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AIX：需要执行2次才能看到真实结果

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AIX：需要执行2次才能看到真实结果

为什么每次我执行脚本时，我都要执行两次才能看到真正的结果，

script_dir=/opt/scripts/x186299

smo_log=/oracle/bxcons21/base/admin/bxcons21/scripts/logs

error_txt=$script_dir/logs/SMO_error.txt

count_error=$script_dir/logs/count_error.txt

yest=$script_dir/logs/yesterday.txt

ECount=`cat $count_error`

rm $error_txt $count_error $yest

echo `TZ=MDT+48 date +%Y%m%d` > $yest

cd $smo_log

#grep -i "Error invoking command" $yest_* >> $error_txt

echo "Error invoking command" >> $error_txt

cat $error_txt | wc -l >> $count_error

if [ $ECount = 0 ];

then

    echo "SMO Backup is failed"

    exit

else

    echo "SMO Backup is succesful"

    exit

fi


if I ran this, it will show succesfull, then if I changed this part to

old

"#grep -i "Error invoking command" $yest_* >> $error_txt

"echo "Error invoking command" >> $error_txt

new

"grep -i "Error invoking command" $yest_* >> $error_txt

"#echo "Error invoking command" >> $error_txt

我希望看到正确的失败消息，但不幸的是它显示成功，然后如果我再次执行脚本而不进行更改，它将显示失败消息。

如果只使用 # 就会变得粗体，所以我把 " 放在 grep 和 echo 行上

答案1

你设置得ECount太早了，此时count_error脚本第一次执行时文件还是空的：

ECount=`cat $count_error`
...
grep -i "Error invoking command" $yest_* >> $error_txt
...
cat $error_txt | wc -l >> $count_error
if [ $ECount = 0 ];

使用

grep -i "Error invoking command" $yest_* >> $error_txt
...
wc -l $error_txt >> $count_error
ECount=`cat $count_error`
if [ $ECount = 0 ];

相反，或者（如果您以后不需要该文件）

ECount=`wc -l $error_txt`
if [ $ECount = 0 ];

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