小编use*_*173的帖子

我只有一个名为" Test"的模式.它显示在数据库名称旁边的仪表板中(在下图中为黄色):

我通过My​​SQLWorkbench向我的实例添加了另一个Schema,让我们说它叫做" NewDB".

如何在我的仪表板中将数据库名称更改为我的新架构" NewDB"？

我正在尝试CouchbaseKit在Swift中创建一个名为(Xcode中的新目标)的自定义框架.在我的内部CouchbaseKit,我需要访问CouchBaseLite Framework完全用Obj-C编写的内容.我正在使用Cocoapods管理CouchBaseLite和其他几个框架.下面是我的podfile.

Podfile

# Uncomment this line to define a global platform for your project
link_with ['CouchbaseKit']
# platform :ios, '8.0'

use_frameworks!
target 'CouchbaseDB' do
link_with ['CouchbaseKit']

pod 'couchbase-lite-ios'
pod 'SwiftyJSON', '~> 2.2.0'
pod 'Alamofire', '~> 1.2'
pod 'XCGLogger', '~> 2.0'
end

target 'CouchbaseDBTests' do
end

target 'CouchbaseKit' do
end

target 'CouchbaseKitTests' do
end

项目内的Pod:

对于我的TARGETS,我在Build Phases中有以下设置.

定义模块 是

允许框架模块中的非模块化包含是

问题: 当我尝试访问CouchbaseKit(我的自定义框架)中的CouchbaseLite框架时,出现错误,"没有这样的模块'CouchbaseLite'不存在.

尝试:

1. 由于项目是在Swift中,我创建了一个Objective-C文件并点击是" 你想配置一个Objective-C桥接头吗？ …

我创建了一个动态可可框架,我希望在我的应用程序中使用它.

在为实际设备构建框架之后,我将其带到了应用程序.我只能在该设备上运行该应用程序.

当我尝试在模拟器上运行它时,框架文件无法识别.我收到错误消息: 'ViewController' is unavailable: cannot find swift for declaration for this class

我尝试构建iPhone 6模拟器的框架并在iPhone 6模拟器上运行应用程序,但同样的问题仍然存在.

如何创建/构建一个我可以带入任何应用程序并能够在模拟器和设备上使用它的框架？

下面是我在 Hooks 中使用的代码来更新updatedAt两个对象的列：

hooks: {
                afterUpdate: (group, options, callback) => {
                    console.log("groudId " + groupId + " options " + options)
                },
                afterCreate: (member, options, callback) => {
                    return new Promise((resolve, reject) => {
                        sequelize.models.Group.findOne({
                            where: {
                                id: member.group_id
                            }
                        }).then((group) => {
                            if (group) {
                                var date = new Date();
                                console.log("BEFORE group.updatedAt " + group.updatedAt)
                                group.dataValues.updatedAt = new Date()
                                console.log("CHANGED group.updatedAt " + group.updatedAt)
                                group.save().then((Group) => {
                                    if (Group) {
                                        console.log("UPDATED Group.updatedAt " + Group.updatedAt)
                                        console.log("UPDATED group.updatedAt …

Core Data允许我保存,但是当我尝试删除对象时,我收到以下错误:

*** Terminating app due to uncaught exception 'NSObjectInaccessibleException', reason: 'CoreData could not fulfill a fault for '0xa6b7e00 <x-coredata://8A687ECB-03F8-47C0-8470-087B9CF032B1-2801-00000A8C9A09CDB7/Notification/p2774005D-4B49-4378-A109-949F15D37032>'"

[self.managedObjectContext executeFetchRequest:allFetchRequest onSuccess:^(NSArray *results) {
                NSLog(@"Fetch for Objects with origial_id");
                if (results.count>0)
                {
                    NSLog(@"original_id =  %@", dlPlanDate.original_id);
                    NSLog(@"Results Recieved %i", results.count);
                    for (PlanDate *pd in results)
                    {
                        PlanDateResolved *pdResolved = [NSEntityDescription insertNewObjectForEntityForName:@"PlanDateResolved" inManagedObjectContext:self.managedObjectContext];
                        NSArray *keys = [[[pd entity] attributesByName] allKeys];
                        NSDictionary *dict= [pd dictionaryWithValuesForKeys:keys];
                        [pdResolved setValuesForKeysWithDictionary:dict];
                        [pdResolved setValue:[pd plandate_id] forKey:[pdResolved primaryKeyField]];
                        pdResolved.senderUser = pd.senderUser;
                        pdResolved.objectCreator = pd.objectCreator;
                        pdResolved.users = …

有没有办法从Github网站的问题创建一个分支？

一旦我选择了一个问题,有没有办法从网站创建一个分支然后克隆它？还是另类？

刚移到Xcode 11并在启动时遇到以下崩溃：

CoreData: fault: One or more models in this application are using transformable properties with transformer names that are either unset, or set to NSKeyedUnarchiveFromDataTransformerName. Please switch to using "NSSecureUnarchiveFromData" or a subclass of NSSecureUnarchiveFromDataTransformer instead. At some point, Core Data will default to using "NSSecureUnarchiveFromData" when nil is specified, and transformable properties containing classes that do not support NSSecureCoding will become unreadable.
CoreData: warning: Property 'color' on Entity 'Group' is using nil or an insecure NSValueTransformer.  Please switch …

我有一个UISearchBar.当我选择单元格时,我希望整个单元格都有[UIColor grayColor];

使用下面的代码,contentView颜色显示为Gray; 但是,背景accessoryType颜色显示为蓝色:

- (void)tableView:(UITableView *)tableView didSelectRowAtIndexPath:(NSIndexPath *)indexPath {     

UITableViewCell *cell = [self.searchDisplayController.searchResultsTableView cellForRowAtIndexPath:indexPath];
    cell.contentView.backgroundColor = [UIColor grayColor];

    if (self.lastSelected && (self.lastSelected.row == indexPath.row))
    {
        cell.accessoryType = UITableViewCellAccessoryNone;
        [cell setSelected:NO animated:TRUE];
        self.lastSelected = nil;
    } else {
        cell.accessoryType = UITableViewCellAccessoryCheckmark;
        cell.accessoryView.backgroundColor = [UIColor grayColor]; // Not working
        [cell setSelected:TRUE animated:TRUE];

        UITableViewCell *old = [self.searchDisplayController.searchResultsTableView cellForRowAtIndexPath:self.lastSelected];
        old.accessoryType = UITableViewCellAccessoryNone;
        [old setSelected:NO animated:TRUE];
        self.lastSelected = indexPath;
    }

如何使蓝色也显示为[UIColor grayColor]？

我试图解析以下JSON:

{ 
  "String For Key 1": [
    "Some String A",
    "Some String B",
    "Some String C",
  ],
  "String For Key 2": [
    "Some String D",
    "Some String E",
    "Some String F"
  ],
  "String For Key 3": [
    "Some String G",
    "Some String H",
    "Some String I",
  ],
  "String For Key 4": [
    "Some String J",
    "Some String K",
    "Some String L"
  ],
  "String For Key 5": [
    "Some String M",
    "Some String N",
    "Some String O"
  ],
  "String For Key …

我正在学习教程，下面是代码：

var Sequelize = require('sequelize')
var sequelize = new Sequelize('basic-mysql-database.mysql', 'root', 'password', {
    'dialect': 'mysql',
    'host': "localhost",
    "port": "3306"
});

var Todo = sequelize.define('todo', {
    description: {
        type: Sequelize.STRING
    },
    completed: {
        type: Sequelize.BOOLEAN

    }
})

sequelize.sync().then(function(){
    console.log('Everything is synced')

    Todo.create({
        description: 'Walking my dog',
        completed: false
    }).then(function (todo){
        console.log('Finished!')
        console.log(todo)
    })
});

我已经安装了 MySQL。当我进入Settings> 时MySQL，它说MySQL Server instance is running

当我运行时node testDB.js，出现以下错误：

Unhandled rejection SequelizeConnectionError: ER_BAD_DB_ERROR: Unknown database 'basic-mysql-database.mysql'
    at Handshake._callback (/Users/Kausi/Documents/Development/todo-api/node_modules/sequelize/lib/dialects/mysql/connection-manager.js:63:20)
    at …