更新用戶信息使用SQL查詢

-1

更新代碼，信息輸入及提交按鈕點擊運行此代碼後，它可以追溯到帳戶頁面，但犯規更新數據庫：更新用戶信息使用SQL查詢

<font face="ClearSans-Thin"> 
<font color="lightgray"> 

<?php 

include 'editaccount.php'; 
include 'connection.php'; 
?> 

<center> 
<?php 
if (isset($_POST['uregsubmit'])) { 
    $firstname = $_POST['ufirstname']; 
    $lastname = $_POST['ulastname']; 
    $email = $_POST['uemail']; 
    $dob = $_POST['udob']; 
    $user = $_POST['uregisterusername']; 

    $pass = $_POST['uregisterpassword']; 

    } 

    //the query 
$query = "UPDATE Users SET FirstName='$firstname', LastName='$lastname' WHERE Username='$user'"; 
//execute the query 
$result = mysqli_query($connection, $query) 
or die("Error: ".mysqli_error($connection)); 
//check and see if any data returned 





?> 
</center>

來源

2015-01-09 mattttttttt

我不認爲逗號應該是前'WHERE' –

我沒有它之前不知道之前額外的逗號爲什麼現在在那裏。但我刪除了，當我點擊提交它不會更新信息。 – mattttttttt

寫SQL查詢中的if語句

<?php 
if (isset($_POST['uregsubmit'])) { 
    $firstname = $_POST['ufirstname']; 
    $lastname = $_POST['ulastname']; 
    $email = $_POST['uemail']; 
    $dob = $_POST['udob']; 
    $user = $_POST['uregisterusername']; 
    $pass = $_POST['uregisterpassword']; 

    //the query 
    $query = "UPDATE Users SET FirstName='$firstname', LastName='$lastname' WHERE Username='$user'"; 
    //execute the query 
    $result = mysqli_query($connection, $query) 
    or die("Error: ".mysqli_error($connection)); 
    //check and see if any data returned 
} 
?>

來源

2015-01-09 04:20:45

仍然不會改變信息！ :( – mattttttttt

裏面如果語句做'print_r（$ _ POST）'並檢查你是否得到所有數據 –

我得到這個輸出：Array（[ufirstname] => test [ulastname] => test [uemail] => [udob ] => [uregisterusername] => [uregisterpassword] => [uregsubmit] =>提交） – mattttttttt

你有一個WHERE

來源

2015-01-09 02:24:33

很酷，你可以關閉它 –

，但它不能解決我的問題！ :) – mattttttttt

任何error_log ?? –

更新用戶信息使用SQL查詢

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