您需要在值本身中设置它，而不是在准备好的语句sql字符串中。
因此，对于前缀匹配，应该这样做：

notes = notes
    .replace("!", "!!")
    .replace("%", "!%")
    .replace("_", "!_")
    .replace("[", "![");
PreparedStatement pstmt = con.prepareStatement(
        "SELECT * FROM analysis WHERE notes LIKE ? ESCAPE '!'");
pstmt.setString(1, notes + "%");

或后缀匹配：

pstmt.setString(1, "%" + notes);

或全球匹配：

pstmt.setString(1, "%" + notes + "%");

编码如下：

PreparedStatement pstmt = con.prepareStatement(
    "SELECT * FROM analysis WHERE notes like ?");
pstmt.setString(1, notes + "%");`

请确保不要包含下面这样的引号“”，因为它们会导致异常。

pstmt.setString(1,"'%"+ notes + "%'");

我们可以使用 CONCAT sql函数。

PreparedStatement pstmt = con.prepareStatement(
      "SELECT * FROM analysis WHERE notes like CONCAT( '%',?,'%')";
pstmt.setString(1, notes);
ResultSet rs = pstmt.executeQuery();

这对我的案子很有效。

PreparedStatement ps = cn.prepareStatement("Select * from Users where User_FirstName LIKE ?");
ps.setString(1, name + '%');

试试这个。

String fname = "Sam\u0025";

PreparedStatement ps= conn.prepareStatement("SELECT * FROM Users WHERE User_FirstName LIKE ? ");

ps.setString(1, fname);

String query="select * from test1 where "+selected+" like '%"+SelectedStr+"%';";

PreparedStatement preparedStatement=con.prepareStatement(query);

// where seleced and SelectedStr are String Variables in my program