CODEWITHSUNDEEP
javajdbc
Bob
Bob

Reputation: 7

JDBC Error for prepared Statement giving SQL exception. Why?

I am trying to compile a java file in the format:

PreparedStatement var = con.prepareStatement("SELECT * FROM table WHERE column LIKE '%?%';");

var.setString(1, string);
var.executeQuery();

However, it is not compiling, with the following errors. Can anyone explain why I am getting these errors?

BookPurchase.java:97: error: unreported exception SQLException; must be caught or declared to be thrown
            con.prepareStatement(
                                ^
BookPurchase.java:100: error: unreported exception SQLException; must be caught or declared to be thrown
            searchResults.setString(1, keyword);
                                   ^
BookPurchase.java:101: error: unreported exception SQLException; must be caught or declared to be thrown
            searchResults.executeQuery();

Upvotes: 1

Views: 5886

Answers (2)

MaVRoSCy
MaVRoSCy

Reputation: 17859

That is because the prepareStatement is declared to throw an Exception.

To overcome your problem enclose your code in a try / catch block:

try {
    PreparedStatement prepStmt = con.prepareStatement(mySql);
    var.setString(1, string);
    var.executeQuery();
}catch(SQLException ex){
    //deal with exception
}

or you can make your method that executes this code throw an Exception:

public void myMethod() throws SQLException {
    ...
    ...
    PreparedStatement prepStmt = con.prepareStatement(mySql);
    var.setString(1, string);
    var.executeQuery();
    ...
    ...
}

This is the most basic on exception-handling in java. You can refer here for more info

Upvotes: 4

m.aibin
m.aibin

Reputation: 3603

Try that:

SELECT * FROM table WHERE column LIKE = ?;

then, set String as: var.setString(1, "%"+string+"%");

It should work fine.

Look also here: Using "like" wildcard in prepared statement

Upvotes: -1

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