2013-05-14 84 views
-1

這是我的我的主要代碼PHP MySQL查詢不能執行

if (isset($_POST)) 
{ 
    include('connection.php'); 

    $query = "call pgetuseridfromemail('" . $_SESSION['email']. "')"; 
    $result = mysqli_query($conn, $query); 
    $row = mysqli_fetch_assoc($result); 
    $userid = $row['idusers']; 

    $query = "SELECT * FROM USERS"; 
    $result = mysqli_query($conn, $query) or die("query no executed"); 
    while($row = mysqli_fetch_assoc($result)) 
    { 
     echo $row['USERNAME']; 
    } 
} 

這是我的數據庫連接

session_start(); 

$user = "root"; 
$pass = ""; 
$hostname = "localhost"; 
$database = "faana final"; 
//conection to database 
$conn = mysqli_connect($hostname, $user, $pass, $database) or die(mysql_error()); 

這是我的存儲過程

BEGIN 
     select idusers from users where email = email1; 
END 

我無法執行查詢,因爲它迴響query not executed

+2

請不要輸入全部使用大寫字母。它從未*適當。 – meagar

+2

你在哪裏定義$ query1? – redburn

+0

http://php.net/manual/en/mysqli.error.php – Kevin

回答

1

表名稱是區分在MySQL敏感,取決於Operating System

$query = "SELECT * FROM USERS"; 

select idusers from users where email = email1; 

你可以打印MySQL錯誤像下面幾行:

$result = mysqli_query($conn, $query1) or die(mysql_error()); //die("query no executed"); 
+0

mysql_error返回空字符串 –

+0

mysql_query()期望參數1是字符串, A:\ wamp \ www \ AddFormData.php在第22行 –

+0

你混淆了mysql_query()和mysqli_query()。 – redburn

0

在代碼行$query = "SELECT * FROM USERS";中使用$query1而不是$query,反之亦然。

+0

我更正了$ query1到$ query1這是一個輕微的錯誤仍然無法執行查詢,當我評論上面查詢下一個查詢成功運行 –