我们可以更改sqlldr命令的返回输出吗？

Question

使用

    $command = "sqlldr userid=abc/abc_def@defghijk CONTROL= $controlFileName DATA=\"$csvFilePath\" LOG=$logFile1 BAD=$logFile2";

    exec($command,$output,$returnValue);

得到了输出：

       SQL*Loader: Release 11.2.0.1.0 - Production on Wed Feb 19 15:53:58 2014

       Copyright (c) 1982, 2009, Oracle and/or its affiliates.  All rights reserved.

       Load completed - logical record count 196.

而不是这个输出，是否可以从日志文件到PHP的结果的下面部分：

  Table table_name:
  195 Rows successfully loaded.
  1 Row not loaded due to data errors.
  0 Rows not loaded because all WHEN clauses were failed.
  0 Rows not loaded because all fields were null.

  Partition SYS_P42102: 195 Rows loaded.

  Bind array size not used in direct path.
  Column array  rows :    5000
  Stream buffer bytes:  256000
  Read   buffer bytes: 1048576

  Total logical records skipped:          2
  Total logical records read:           196
  Total logical records rejected:         1
  Total logical records discarded:        0
  Total stream buffers loaded by SQL*Loader main thread:        2
  Total stream buffers loaded by SQL*Loader load thread:        0

即，我想从sqlldr命令返回值，即读取了多少行，拒绝了多少行以及插入了多少行？

感谢。