PHP/where子句使用的代碼的PHP/MySQL Where子句返回一個空值

Question

我有一個小問題如下所示的MySQL：

<?php 
$con=mysqli_connect("host","username","password","database"); 
//Note that I have replaced my parametres just for this question 
// Check connection 
if (mysqli_connect_errno()) 
{ 
echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
} 
//Request Session ID 
$id = $_SESSION['uid']; 
echo "Session ID = ' . $id; 
echo "<br>"; 
//Request Username 
echo "Username = " . mysqli_query($con,"SELECT username FROM 'users' WHERE id = 4"); 

?> 

的這個輸出是（在我的單獨的登錄後登錄頁面）：

會話ID = 4

用戶名=

所以很明顯，存在的問題與where子句。 據我所知，連接參數沒有問題。 當我在PHPmyadmin中運行MySQL命令時，我得到了'Admin'的預期結果。 我的輸入被正確命名。

我不知道是什麼原因造成的，我在論壇上找不到任何類似的問題。任何幫助，將不勝感激。謝謝。

UPDATE

我已經適應了下面的答案，使這個代碼：

<?php 
// Only run this script if the sendRequest is from my flash application 
if ($_POST['sendRequest'] == "parse") { 
//conection: 
$con=mysqli_connect("mysqlXX.000webhost.com","a4935911_***","***","a4935911_***"); 
$id = $_SESSION['uid']; 
$getuname = mysqli_fetch_assoc(mysqli_query($con, "SELECT username FROM users WHERE id = $id")); 
$uname = $getuname ['username']; 

// Print 1 var to flash 
print "var1=The username of this user is $uname."; 

} 

?> 

這是我的Flash應用程序觸發。如果我不使用會話ID變量，並且只使用例如4作爲我的身份證號碼值，但我需要使用此會話ID。任何想法是什麼與此？

Answer 1

你應該使用下面的代碼： -

//conection: 
$con=mysqli_connect("host","username","password","database"); 

if (mysqli_connect_errno()) 
{ 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
} 

$getuname = mysqli_fetch_assoc(mysqli_query($con, "SELECT username FROM users WHERE id = 4")); 
$username = $getuname ['username']; 

//Request Username 
echo "Username = " .$username ; 

Answer 2

你不應該用普通的單引號括起表名（'）;使用反引號（或者，如果您不使用保留字，則不使用）。所以：

SELECT username FROM `users` WHERE id = 4 

或

SELECT username FROM users WHERE id = 4 

此外，mysqli_query並不在現場返回的結果，而是一個mysqli_result對象。您必須使用獲取命令來獲取結果，例如

$result = mysqli_query($con,"SELECT username FROM 'users' WHERE id = 4"); 
$row = mysqli_fetch_assoc($result); 
echo 'Username = ' . $row['username'];