hi every body(like simulation)

Question

0.00/5 (No votes)

See more:

can i use "like" in datatable.select metheod in C#10?
e.g : datatable1.select("name like 'A__Z'") for records that have 4 lenght size and start with "A" and ends with "Z" ?
thanks.

Posted 3-Dec-12 20:17pm

javad naroogha

Add a Solution

3 solutions

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

yaser shadmehr · Accepted Answer · 2012-12-03T20:50:00

Solution 2

if you considered .Net framework 1.0 or 1.1 instead of "C#10", you should upgrade to .Net framework 2 or later and use Linq to SQL.

Please find difference between .Net Framework 1.1 , 2.0 , 3.5

From C# 2.0 to 4.0

example of :

C#

from u in datatable1
where 
    u.name.Length == 4 &&
    u.name.StartsWith("A") &&
    u.name.EndsWith("Z")
select u

Posted 3-Dec-12 20:50pm

yaser shadmehr

Updated 3-Dec-12 20:54pm

v3

Comments

javad naroogha 4-Dec-12 3:19am

thanks

Mohamed Mitwalli 4-Dec-12 6:09am

+5

__TR__ · Accepted Answer · 2012-12-03T20:51:00

Solution 3

Try

C#

datatable1.Select("Name LIKE 'A%' AND Name LIKE '%Z' AND LEN(Name) = 4");

Posted 3-Dec-12 20:51pm

__TR__

Comments

javad naroogha 4-Dec-12 3:19am

thanks

Mohamed Mitwalli 4-Dec-12 6:09am

5+

__TR__ 7-Dec-12 11:38am

Thank you :)

Expert Coming · Accepted Answer · 2012-12-03T20:21:00

Solution 1

That is going to take either LINQ or RegEx.

Posted 3-Dec-12 20:21pm

Expert Coming

Comments

javad naroogha 4-Dec-12 2:26am

thanks,I don't use LINQ.

MT_ 4-Dec-12 2:26am

This is at the most can be a comment and not solution.