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curl -c命令powershell - cookie文件位置(curl -c command powershell - cookie file location)

我正在使用powershell curl命令来进行REST API调用。 当我使用-c选项运行curl命令时,

curl https://sampleurl/api/sessions -i -X POST -H X-Api-Version:xx -c mycookie -d email=$email -d password=$passwd

在这种情况下,“mycookie”文件存储在哪里。 出于安全考虑,我需要确保它得到保护。

该文档没有提供太多信息,请访问https://curl.haxx.se/docs/manpage.html

谢谢,


I'm using powershell curl commands to make REST API Calls. When I run the curl command with -c option,

curl https://sampleurl/api/sessions -i -X POST -H X-Api-Version:x.x -c mycookie -d email=$email -d password=$passwd

Where is the "mycookie" file stored in this case. For security purposes, I need to make sure it is safeguarded.

The documentation doesn't give much information here https://curl.haxx.se/docs/manpage.html

Thanks,


原文:https://stackoverflow.com/questions/40620032
更新时间:2023-10-27 19:10

最满意答案

像这样的东西会工作。 它是你的方法的变体,但它不使用MM / DD / YYYY文字格式,并且它不会针对不好的输入(无论好坏)发生爆炸。

declare @month tinyint
declare @day tinyint
set @month = 9
set @day = 1

declare @date datetime

-- this could be inlined if desired
set @date = convert(char(4),year(getdate()))+'0101'
set @date = dateadd(month,@month-1,@date)
set @date = dateadd(day,@day-1,@date)

if @date <= getdate()-1
  set @date = dateadd(year,1,@date)

select @date

或者,以YYYYMMDD格式创建一个字符串:

set @date = 
  right('0000'+convert(char(4),year(getdate())),4)
+ right('00'+convert(char(2),@month),2)
+ right('00'+convert(char(2),@day),2)

另一种方法,它可以避免所有的文字:

declare @month tinyint
declare @day tinyint
set @month = 6
set @day = 24

declare @date datetime
declare @today datetime

-- get todays date, stripping out the hours and minutes
-- and save the value for later
set @date = floor(convert(float,getdate()))
set @today = @date

-- add the appropriate number of months and days
set @date = dateadd(month,@month-month(@date),@date)
set @date = dateadd(day,@day-day(@date),@date)

-- increment year by 1 if necessary
if @date < @today set @date = dateadd(year,1,@date)

select @date

something like this would work. It's variation on your method, but it doesn't use the MM/DD/YYYY literal format, and it won't blowup against bad input (for better or for worse).

declare @month tinyint
declare @day tinyint
set @month = 9
set @day = 1

declare @date datetime

-- this could be inlined if desired
set @date = convert(char(4),year(getdate()))+'0101'
set @date = dateadd(month,@month-1,@date)
set @date = dateadd(day,@day-1,@date)

if @date <= getdate()-1
  set @date = dateadd(year,1,@date)

select @date

Alternatively, to create a string in YYYYMMDD format:

set @date = 
  right('0000'+convert(char(4),year(getdate())),4)
+ right('00'+convert(char(2),@month),2)
+ right('00'+convert(char(2),@day),2)

Another method, which avoids literals all together:

declare @month tinyint
declare @day tinyint
set @month = 6
set @day = 24

declare @date datetime
declare @today datetime

-- get todays date, stripping out the hours and minutes
-- and save the value for later
set @date = floor(convert(float,getdate()))
set @today = @date

-- add the appropriate number of months and days
set @date = dateadd(month,@month-month(@date),@date)
set @date = dateadd(day,@day-day(@date),@date)

-- increment year by 1 if necessary
if @date < @today set @date = dateadd(year,1,@date)

select @date

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