2016-04-14 77 views
-2

我試圖做問卷調查,但它只是不會連接我的數據庫,這很奇怪,因爲在其他PHP幾乎相同的連接工作。你能幫忙嗎?不能連接數據庫

<?php 
if (isset($_POST['pohlavi'])) { 
    if (!isset($_POST['platforma']) || !isset($_POST['ano']) || !isset($_POST['nazor']) || $_POST['platforma'] = "" || $_POST['nazor'] = "" || $_POST['ano'] = "") { 
     die("Nastala chyba, musíte vyplnit všechny pole! Dotazník nebyl odeslán."); 
    } else { 
     $conn = new mysqli('localhost', 'adress', 'password', 'czseznamka'); 
     $conn->query("INSERT INTO dotaznik (pohlavi, platforma, ano, nazor) VALUES ('" . $_POST['pohlavi'] . "', '" . $_POST['platforma'] . "', '" . $_POST['ano'] . "' , '" . $_POST['nazor'] . "');"); 
     echo '<span style="text-align:center;color:green;font-size:30px;"> Dotazník odeslán</span>'; 
    } 
} 
?> 
+0

是什麼的var_dump'輸出($康恩)' – uzaif

+0

你需要告訴mysqli的拋出異常: –

+0

什麼輸出現在越來越 –

回答

0

嘗試用以下啓動錯誤

if (isset($_POST['pohlavi'])) { 
    if (!isset($_POST['platforma']) || !isset($_POST['ano']) || !isset($_POST['nazor']) || $_POST['platforma'] = "" || $_POST['nazor'] = "" || $_POST['ano'] = "") { 
     mysqli_report(MYSQLI_REPORT_STRICT); 

    try { 
     $connection = new mysqli('localhost', 'root', '', 'dbname') ; 
     $connection->query("INSERT INTO dotaznik (pohlavi, platforma, ano, nazor) VALUES ('" . $_POST['pohlavi'] . "', '" . $_POST['platforma'] . "', '" . $_POST['ano'] . "' , '" . $_POST['nazor'] . "');"); 

     } catch (Exception $e) { 
     echo "Service unavailable"; 
     echo "message: " . $e->message; 
     exit; 
     } 
    } 
} 
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希望沒有人足夠愚蠢的離開那裏沒有任何密碼的root超級用戶帳戶。你不知道有人愚蠢嗎? – RiggsFolly