How to solve this error message?

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Hii,everyone
there is aproblem in query expression when search by student id's
and error messing about missing operator (Syntax error(missing operator)in query expression'student.id_civil_st like123456'
as in this code :

C#

string conString = @"Provider=Microsoft.ACE.OLEDB.12.0;Data Source=..\\..\\org.accdb";
           OleDbConnection con = new OleDbConnection(conString);
           DataTable dt = new DataTable();
           OleDbDataAdapter oda = new OleDbDataAdapter("SELECT * FROM student WHERE student.id_civil_st like" + int.Parse(textBox1.Text), con);
           oda.Fill(dt);
           dataGridView1.DataSource = dt;

Posted 19-Sep-12 21:48pm

mhassan083

Updated 19-Sep-12 21:52pm

AmitGajjar

v2

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Comments

Kuthuparakkal 20-Sep-12 3:54am

You dont even know SQL & how to use it. Learn SQL queries before you jump in

4 solutions

Solution 2

LIKE determines whether or not a given character string matches a specified pattern.

Posted 19-Sep-12 21:54pm

Kuthuparakkal

Solution 3

C#

OleDbDataAdapter oda = new OleDbDataAdapter("SELECT * FROM student WHERE student.id_civil_st like % '"+ int.Parse(textBox1.Text)+"' ", con);

Posted 19-Sep-12 21:57pm

Kamalkant(kk)

Updated 19-Sep-12 22:17pm

v3

Solution 4

student.id_civil_st like123456'

As you watch the error message closely you see you miss a space between your 'like' keyword and the value after that.
However as stated above in other suggestions, like is supposed to determine whether or not a given character string matches a pattern. The variables in this pattern are often noted by '%' as such:

Select * from student where student.id_civil_st like '%123456';

Otherwise if you want to compare a fixed number it's just fine to use the equality operator (=)

Posted 19-Sep-12 23:31pm

Michael Siroen

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**AmitGajjar** · Accepted Answer · 2012-09-19T21:54:00

Hi,

Use quotation mark when searching for ID,
Query should be like,

OleDbDataAdapter oda = new OleDbDataAdapter("SELECT * FROM student WHERE student.id_civil_st like '" + int.Parse(textBox1.Text), con) + "'";

Another Sample,

SQL

Select * from student where student.id_civil_st like '123456';

Hope this resolve your issue