PHP / MySQL where子句返回空值

debugcn 发表于 Dev

用户名

我在下面所示的代码中使用PHP / MySQL的Where子句时遇到了一个小问题：

<?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");

?>

输出为（在我单独的登录页面上登录后）：

会话ID = 4

用户名=

因此很明显，where子句存在问题。据我所知，连接参数没有问题。当我在PHPmyadmin中运行MySQL命令时，会得到“ Admin”的预期结果。我的输入正确命名。

我不知道是什么原因造成的，并且在论坛上找不到任何类似的问题。任何帮助，将不胜感激。谢谢。

更新

我改编了以下答案以编写此代码：

<?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.";

}

?>

这是由我的Flash应用程序触发的。如果我不使用会话ID变量，而仅使用例如4作为我的ID值，则可以正常工作，但是我需要为此使用会话ID。有什么想法吗？

山姆·极客

您应该使用以下代码：

//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 ;

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