小编Rel*_*rin的帖子

无法使用表输出参数创建存储过程

我有这个代码:

IF  EXISTS (SELECT * FROM sys.objects WHERE object_id = OBJECT_ID(N'[dbo].[GetProfitDeals]') AND type in (N'P', N'PC'))
DROP PROCEDURE [dbo].[GetProfitDealsVar2]
GO

IF EXISTS(SELECT 1 FROM sys.types WHERE name = 'TableOrderType' AND is_table_type = 1 AND SCHEMA_ID('dbo') = schema_id)
DROP TYPE [dbo].[TableOrderType];

CREATE TYPE TableOrderType AS TABLE(
    Order_ID int NOT NULL,
    Order_AccNumber int NOT NULL,
    Order_OpenDate datetime NULL,
    Order_CloseDate datetime NULL,
    Order_Profit float NULL
);
GO

CREATE PROCEDURE [dbo].[GetProfitDeals](@OpenDate datetime = NULL, @CloseDate datetime  = NULL, @MinProfit float = NULL, @out TableOrderType …
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sql sql-server sql-server-2012 sql-server-2012-express

7
推荐指数
1
解决办法
2万
查看次数

逐行读取大文件

我正在尝试根据Clojure中的迭代为大文件编写阅读器.但是我怎么能在Clojure中逐行返回?我想做那样的事情:

(println(do_something(readFile(:file opts)));处理并打印第一行
(println(do_something(readFile(:file opts)));处理并打印第二行

码:

(ns testapp.core
  (:gen-class)
  (:require [clojure.tools.cli :refer [cli]])
  (:require [clojure.java.io]))


(defn readFile [file, cnt]
  ; Iterate over opened file (read line by line)
  (with-open [rdr (clojure.java.io/reader file)]
    (let [seq (line-seq rdr)]
      ; how return only one line there? and after, when needed, take next line?
    )))

(defn -main [& args]
  ; Main function for project 
  (let [[opts args banner] 
        (cli args
          ["-h" "--help" "Print this help" :default false :flag true]
          ["-f" "--file" "REQUIRED: File with …
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iteration io iterator file clojure

3
推荐指数
3
解决办法
5714
查看次数

将字符串作为格式化输出中的值

我正试图通过Rust by Example网站传递"Tuple课程" ,但我仍然坚持格式化的输出实现.我有这个代码,它打印传递的矩阵:

#[derive(Debug)]
struct Matrix{
  data: Vec<Vec<f64>>   // [[...], [...],] 
}

impl fmt::Display for Matrix {
    fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
        let output_data = self.data
            // [[1, 2], [2, 3]] -> ["1, 2", "2, 3"]
            .into_iter()
            .map(|row| {
                row.into_iter()
                    .map(|value| value.to_string())
                    .collect::<Vec<String>>()
                    .join(", ")
            })
            .collect::<Vec<String>>()
            // ["1, 2", "2, 3"] -> ["(1, 2)", "(2, 3)"]
            .into_iter()
            .map(|string_row| { format!("({})", string_row) })
            // ["(1, 2)", "(2, 3)"] -> "(1, 2),\n(2, 3)"
            .collect::<Vec<String>>()
            .join(",\n"); …
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string pointers rust borrowing

3
推荐指数
1
解决办法
82
查看次数

无法移出结构的借用内容

我正在尝试为来自另一个程序的BERT数据实现解串器.对于以下代码:

use std::io::{self, Read};

#[derive(Clone, Copy)]
pub struct Deserializer<R: Read> {
    reader: R,
    header: Option<u8>,
}

impl<R: Read> Read for Deserializer<R> {
    #[inline]
    fn read(&mut self, buf: &mut [u8]) -> io::Result<usize> {
        self.reader.read(buf)
    }
}

impl<R: Read> Deserializer<R> {
    /// Creates the BERT parser from an `std::io::Read`.
    #[inline]
    pub fn new(reader: R) -> Deserializer<R> {
        Deserializer {
            reader: reader,
            header: None,
        }
    }

    #[inline]
    pub fn read_string(&mut self, len: usize) -> io::Result<String> {
        let mut string_buffer = String::with_capacity(len);
        self.reader.take(len as …
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rust deserialization borrowing

0
推荐指数
1
解决办法
1587
查看次数