1
以下方法,當用String val = getCell("SELECT col FROM table WHERE LIKE(other_col,'?')", new String[]{"value"});
(這是SQLite)調用時,會拋出一個java.lang.ArrayIndexOutOfBoundsException: 0 at org.sqlite.PrepStmt.batch(PrepStmt.java:131)
。任何人都可以憐惜我可憐的笨蛋在這裏,幫我爲什麼?爲什麼此Java PreparedStatement在parameterIndex = 1時拋出ArrayIndexOutOfBoundsException 0?
/**
* Get a string representation of the first cell of the first row returned
* by <code>sql</code>.
*
* @param sql The SQL SELECT query, that may contain one or more '?'
* IN parameter placeholders.
* @param parameters A String array of parameters to insert into the SQL.
* @return The value of the cell, or <code>null</code> if there
* was no result (or the result was <code>null</code>).
*/
public String getCell(String sql, String[] parameters) {
String out = null;
try {
PreparedStatement ps = connection.prepareStatement(sql);
for (int i = 1; i <= parameters.length; i++) {
String parameter = parameters[i - 1];
ps.setString(i, parameter);
}
ResultSet rs = ps.executeQuery();
rs.first();
out = rs.getString(1);
rs.close();
ps.close();
} catch (SQLException e) {
e.printStackTrace();
}
return out;
}
的setString()
會,在這種情況下,是ps.setString(1, "value")
,不應該是一個問題。顯然我錯了。
許多在此先感謝。
我覺得SQLite的同時做兩... X LIKE y等價於調用標量函數LIKE(X,Y) ,它允許您重新定義操作符如何重新定義函數... – Stobor 2009-07-09 04:19:50