小编Fre*_*nck的帖子

当我正在构建我的Xcode 4应用程序时,我收到此警告:

ld: warning: directory not found for option '-L/Users/frenck/Downloads/apz/../../../Downloads/Google Analytics SDK/Library'
ld: warning: directory not found for option '-L/Users/frenck/Downloads/apz/../Google Analytics SDK/Library'

但我的应用程序中没有Google Analytics,我删除了所有内容,如何删除错误？归档应用程序会给我错误:

clang: error: no such file or directory: 'armv6'
Command /Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/bin/clang failed with exit code 1

我尝试了很多东西,但是当我尝试制作.IPA文件时,我仍然会收到此错误

Tweetbot和Clear show在应用程序的第一次启动时,一个小教程屏幕应用程序如何工作.带有小教程的屏幕仅在应用程序首次启动时弹出(1次)

如何以及我能做出类似的事情？任何人都可以把我推向正确的方向吗？

查看我的意思是:

我正在尝试为我的iOS(iPhone)应用程序制作存档(ipa),但我收到一个奇怪的错误:

Apple LLVM Compiler 3.1 error clang: error: no such file or directory: 'armv6'

Command /Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/bin/clang failed with exit code 1

有谁知道我怎么解决这个问题？

http://i.imgur.com/PKOHx.png

ProcessPCH /Users/wazapi/Library/Developer/Xcode/DerivedData/myapp-hjzccpnbaymgcufalpjtfeuqvjvu/Build/Intermediates/ArchiveIntermediates/myapp/PrecompiledHeaders/myapp-Prefix-cgigtilzpjcadpbgpxbabnqczfxd/myapp-Prefix.pch.pth

     myapp/myapp-Prefix.pch normal armv7 objective-c
     com.apple.compilers.llvm.clang.1_0.compiler
             cd /Users/wazapi/Downloads/myapp
             setenv LANG en_US.US-ASCII
             setenv PATH "/Applications/Xcode.app/Contents/Developer/Platforms/iPhoneOS.platform/Developer/usr/bin:/Applications/Xcode.app/Contents/Developer/usr/bin:/usr/bin:/bin:/usr/sbin:/sbin"
             /Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/bin/clang
     -x objective-c-header -arch armv7 -fmessage-length=0 -std=gnu99 -fobjc-arc -Wno-trigraphs -fpascal-strings -Os -Wno-missing-field-initializers -Wmissing-prototypes -Wreturn-type -Wno-implicit-atomic-properties -Wformat -Wno-missing-braces -Wparentheses -Wswitch -Wno-unused-function -Wno-unused-label -Wno-unused-parameter -Wunused-variable -Wunused-value -Wno-uninitialized -Wno-unknown-pragmas -Wno-shadow -Wno-four-char-constants -Wno-sign-compare -Wno-shorten-64-to-32 -Wpointer-sign -Wno-newline-eof -Wno-selector -Wno-strict-selector-match -Wno-undeclared-selector -Wno-deprecated-implementations -isysroot /Applications/Xcode.app/Contents/Developer/Platforms/iPhoneOS.platform/Developer/SDKs/iPhoneOS5.0.sdk
     -Wprotocol -Wdeprecated-declarations -g -fvisibility=hidden …

我是SQLite的新手.我有一个UITableview包含不同的日子.(星期一至星期日).当我点击例如星期一时,另一个viewcontroller包含一个UITableview.在同一个viewcontroller我有一个UIButton当我点击它我可以添加数据到我的SQLite数据库[ A ],我插入名称和星期几(星期几是在这个例子'星期一'这是因为我点击星期一视图控制器).

当我插入一个名字时,它出现在我的tableview中.但是当我回到我的第一个视图控制器时,我在周三点击了例如我添加的数据也出现在那里.

所以我的问题是; 如何显示我在星期一插入的名称,仅在星期一的tableview中,而不是其他日子(tableviews)

更多信息:

因此,当用户在'星期一'添加名称时,我将带有添加名称的dayoftheweek发送到SQLite数据库,当用户在星期三添加名称时,我发送'dayoftheweek'星期三等.

数据库咖啡看起来像=

CoffeeName    | dayoftheweek
-------------------------
Hello world   | Monday
Hello Planet  | Wednesday
Hello Animal  | Monday
Hello STOVW   | Friday

[一个] const char *sql = "insert into Coffee(CoffeeName, dayoftheweek) Values(?, ?)";

我需要检查一天(例如)星期一是否与dayoftheweek(星期一)相同,然后显示包含'dayoftheweek monday'的项目

我的sqlite看起来像:

+ (void) getInitialDataToDisplay:(NSString *)dbPath {


    if (sqlite3_open([dbPath UTF8String], &database) == SQLITE_OK) {


        const char *sql = "select coffeeID, coffeeName from coffee";
        sqlite3_stmt *selectstmt;
        if(sqlite3_prepare_v2(database, sql, -1, &selectstmt, NULL) == SQLITE_OK) {

            while(sqlite3_step(selectstmt) == …

Viewdeck遇到了一个小问题.

我想用导航栏推送我的导航视图.我现在正在做(下面的代码),但它没有导航栏.

    [self.viewDeckController closeLeftViewBouncing:^(IIViewDeckController *controller)
     {  
    DataViewController *DataController = [[DataViewController
                                           alloc]  initWithNibName:@"DataViewController" bundle:nil];
    DataController.modalPresentationStyle = UIModalPresentationFormSheet;


    DataController.ID = @"hello";

     self.viewDeckController.centerController = DataController;
}];

还尝试过:

        UIViewController *viewController = [[DataViewController alloc] init];
        UINavigationController *navigationController = [[UINavigationController alloc] initWithRootViewController:viewController];
        navigationController.modalPresentationStyle = UIModalPresentationFormSheet;


            DataViewController *DataController = [[DataViewController
                                                   alloc]  initWithNibName:@"DataViewController" bundle:nil];
            DataController.modalPresentationStyle = UIModalPresentationFormSheet;


            DataController.ID = @"hello";

             self.viewDeckController.centerController = DataController;

很难理解(json_encode)我正在使用代码:

<?php

    $query = mysql_query("SELECT * FROM messages ORDER BY ID");
    while($fetch = mysql_fetch_assoc($query))
    {
    $titel = $fetch[title];
    $post = array('items' => array( 0 => array('title' => "$title", 'description' => "$title")));


    echo json_encode($post);

}
?>

输出:

  {"items":[{"title":"title","description":"title"}]}
{"items":[{"title":"title","description":"title"}]}

但我想要一个输出,如:

{

"items": [

{
"title":"title",
"description":"title"
},
{
"title":"title",
"description":"title"
},
{
"title":"title",
"description":"title"
}

]

}

有人可以帮助我得到像上面的代码一样的输出？

抱歉,

我现在感觉很蠢,但我怎么能以正确的方式做到这一点:

'.str_replace(' ', '+','$fetch['title']).'

没有'它很容易但我不知道如何在这个代码行上放置str_replace

'.$fetch['title'].'

我试过了:

 '.str_replace(' ', '+','$fetch['title']).'

//

str_replace(' ', '+',''.$fetch['title']).'