Doubt about SQL query with LIKE

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Hello,

I have a doubt about a sql query that i need if anyone can help thanks in advance.

I want to query a products database created in MS Acess through an application in C #. I need to consult the database looking for the product by reference, this reference may contain various names and spaces, for example: "ABC FFF HHH XXXXXXXX" and i want for example to introduce in textbox part of a string to search, for example: "HHH" and the query will return me all references containing "HHH" regardless of being at the beginning, middle or end of the reference.

The query that i have does not return anything:

SQL

Select * from product where product_ref like '% + TextBox_ref.Text +%';

What should i change in the query to get all the references containing the characters entered in the TextBox regardless of spaces that the reference may have in the database?

Thanks,

What I have tried:

SQL

Select * from product where product_ref like '% + TextBox_ref.Text +%';

Posted 18-Oct-16 12:21pm

Member 12801143

Updated 18-Oct-16 17:17pm

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[no name] 18-Oct-16 19:31pm

You should learn how string concatenation works.
"Select * from product where product_ref like '%" + TextBox_ref.Text + "%'";

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Wendelius · Answer 1 · 2016-10-18T17:17:00

As already pointed out, you should concatenate the value from the text box to separate strings, for example

C#

Select * from product where product_ref like '%'" + TextBox_ref.Text + "'%'

However, what you really should do is to use paramters, OleDbParameter Class (System.Data.OleDb)[^] in this case. The query text would then look like

C#

Select * from product where product_ref like '%' + @searchString + '%'

Now you would add a parameter for the OleDbCommand and set it's value. This would keep you safe from for example SQL injections.

Not sure if you wanted also to find rows having HHH even if the data in the database would be H HH. If that is the case, you can use REPLACE function to remove whitespace. For example

C#

Select * from product where REPLACE(product_ref, ' ', '') like '%' + @searchString + '%'