Xcode連接錯誤

Question

我有Xcode 4.6。 我想連接mysql數據庫與我的應用程序。 所以我發現這個教程

http://www.youtube.com/watch?v=ipppykYUzh4#at=104,http://www.youtube.com/watch?v=tvv1KlZ-594

我是繼續循序漸進，但我不知道哪裏是我的錯誤。

這是我的xcode項目。請看看這個，告訴我什麼是錯的。 https://www.dropbox.com/s/4pj1xj4f736l42m/mysql.zip

感謝您的幫助。

這是PHP代碼

<?php 
header('Content-type: application/json'); 
$DB_HostName = '127.0.0.1'; 
$DB_Name = 'test'; 
$DB_User = 'root'; 
$DB_Pass = ''; 

$con = mysql_connect($DB_HostName,$DB_User,$DB_Pass) or die(mysql_error()); 
mysql_select_db($DB_Name,$con) or die(mysql_error()); 

$sql = 'SELECT * FROM phpmysql'; 

$result = mysql_query($sql,$con) or die(mysql_error()); 

$num = mysql_numrows($result); 

mysql_close(); 

$rows =array(); 
while ($r = mysql_fetch_assoc($result)){ 

    $rows[] = $r; 
} 

echo json_encode($rows); 



?> 

THIS IS MY錯誤

2013-07-30 10:27:22.335 mysql[4095:c07] *** Assertion failure in -[UITableView dequeueReusableCellWithIdentifier:forIndexPath:], /SourceCache/UIKit_Sim/UIKit-2380.17/UITableView.m:4460 
2013-07-30 10:27:22.336 mysql[4095:c07] *** Terminating app due to uncaught exception 'NSInternalInconsistencyException', reason: 'unable to dequeue a cell with identifier Cell - must register a nib or a class for the identifier or connect a prototype cell in a storyboard' 
*** First throw call stack: 
(0x1c91012 0x10cee7e 0x1c90e78 0xb64665 0xc46c4 0x2c88 0xcd8fb 0xcd9cf 0xb61bb 0xc6b4b 0x632dd 0x10e26b0 0x228dfc0 0x228233c 0x228deaf 0x1022bd 0x4ab56 0x4966f 0x49589 0x487e4 0x4861e 0x493d9 0x4c2d2 0xf699c 0x43574 0x4376f 0x43905 0x4c917 0x1096c 0x1194b 0x22cb5 0x23beb 0x15698 0x1becdf9 0x1becad0 0x1c06bf5 0x1c06962 0x1c37bb6 0x1c36f44 0x1c36e1b 0x1117a 0x12ffc 0x243d 0x2365) 
libc++abi.dylib: terminate called throwing an exception 
(lldb) 

Answer 1

好吧，這裏的問題是，你的PHP腳本首先返回json_encode($rows)，然後var_dump($rows)。所以，顯然，這不是返回的JSON，而您的Objective-C代碼需要JSON。

嘗試在文件的開頭添加一個header('Content-type: application/json');，並在最後刪除var_dump。

編輯：這是新的PHP腳本

<?php 

header('Content-type: application/json'); // Specify that the result of your script is a JSON 

$DB_HostName = '127.0.0.1'; 
$DB_Name = 'test'; 
$DB_User = 'root'; 
$DB_Pass = ''; 

$con = mysql_connect($DB_HostName,$DB_User,$DB_Pass) or die(mysql_error()); 
mysql_select_db($DB_Name,$con) or die(mysql_error()); 

$sql = 'SELECT * FROM phpmysql'; 

$result = mysql_query($sql,$con) or die(mysql_error()); 

$num = mysql_numrows($result); 

mysql_close(); 

$rows =array(); 
while ($r = mysql_fetch_assoc($result)){ 

    $rows[] = $r; 
} 

echo json_encode($rows); 
?> 

編輯2：Here是一個類似的問題，以你的，一看便知。