HttpPost 500錯誤。（SimpleCode）

Question

任何人都可以告訴我這段代碼有什麼問題嗎？ 基本上我試圖使用Android和PHP輸入數據到表中。 但我甚至不能得到PHP方面的工作，因爲我得到HTTP錯誤500，有沒有人看到這裏的錯誤？ 謝謝！

>

<?php 
/* Connect to MySQL and select the database. */ 
$con = mysql_connect($DB_SERVER,$DB_USERNAME,$DB_PASSWORD); 

mysql_select_db ($DB_DATABASE, $con) or die (「Database Selection Failed」 ); 




//$username = $_REQUEST['username']; 
//$password = $_REQUEST['password']; 
//$givenname = $_REQUEST['givenname']; 
//$email = $_REQUEST['email']; 
//$phonenumber = $_REQUEST['phonenumber']; 

$username = isset($_POST[‘username’]) ? $_POST[‘username’] : 」; 
$password = isset($_POST[‘password’]) ? $_POST[‘password’] : 」; 
$givenname = isset($_POST[‘givenname’]) ? $_POST[‘givenname’] : 」; 
$email = isset($_POST[‘email’]) ? $_POST[‘email’] : 」; 
$phonenumber = isset($_POST[‘phonenumber’]) ? $_POST[‘phonenumber’] : 」; 



$flag['code']=0; 
if ($result = mysql_query (「INSERT INTO demo VALUES(‘$username’, ‘$password’, `$givenname`, `$email`, ‘$phonenumber’)」, $con)) { 
$flag [‘code’] = 1; 
echo 「hi」; 
} 

echo (json_encode ($flag)) ; 
mysql_close ($con); 

?> 

Answer 1

是的。修復那些「智能」引號：

<?php 
/* Connect to MySQL and select the database. */ 
$con = mysql_connect($DB_SERVER,$DB_USERNAME,$DB_PASSWORD); 

mysql_select_db ($DB_DATABASE, $con) or die ("Database Selection Failed" ); 




//$username = $_REQUEST['username']; 
//$password = $_REQUEST['password']; 
//$givenname = $_REQUEST['givenname']; 
//$email = $_REQUEST['email']; 
//$phonenumber = $_REQUEST['phonenumber']; 

$username = isset($_POST['username']) ? $_POST['username'] : ''; 
$password = isset($_POST['password']) ? $_POST['password'] : ''; 
$givenname = isset($_POST['givenname']) ? $_POST['givenname'] : ''; 
$email = isset($_POST['email']) ? $_POST['email'] : ''; 
$phonenumber = isset($_POST['phonenumber']) ? $_POST['phonenumber'] : ''; 



$flag['code']=0; 
if ($result = mysql_query ("INSERT INTO demo VALUES('$username', '$password', `$givenname`, `$email`, '$phonenumber')", $con)) { 
$flag ['code'] = 1; 
echo "hi"; 
} 

echo (json_encode ($flag)) ; 
mysql_close ($con); 

?>