小编the*_*er7的帖子

这更像是一个通用问题，我正在构建一个报告生成器：

public ReportGenerator(IReportGeneratorConfig config)
{
    Configuration = config;
    ImagePaths = new Dictionary<string, string>();
    ReportErrors = new StringBuilder();

    DatabaseForReportQueries = DatabaseFactory.CreateDatabase(Configuration.DatabaseName);

    using (System.Data.Common.DbConnection conn = DatabaseForReportQueries.CreateConnection())
    {
        //get the connection string for use with the report directly.
        ReportConnectionString = conn.ConnectionString + "Provider=SQLOLEDB;";
        conn.Close();
    }
}

这是我在尝试运行报告生成按钮时收到的错误消息

Cannot open data source, please check ConnectionString and RecordSource properties. 
ConnectionString:   Database=Test80;Server=SQL01;uid=mcadmin;pwd=password;Provider=SQLOLEDB; 
RecordSource: 

除了我在代码中发送的提供者之外，所有信息都完好无损。我不知道如何在 Web 应用程序中找出提供者名称。这与 中的providerName标签不同connectionStrings.config。它需要是Provider=something;

连接字符串有一个 providerName = System.Data.SqlClient

我运行了命令autotest,这是我得到的错误.我正在关注Hartl的书,并想知道这场冲突是否正在发生,因为rails现在带有ZenTest或其他东西？我如何解决此错误.我是RoR的新手

Invalid gemspec in [/Users//.rvm/gems/ruby-1.9.2-p320@twitclone/specifications/ZenTest-4.8.4.gemspec]: Illformed requirement ["< 2.1, >= 1.8"]
Invalid gemspec in [/Users//.rvm/gems/ruby-1.9.2-p320@twitclone/specifications/ZenTest-4.8.4.gemspec]: Illformed requirement ["< 2.1, >= 1.8"]
/Users//.rvm/rubies/ruby-1.9.2-p320/lib/ruby/site_ruby/1.9.1/rubygems/dependency.rb:247:in `to_specs': Could not find ZenTest (>= 0) amongst [abstract-1.0.0, actionmailer-3.2.8, actionmailer-3.2.8.rc1, actionmailer-3.0.1, actionmailer-3.0.0, actionpack-3.2.8, actionpack-3.2.8.rc1, actionpack-3.0.1, actionpack-3.0.0, activemodel-3.2.8, activemodel-3.2.8.rc1, activemodel-3.0.1, activemodel-3.0.0, activerecord-3.2.8, activerecord-3.2.8.rc1, activerecord-3.0.1, activerecord-3.0.0, activeresource-3.2.8, activeresource-3.2.8.rc1, activeresource-3.0.1, activeresource-3.0.0, activesupport-3.2.8, activesupport-3.2.8.rc1, activesupport-3.0.1, activesupport-3.0.0, addressable-2.3.2, arel-3.0.2, arel-1.0.1, autotest-4.4.6, builder-3.0.4, builder-3.0.3, builder-2.1.2, bundler-1.2.1, bundler-1.2.0, bundler-1.0.22, coffee-rails-3.2.2, coffee-script-2.2.0, coffee-script-source-1.4.0, coffee-script-source-1.3.3, diff-lcs-1.1.3, erubis-2.7.0, erubis-2.6.6, excon-0.16.2, execjs-1.4.0, heroku-2.31.2, heroku-api-0.3.5, hike-1.2.1, i18n-0.6.1, i18n-0.4.2, journey-1.0.4, …

我是python的新手,我有这个c ++的代码片段:

do
    {
    cout << "Make sure the number of digits are exactly 12 : ";
    cin >> input ;
    } while((input.length()) != 12 );

如何将此部分更改为python？到目前为止,我已经尝试过这个,但是我没有得到正确的语法或逻辑流程.这就是我所拥有的:

while True:
  print("Make sure the number of digits are exactly 12 : ")
  input = raw_input()
  check = len(input)
  if check != 12
  break

以上部分解决了!

另外,另一个c ++片段是:input是string

for (int i = 0; i < 12 ; i++)
    {
     code[i] = input.at(i) - '0';
    }

我无法弄清楚如何将此部分更改为python代码

code[i] = input.at(i) - '0';

所以,我遇到的问题是我无法弄清楚如何初始化数组

int …

基本上,我有2个文件(.adb和.ads).我是Ada的新手,也是如何编译2个文件的.该程序是一个基本的堆栈实现.编译.adb文件时出现此编译错误.

$ gcc -c test_adt_stack.adb
abstract_char_stack.ads:22:01: end of file expected, file can have only one compilation unit

我拥有的2个文件是: abstract_char_stack.ads

-----------------------------------------------------------
package Abstract_Char_Stack is
  type Stack_Type is private;
  procedure Push(Stack : in out Stack_Type;
                 Item  : in Character);
  procedure Pop (Stack : in out Stack_Type;
                 Char  : out Character);
private
  type Space_Type is array(1..8) of Character;
  type Stack_Type is record
    Space : Space_Type;
    Index : Natural := 0;
  end record;
end Abstract_Char_Stack;
-----------------------------------------------------------
package body Abstract_Char_Stack is
----------------------------------------------
  procedure Push(Stack : …

我们如何编写一个包装器词法转换函数来实现如下行:

int value = lexical_cast<int> (string)

我对编程很陌生,并且想知道我们如何编写函数.我不知道如何找出一个模板.我们也可以为double编写一个包装函数吗？喜欢

double value = lexical_cast2<double> (string)

??