也许这只是我遇到这样一个恼人的"功能":
从Xcode 6.0.1升级到Xcode 6.1后,情况发生了变化.Xcode 6.1永远索引项目或编译源文件.该项目不是一个巨大的项目.它只包含工作区中的一堆Swift文件和AWS SDK 2.0 Cocoapods.我不认为它应该阻止整个索引和编译顺利.我尝试了一些aws-sdk-ios-samples,只是为了看看Xcode 6.1如何在它们上工作,它最终在同一个永远等待.
到目前为止我尝试了哪些解决方案:
不幸的是,他们都没有工作.
PS也许我应该尝试重新创建项目?我的电脑设置:MacBook Pro(Retina,13英寸,2014年中),内存8 GB 1600 MHz DDR3,优胜美地.(我认为这足以运行这个小项目.)
升级到Mac OS Yosemite(10.10)后,我的Eclipse将不再启动.相反,我得到了
JVM共享库/Library/Java/JavaVirtualMachines/jdk1.8.0_05.jdk/Contents/Home/bin/../jre/lib/server/libjvm.dylib"不包含JNI_CreateJavaVM符号.
我检查java -version并获得
java version "1.8.0_05"
Java(TM) SE Runtime Environment (build 1.8.0_05-b13)
Java HotSpot(TM) 64-Bit Server VM (build 25.5-b02, mixed mode)
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我搜索并发现一个类似的问题在OSX上重新安装Java 7 JDK.它建议重新安装Java.我不确定我的问题是否与他的问题属于同一类别.那么,重新安装是解决问题的好方法吗?还有其他建议吗?谢谢!
叹了口气,整个下午我一直在努力......这是我的噩梦:
我正在尝试使用Apple制作的KeychainItemWrapper.但我将其Objective-C代码"翻译"为Swift:
import Foundation
import Security
class MyKeychainItemWrapper: NSObject {
var keychainItemData: NSMutableDictionary?
var genericPasswordQuery: NSMutableDictionary = NSMutableDictionary()
init(identifier: String, accessGroup: String?) {
super.init()
// Begin Keychain search setup. The genericPasswordQuery leverages the special user
// defined attribute kSecAttrGeneric to distinguish itself between other generic Keychain
// items which may be included by the same application.
genericPasswordQuery.setObject(kSecClassGenericPassword, forKey: kSecClass)
genericPasswordQuery.setObject(identifier, forKey: kSecAttrGeneric)
// The keychain access group attribute determines if this item can be shared
// amongst multiple apps whose …Run Code Online (Sandbox Code Playgroud) 刚才,我更新了我的AWS SDK pod,我后悔了.当pod正在更新时我第一次看到这个:
dyld:警告,LC_RPATH @executable_path /../../../../../../../ SharedApprameworks in /Applications/Xcode.app/Contents/SharedFrameworks/DVTFoundation.framework/Versions/A/ ../../..//DVTInstrumentsFoundation.framework/Versions/A/DVTInstrumentsFoundation在受限制的程序中被忽略,因为@executable_path dyld:warning,LC_RPATH @executable_path /../ lib in /Applications/Xcode.app/Contents/由于@executable_path,在受限程序中忽略框架/ IDEFoundation.framework/Versions/A /../../../../ Developer/Toolchains/XcodeDefault.xctoolchain/usr/lib/libclang.dylib
dyld:警告,LC_RPATH @executable_path /../ lib在/Applications/Xcode.app/Contents/Frameworks/IDEFoundation.framework/Versions/A/../../../../Developer/Toolchains/XcodeDefault.由于@executable_path,在受限程序中忽略xctoolchain/usr/lib/libLTO.dylib
dyld:警告,LC_RPATH @executable_path /../框架/Applications/Xcode.app/Contents/PlugIns/Xcode3Core.ideplugin/Contents/MacOS/Xcode3Core由于@executable_path而在受限程序中被忽略
然后当我打开我的项目并构建时,我得到了这么多构建失败,抱怨同样的事情XXX类的重复接口定义:
我尝试了很多解决方案,重新安装AWS SDK pod,重新安装cocoapods,将cocoapods降级到0.38.2,将AWS SDK pod降级到2.2.6,但没有任何帮助.如果我在Objective-C桥接头中注释掉导入,我没有这些错误,但Xcode只是抱怨没有找到代码中使用的某些类.但我不认为评论它们是正确的解决方案.

好吧,我搜索了https://github.com/CocoaPods/CocoaPods/issues/4302,它似乎可能是cocoapods的问题?
有人对此有任何想法吗?谢谢.
我从iOS教程中学到了这一点,但它不起作用
func imagePickerController(picker: UIImagePickerController!, didFinishPickingMediaWithInfo info: NSDictionary!) {
var mediaType = info.objectForKey(UIImagePickerControllerMediaType) as String
var originalImage, editedImage, imageToUse: UIImage
// Handle a still image picked from a photo album
if (CFStringCompare(CFStringRef(mediaType), kUTTypeImage, 0) == CFComparisonResult.CompareEqualTo) {
editedImage = info.objectForKey(UIImagePickerControllerEditedImage) as UIImage
originalImage = info.objectForKey(UIImagePickerControllerOriginalImage) as UIImage
if (editedImage) {
imageToUse = editedImage
} else {
imageToUse = originalImage
}
// Do something with imageToUse
}
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它经常提醒我
CFStringRef不能用'@lvalue String'构造
所以我尝试了这个:
// Handle a still image picked from a photo album
var …Run Code Online (Sandbox Code Playgroud) 我一直在寻找解决方案一整天,我几乎已经完成了Stackoverflow上的所有答案,仍然无法解决我的问题.所以,我必须提出这个问题.
在models.py中,我有:
from django.db import models
import datetime
from django.utils import timezone
class Cap(models.Model):
title = models.CharField(max_length=50)
user_id = models.CharField(max_length=100)
dest_id = models.CharField(max_length=100)
cap_id = models.CharField(max_length=500)
text_part = models.FileField(upload_to='texts/%Y/%m/%d')
time_to_live = models.IntegerField()
pub_date = models.DateField()
def __unicode__(self):
return self.title
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然后在forms.py中,我有:
from django.forms import ModelForm
from latercap.models import Capsule
class CapForm(ModelForm):
class Meta:
model = Cap
fields = '__all__'
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然后在views.py中,我有:
from django.shortcuts import render
from django.http import HttpResponseRedirect
from myapp.forms import CapForm
def upload(request):
# Handle file upload
if request.method == …Run Code Online (Sandbox Code Playgroud) 我正在开发一个使用AWS作为其后端的iOS应用程序.由于我希望用户能够在全球范围内使用,因此我需要切换应用下载照片的位置,以提高延迟性能.我在EUWest1中创建了一个S3存储桶,在APNorthEast1中创建了另一个存储桶.正如Yosuke在此处所述(https://forums.aws.amazon.com/message.jspa?messageID=586581#586581),您可以将AWSServiceManager.defaultServiceManager()的默认配置从一个地区更改为另一个地区,以防止错误.
但是,即使我按照他的提示,我仍然有相同的错误结果.错误是:
Error Domain = com.amazonaws.AWSS3ErrorDomain Code = 0"无法完成操作.(com.amazonaws.AWSS3ErrorDomain error 0.)"UserInfo = 0x14fb7720 {RequestId = D708XXX8E1EDBXXX,Endpoint = bucket-ap-northeast-1.s3 -ap-northeast-1.amazonaws.com,Message =您尝试访问的存储区必须使用指定的端点进行寻址.请将以后的所有请求发送到此端点.,HostId = xxxY31S3yFhpssQe1EsfipWWLw =,Code = PermanentRedirect,Bucket = bucket-ap-northeast-1}
这是我的代码:
var credentialsProvider = AWSCognitoCredentialsProvider.credentialsWithRegionType(kCognitoRegionType, accountId: kAWSAccountID, identityPoolId: kCognitoPoolID, unauthRoleArn: kCognitoRoleUnauth, authRoleArn: kCognitoRoleAuth)
var serviceConfiguration = AWSServiceConfiguration(region: AWSRegionType.APNortheast1, credentialsProvider: credentialsProvider)
AWSServiceManager.defaultServiceManager().setDefaultServiceConfiguration(serviceConfiguration)
var transferManager = AWSS3TransferManager.defaultS3TransferManager()
transferManager.download(downloadRequest)
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欢迎任何建议!谢谢!
我知道我们可以通过向AWSCognitoCredentialProvider提供"登录"来将未经身份验证的用户更改为经过身份验证.我已经实现了FB登录,效果很好.
现在问题是,如果用户想要在没有FB凭证的情况下再次登录该怎么办?就像他之前将FB帐户链接到应用程序之前的方式一样.因此,我需要清除与其FB凭证关联的凭据提供程序,并将其返回给未经身份验证的角色.
我试过的是使用清除缓存凭证方法,但是一旦我初始化没有登录的新凭据提供程序,就会发生错误.错误是关于刷新失败,openIDGet失败.
有人可以帮忙解决这个问题吗?
谢谢!
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