編輯:加'mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);'當處理DB – OldPadawan