SQL Server LIKE statement with multiple values?

Question

Forgive if the question is repeated, I'm new to this site.

Been trying to create a search text box for a program that saved and organized documents. The software is database-based, and it can search for only one column at a time. I need it to look for any info in any column (expect date).

Here is the code:

private void textBox2_TextChanged(object sender, EventArgs e)
{
        DataView vista = new DataView(tablaSql);
        vista.RowFilter = string.Format("asunto_corres LIKE'%{0}%'", textBox2.Text);
        dgTodo.DataSource = vista;
}

It works but only with the name of the column specified.

Any help to make this textbox look for info in any field/column.

Thanks

You have to build that SQL-like column by column (with a long OR). BUT: 1) you can do it inspecting schema (no need to hard code all columns) and 2) use SQL parameters, do not build SQL commands with user inputs (what about SQL injection?). — Adriano Repetti
– Adriano Repetti, Commented Feb 19, 2014 at 14:52
@Adriano in this context the SQL Injection is not a concern (the RowFilter works on the in memory view) — Steve
– Steve, Commented Feb 19, 2014 at 14:56
You can use CONTAINS and FREETEXT clauses of Full-Text Search engine, but that can be overhead if your search is a small piece of app. — veljasije
– veljasije, Commented Feb 19, 2014 at 14:56
@Steve the point here is not (just) malicious SQL code but text that will make code fail (for example a search for a string like Andrew's cat). I mentioned SQL Injection because it's an easy search topic for Google if he need to search for that. — Adriano Repetti
– Adriano Repetti, Commented Feb 19, 2014 at 14:58

mnieto · Accepted Answer · 2014-02-19 14:53:00Z

2

try:

vista.RowFilter = string.Format("{0} LIKE'%{1}%'", fieldName, textBox2.Text);

answered Feb 19, 2014 at 14:53

mnieto

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Steve · Accepted Answer · 2014-02-19 14:55:14Z

0

You can do it by adding some "OR" operators:

vista.RowFilter = string.Format("asunto_corres LIKE'%{0}%' OR Column2 LIKE'%{0}%' OR ...", textBox2.Text);

answered Feb 19, 2014 at 14:55

Steve

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