[SOLVED] Help with "sed"

wh33t · 12-10-2014, 08:31 PM

Hey LQ,

I just followed this example here: http://www.isaacsukin.com/news/2013/...iles-directory

I'm using this command to update some PHP scripts and I need to find a pattern and replace it with a $var. Sed doesn't seem to want to work with the $ sign as it does everything else I expect it to except insert $var.

Any ideas?

Code:

grep -lr --exclude-dir=".git" -e "oldword" . | xargs sed -i "s/oldword/newword/g"

grail · 12-10-2014, 08:34 PM

Doesn't seem to be any $ signs in your example??

wh33t · 12-10-2014, 08:46 PM

Quote:

Originally Posted by grail

Doesn't seem to be any $ signs in your example??

I was just pasting the generic command. Would it actually help if I showed you I wanted a dollar sign in it?

Code:

grep -lr --exclude-dir=".git" -e "someword" . | xargs sed -i "s/someword/$var,someword/g"

SAbhi · 12-10-2014, 11:35 PM

WARNING: Do not use any copied code untill you are quite sure what it is doing.
One more thing, Please show us some effort of your own instead of copying code and asking us to correct it.

Quote:

Originally Posted by wh33t

Code:

grep -lr --exclude-dir=".git" -e "someword" . | xargs sed -i "s/someword/$var,someword/g"

The grep will list all files having the someword in your curent directory unless you exclude, so any mistake of yours can give you a bad day at work.
I would prefer wrapping the variables

Code:

${variable}
------------

$var,someword
would be treated as a variable name unless you separate the variable to be identified by the nmae you gave it.

wh33t · 12-11-2014, 02:13 AM

Quote:

Originally Posted by SAbhi

WARNING: Do not use any copied code untill you are quite sure what it is doing.
One more thing, Please show us some effort of your own instead of copying code and asking us to correct it.

The grep will list all files having the someword in your curent directory unless you exclude, so any mistake of yours can give you a bad day at work.
I would prefer wrapping the variables

Code:

${variable}
------------

$var,someword
would be treated as a variable name unless you separate the variable to be identified by the nmae you gave it.

I apologize but I do not administer linux systems often. This is a random happenstance that comes up from time to time. Isn't the point of forums to get help? Perhaps you don't realize how I like to learn. I need examples to build a bit of confidence and then I go from there. Linux system administration to me, is a very daunting task. I'm always intimidated by it.

Would you mind explaining how I can wrap "$var" I tried with single quotes and it didn't work.

astrogeek · 12-11-2014, 02:19 AM

Quote:

Originally Posted by wh33t

I apologize but I do not administer linux systems often. This is a random happenstance that comes up from time to time. Isn't the point of forums to get help? Perhaps you don't realize how I like to learn. I need examples to build a bit of confidence and then I go from there. Linux system administration to me, is a very daunting task. I'm always intimidated by it.

Would you mind explaining how I can wrap "$var" I tried with single quotes and it didn't work.

He means to "wrap" it in the curlybraces as shown...

Code:

$variable
...when wrapped becomes...
${variable}

... where the variable name goes inside the {...}.

UPDATE ****

But this is really only relevant if it goes into a shell script - potentially a problem inside a PHP script which is your actual target! So see next reply...

astrogeek · 12-11-2014, 02:29 AM

But now that I have read the whole thread (I had only read the last post previously...) I see that no one actually answered your question...

Try this...

Code:

grep -lr --exclude-dir=".git" -e "someword" . | xargs sed -i "s/someword/\$var,someword/g"

Note the backslash before the $.

In the context of the resultant php file the wrapping is potentially irrelevant or even an error, so do it as shown above.

As cautioned by others - test this first on an example file - do not copy, paste and use on your live files!

If you could post a snippet of an actual PHP file that you want to change, including an example of exactly what you want it changed to, we could give you a working example much more easily.

grail · 12-11-2014, 07:56 AM

Could you not just use single quotes to stop the variable from expanding ... or have I missed something?

SAbhi · 12-11-2014, 09:29 PM

Quote:

Originally Posted by wh33t

with the $ sign as it does everything else I expect it to except insert $var.

Code:

grep -lr --exclude-dir=".git" -e "someword" . | xargs sed -i "s/someword/$var,someword/g"

I think i misunderstood the question, Sorry if it is so. You want $var to be inserted in the file but not its value, right ?

you can follow @astrogeek or @grail's solution.

wh33t · 12-11-2014, 09:33 PM

Quote:

Originally Posted by astrogeek

But now that I have read the whole thread (I had only read the last post previously...) I see that no one actually answered your question...

Try this...

Code:

grep -lr --exclude-dir=".git" -e "someword" . | xargs sed -i "s/someword/\$var,someword/g"

Note the backslash before the $.

In the context of the resultant php file the wrapping is potentially irrelevant or even an error, so do it as shown above.

As cautioned by others - test this first on an example file - do not copy, paste and use on your live files!

If you could post a snippet of an actual PHP file that you want to change, including an example of exactly what you want it changed to, we could give you a working example much more easily.

Ahh yes, I knew there would be a way to escape characters somehow. This is it. Thank you.

astrogeek · 12-12-2014, 12:39 AM

Quote:

Originally Posted by wh33t

Ahh yes, I knew there would be a way to escape characters somehow. This is it. Thank you.

You are welcome, glad that worked!