急,如何用grep命令实现成功或失败的输出

jack7527

成功的样例
wld.log
Table CIN_CAPACITY_INFORMATION_NEW:
  312 Rows successfully loaded.
  0 Rows 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.


如果下面这三行都是以0开头,表示没有失败的记录,即全部成功,如何用grep命令表示
  0 Rows 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.



失败的样例
wld.log
Table CIN_CAPACITY_INFORMATION_NEW:
  310 Rows successfully loaded.
  2 Rows 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.


如果下面这三行只要有一行不是以0开头,就表示有失败的记录,即表示失败
  0 Rows not loaded due to data errors.
  2 Rows not loaded because all WHEN clauses were failed.
  0 Rows not loaded because all fields were null.
可以用表示 grep -i "[1-9] rows not loaded"

请问我如何用grep命令动态地表示成功或是失败呢? 即要输出表示这个日志是成功了还是失败了,不知如何实现,请指教

bjgirl

1. [[ $(cat file|wc -l) -eq $(grep -wc ^0 file) ]]&&echo "ok"||echo "false"

复制代码

jack7527

test.sh内容

#!/bin/csh -f
cp current_`date +%d-%b-%Y|tr -s "[]" "[]"`.csv /ue21/as52app/OUT/batch/import
cp current_2`date +%d-%b-%Y|tr -s "[]" "[]"`.csv /ue21/as52app/OUT/batch/import
cd import
mv current_`date +%d-%b-%Y|tr -s "[]" "[]"`.csv test1.csv
mv current_2`date +%d-%b-%Y|tr -s "[]" "[]"`.csv test2.csv
cd ..


sqlldr userid=tpas_user/tpas_user control=import1.ctl  log=`date +%Y%m%d-1` skip=1
sqlldr userid=tpas_user/tpas_user control=import2.ctl  log=`date +%Y%m%d-2` skip=1

set log1=`date +%Y%m%d-1.log`
set log2=`date +%Y%m%d-2.log`

set file1=`date +%d-%b-%Y|tr -s "[]" "[]"`.csv
set file2=`date +%d-%b-%Y|tr -s "[]" "[]"`.csv


if grep -i "[1-9] rows not loaded" $log1 >/dev/null
then
echo "NOT OK"
else
echo "OK"
fi


if grep -i "[1-9] rows not loaded" $log2 >/dev/null
then
echo "NOT OK"
else
echo "OK"
fi

运行test.sh
错误提示:
if :Expression syntax

jack7527

test.sh

#!/bin/csh -f
cp current_`date +%d-%b-%Y|tr -s "[]" "[]"`.csv /ue21/as52app/OUT/batch/import
cp current_2`date +%d-%b-%Y|tr -s "[]" "[]"`.csv /ue21/as52app/OUT/batch/import
cd import
mv current_`date +%d-%b-%Y|tr -s "[]" "[]"`.csv test1.csv
mv current_2`date +%d-%b-%Y|tr -s "[]" "[]"`.csv test2.csv
cd ..


sqlldr userid=tpas_user/tpas_user control=import1.ctl  log=`date +%Y%m%d-1` skip=1
sqlldr userid=tpas_user/tpas_user control=import2.ctl  log=`date +%Y%m%d-2` skip=1

set log1=`date +%Y%m%d-1.log`
set log2=`date +%Y%m%d-2.log`

[[ $(cat log1|wc -l) -eq $(grep -wc ^0 log1) ]]&&echo "ok"||echo "false"
[[ $(cat log2|wc -l) -eq $(grep -wc ^0 log2) ]]&&echo "ok"||echo "false"

运行test.sh,错误提示如下:
Variable syntax

注:我的操作系统是sunOS 5.8

gunguymadman

[quote]原帖由 "bjgirl"][[ $(cat file|wc -l) -eq $(grep -wc ^0 file) ]]&&echo "ok"||echo "false"[/quote 发表：
我试验了一下   用得是楼主ok得数据 输出是false   
我不大懂  我觉得cat file| wc -l 是求log得行数   grep -wc ^0 file 是求以0为首得行数   然后看二者相不相等   无论成功失败都有310 Rows successfully loaded.    这样就永远不相等了  输出只会是false
还是和3比好