Question

我有Xcode 4.6。我想用我的应用程序连接mysql数据库。所以我找到了这个教程

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

我是继续一步一步但我不知道我的错误在哪里。

这是我的xcode项目。请看看这个并告诉我出了什么问题。 https://www.dropbox.com/s/4pj1xj4f736l42m/mysql.zip

感谢您的帮助。

这是php代码

<?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);



?>

这是我的错误

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

好的，这里的问题是你的PHP脚本首先返回json_encode($rows)，然后是var_dump($rows)。所以，显然，它不是返回的JSON，而您的Objective-C代码需要JSON。

尝试在文件开头添加header('Content-type: application/json');，然后删除最后的var_dump。

编辑：这是新的PHP脚本

<?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);
?>

编辑2： Here与您的问题类似，请参阅答案。

Xcode连接错误

1 个答案: