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
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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