Pass pattern from variable to grep command

Question

Hi I need help to know how can we pass the vairble 2015-12-16 08:16: to grep command

bash-3.2$ Heartbeat=$(date +%Y-%m-%d$'\t'%H:%M:)                            
bash-3.2$ echo $Heartbeat
2015-12-16 08:16:

bash-3.2$ grep $Heartbeat file.txt is not working correct, but grep '2015-12-16 08:16:' file.txt works correctly.

Please help me how can we do this for variable.

shellter · Accepted Answer · 2015-12-16 13:45:03Z

4

grep -- "$Heartbeat" file

single quotes prevent variable expansion, double quotes allow variable expansion and keep space delimited value together as one argument.

In your case, we also need to add -- (end of arguments) as your $Heartbeat value contains - chars, which maybe being interpreted as grep options.

IHTH

edited Dec 16, 2015 at 13:45

answered Dec 16, 2015 at 13:36

shellter

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6 Comments

user4786572 Over a year ago

No Its not working Heartbeat=$(date +%Y-%m-%d$'\t'%H:%M:);grep "$Heartbeat" LOOPEUR2015-12-16.log;echo $Heartbeat;2015-12-16 08:41:;grep '2015-12-16 08:41:' LOOPEUR2015-12-16.log;2015-12-16 08:41:01,967 INFO in.FD_LOOPE_SSGA - <2220 Heartbeat No output when i use"" but we do get output when we pass as '2015-12-16 08:41:'

shellter Over a year ago

Check my updated answer. Also please be more specific when saying it doesn't work. Just no output, OR error messages?

shellter Over a year ago

thank you for specific details. Reading thru them now.

chepner Over a year ago

-- is only necessary if - were the first character $Heartbeat.

shellter Over a year ago

@user4786572 , I think chepner is right. I added -- as a last shot. I don't know why grep "$var" file doesn't return any output when grep '2015-..' file does. I'll hope that someone else can help you. (It should work). Good luck.

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