我收到错误消息“您的sql语法有错误；请检查与您的MySQL服务器版本相对应的手册,以在第1行的“ orderr”附近使用正确的语法”,因此我认为错误是已经使用了两个’,但是在我的代码中我没有使用任何’.注意,该表实际上被命名为订购者.
public void insertIntoDatabase(String table, Object... entries) { // take a table and
Connection con = connect(); //add entries
PreparedStatement preparedStatement = null;
PreparedStatement preparedStatement2 = null;
ResultSet rs = null;
StringBuffer columnNames = new StringBuffer();
StringBuffer sbEntries = new StringBuffer();
for (int i = 0; i < entries.length; i++) {
if (entries[i] instanceof Integer)
sbEntries.append((Integer) entries[i]);
else if (entries[i] instanceof String)
sbEntries.append((String) entries[i]);
if (i != entries.length - 1)//if not last entry add
sbEntries.append(" ,"); // a ' ,'.
}
try {
preparedStatement = con.prepareStatement("select * from ? ;");
preparedStatement.setString(1, table);
preparedStatement2 = con
.prepareStatement("Insert into ?( ? ) values ( ? );");
ResultSet resultSet = preparedStatement.executeQuery(); // get the
// number of
// columns
ResultSetMetaData rsmd; // for the table
rsmd = resultSet.getMetaData();
int columnCount = rsmd.getColumnCount();
for (int i = 1; i < columnCount + 1; i++) { // get column names, add to
columnNames.append(rsmd.getColumnName(i)); // to sb
if (i != columnCount)
columnNames.append(" ,");
}
columnCount = rsmd.getColumnCount();
preparedStatement2.setString(1, table);
preparedStatement2.setString(2, columnNames.toString()); //add sb's to statement
preparedStatement2.setString(3, sbEntries.toString());
preparedStatement2.executeUpdate();
} catch (sqlException e) {
System.out.println("2" + e.getMessage());
}
finally{
try {
if (rs != null) {
rs.close();
}
if (preparedStatement != null) {
preparedStatement.close();
}
if(preparedStatement2 != null){
preparedStatement2.close();
}
if (con != null) {
con.close();
}
} catch (sqlException e) {
System.out.print("3" +e.getMessage());
}
}
}
解决方法:
在大多数数据库中,您无法参数化表名称之类的对象名称,在MysqL中,理论上您可以做到,因为默认情况下MysqL Connector / J不使用服务器端参数,而是在将查询发送到服务器之前重写查询.但是,该值将作为带引号的字符串插入,并且对象名称不能是带引号的字符串,因此仍然无法使用.
如此插入？还是SELECT … FROM？将不起作用,因为它会产生“插入到’theTable’或SELECT … FROM’theTable’的信息.
