Informations of All Access Users and New Question.

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  • Start date Start date
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Guest

To all users,
Good Day,
1. I was having following problem.
2. I tried my best to find within the previous posts but failed to get.
3. I posted my problem but sad to say it was so easy and simple but I
did not get sollution.
4. I got responses for that, however I am thank full for the people who
spent their valued time and responsed me them.
5. Now I am exploring it for all users. Access users who are already
familier with it and off course we begniners need their attention and request
them please take care. You will not loose any thing and we will get lot.
Here is problem

ParkingDuration: DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<70,"1
Hour" & IIf >70&<130,"2 Hour"

Solution
ParkingDuration:
IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=70,'1
Hour',IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=130,'2
Hours',IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=190,'3
Hours',)))

New Question.
Now I would like to put two conditions.

ParkingDuration:
IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=70&this,'1
Hour',IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=130&this,'2Hours',

Any one if could help me?

Regards,
 
hi,
i think it will be (as i understand you):
IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=70,'1
Hour',IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=130,'2
Hours'))
so it will produce either 1 hour, or 2 hours, or nothing if number of
minutes > 130
 
Alex,
Good Day,
I was waiting to get pointed respose.

Opening, reading, searching and in between that when browsing webs attention
get devote with other points, is long process and requied time.

Could you please revert me on the request.

Regards,

--
Access Keen User



Alex Dybenko said:
hi,
i think it will be (as i understand you):
IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=70,'1
Hour',IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=130,'2
Hours'))
so it will produce either 1 hour, or 2 hours, or nothing if number of
minutes > 130

--
Alex Dybenko (MVP)
http://Alex.Dybenko.com
http://www.PointLtd.com


Access Keen User said:
To all users,
Good Day,
1. I was having following problem.
2. I tried my best to find within the previous posts but failed to
get.
3. I posted my problem but sad to say it was so easy and simple but I
did not get sollution.
4. I got responses for that, however I am thank full for the people
who
spent their valued time and responsed me them.
5. Now I am exploring it for all users. Access users who are already
familier with it and off course we begniners need their attention and
request
them please take care. You will not loose any thing and we will get lot.
Here is problem

ParkingDuration: DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<70,"1
Hour" & IIf >70&<130,"2 Hour"

Solution
ParkingDuration:
IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=70,'1
Hour',IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=130,'2
Hours',IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=190,'3
Hours',)))

New Question.
Now I would like to put two conditions.

ParkingDuration:
IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=70&this,'1
Hour',IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=130&this,'2Hours',

Any one if could help me?

Regards,
Click to expand...
Click to expand...
 
Hi,
so what is wrong with a code i wrote you? is it doing something else then
you expected?

--
Alex Dybenko (MVP)
http://alexdyb.blogspot.com
http://www.PointLtd.com


Access Keen User said:
Alex,
Good Day,
I was waiting to get pointed respose.

Opening, reading, searching and in between that when browsing webs
attention
get devote with other points, is long process and requied time.

Could you please revert me on the request.

Regards,

--
Access Keen User



Alex Dybenko said:
hi,
i think it will be (as i understand you):
IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=70,'1
Hour',IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=130,'2
Hours'))
so it will produce either 1 hour, or 2 hours, or nothing if number of
minutes > 130

--
Alex Dybenko (MVP)
http://Alex.Dybenko.com
http://www.PointLtd.com


Access Keen User said:
To all users,
Good Day,
1. I was having following problem.
2. I tried my best to find within the previous posts but failed to
get.
3. I posted my problem but sad to say it was so easy and simple but
I
did not get sollution.
4. I got responses for that, however I am thank full for the people
who
spent their valued time and responsed me them.
5. Now I am exploring it for all users. Access users who are
already
familier with it and off course we begniners need their attention and
request
them please take care. You will not loose any thing and we will get
lot.
Here is problem

ParkingDuration:
DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<70,"1
Hour" & IIf >70&<130,"2 Hour"

Solution
ParkingDuration:
IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=70,'1
Hour',IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=130,'2
Hours',IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=190,'3
Hours',)))

New Question.
Now I would like to put two conditions.

ParkingDuration:
IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=70&this,'1
Hour',IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=130&this,'2Hours',

Any one if could help me?

Regards,
Click to expand...
Click to expand...
Click to expand...
 
Alex,
Thanks your prompt responst,
Syntexed formula is working. Now I want to put two conditions, <= this and
<=that, then should be this.
Regards,


--
Access Keen User



Alex Dybenko said:
Hi,
so what is wrong with a code i wrote you? is it doing something else then
you expected?

--
Alex Dybenko (MVP)
http://alexdyb.blogspot.com
http://www.PointLtd.com


Access Keen User said:
Alex,
Good Day,
I was waiting to get pointed respose.

Opening, reading, searching and in between that when browsing webs
attention
get devote with other points, is long process and requied time.

Could you please revert me on the request.

Regards,

--
Access Keen User



Alex Dybenko said:
hi,
i think it will be (as i understand you):
IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=70,'1
Hour',IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=130,'2
Hours'))
so it will produce either 1 hour, or 2 hours, or nothing if number of
minutes > 130

--
Alex Dybenko (MVP)
http://Alex.Dybenko.com
http://www.PointLtd.com


message To all users,
Good Day,
1. I was having following problem.
2. I tried my best to find within the previous posts but failed to
get.
3. I posted my problem but sad to say it was so easy and simple but
I
did not get sollution.
4. I got responses for that, however I am thank full for the people
who
spent their valued time and responsed me them.
5. Now I am exploring it for all users. Access users who are
already
familier with it and off course we begniners need their attention and
request
them please take care. You will not loose any thing and we will get
lot.
Here is problem

ParkingDuration:
DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<70,"1
Hour" & IIf >70&<130,"2 Hour"

Solution
ParkingDuration:
IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=70,'1
Hour',IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=130,'2
Hours',IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=190,'3
Hours',)))

New Question.
Now I would like to put two conditions.

ParkingDuration:
IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=70&this,'1
Hour',IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=130&this,'2Hours',

Any one if could help me?

Regards,
Click to expand...
Click to expand...
Click to expand...
 
ahh, ok
then you can use AND:
IIf( (DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=70) AND
(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=130),'true
message','false message')

--
Alex Dybenko (MVP)
http://alexdyb.blogspot.com
http://www.PointLtd.com


Access Keen User said:
Alex,
Thanks your prompt responst,
Syntexed formula is working. Now I want to put two conditions, <= this
and
<=that, then should be this.
Regards,


--
Access Keen User



Alex Dybenko said:
Hi,
so what is wrong with a code i wrote you? is it doing something else then
you expected?

--
Alex Dybenko (MVP)
http://alexdyb.blogspot.com
http://www.PointLtd.com


Access Keen User said:
Alex,
Good Day,
I was waiting to get pointed respose.

Opening, reading, searching and in between that when browsing webs
attention
get devote with other points, is long process and requied time.

Could you please revert me on the request.

Regards,

--
Access Keen User



:

hi,
i think it will be (as i understand you):
IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=70,'1
Hour',IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=130,'2
Hours'))
so it will produce either 1 hour, or 2 hours, or nothing if number of
minutes > 130

--
Alex Dybenko (MVP)
http://Alex.Dybenko.com
http://www.PointLtd.com


message To all users,
Good Day,
1. I was having following problem.
2. I tried my best to find within the previous posts but failed
to
get.
3. I posted my problem but sad to say it was so easy and simple
but
I
did not get sollution.
4. I got responses for that, however I am thank full for the
people
who
spent their valued time and responsed me them.
5. Now I am exploring it for all users. Access users who are
already
familier with it and off course we begniners need their attention
and
request
them please take care. You will not loose any thing and we will get
lot.
Here is problem

ParkingDuration:
DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<70,"1
Hour" & IIf >70&<130,"2 Hour"

Solution
ParkingDuration:
IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=70,'1
Hour',IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=130,'2
Hours',IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=190,'3
Hours',)))

New Question.
Now I would like to put two conditions.

ParkingDuration:
IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=70&this,'1
Hour',IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=130&this,'2Hours',

Any one if could help me?

Regards,
Click to expand...
Click to expand...
Click to expand...
 
What's the point of that?

If DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=130 is true, then
DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=70 will always be true!

If DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=130 is false, then it
doesn't matter whether
DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=70 is true or false: the
result will be false.

--
Doug Steele, Microsoft Access MVP

(no e-mails, please!)


Alex Dybenko said:
ahh, ok
then you can use AND:
IIf( (DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=70) AND
(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=130),'true
message','false message')

--
Alex Dybenko (MVP)
http://alexdyb.blogspot.com
http://www.PointLtd.com


Access Keen User said:
Alex,
Thanks your prompt responst,
Syntexed formula is working. Now I want to put two conditions, <= this
and
<=that, then should be this.
Regards,


--
Access Keen User



Alex Dybenko said:
Hi,
so what is wrong with a code i wrote you? is it doing something else then
you expected?

--
Alex Dybenko (MVP)
http://alexdyb.blogspot.com
http://www.PointLtd.com


message Alex,
Good Day,
I was waiting to get pointed respose.

Opening, reading, searching and in between that when browsing webs
attention
get devote with other points, is long process and requied time.

Could you please revert me on the request.

Regards,

--
Access Keen User



:

hi,
i think it will be (as i understand you):
IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=70,'1
Hour',IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=130,'2
Hours'))
so it will produce either 1 hour, or 2 hours, or nothing if number of
minutes > 130

--
Alex Dybenko (MVP)
http://Alex.Dybenko.com
http://www.PointLtd.com


message To all users,
Good Day,
1. I was having following problem.
2. I tried my best to find within the previous posts but failed
to
get.
3. I posted my problem but sad to say it was so easy and simple
but
I
did not get sollution.
4. I got responses for that, however I am thank full for the
people
who
spent their valued time and responsed me them.
5. Now I am exploring it for all users. Access users who are
already
familier with it and off course we begniners need their attention
and
request
them please take care. You will not loose any thing and we will get
lot.
Here is problem

ParkingDuration:
DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<70,"1
Hour" & IIf >70&<130,"2 Hour"

Solution
ParkingDuration:
IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=70,'1
Hour',IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=130,'2
Hours',IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=190,'3
Hours',)))

New Question.
Now I would like to put two conditions.

ParkingDuration:
IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=70&this,'1
Click to expand...
Click to expand...
Hour',IIf(DateDiff("n",[ParkingDatetime],[DriveOutDateTime])<=130&this,'2Hou
rs',
Click to expand...
 
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