64

So I have a script that, when I give it two addresses, will search two HTML links:

echo "http://maps.google.be/maps?saddr\=$1\&daddr\=$2" | sed 's/ /%/g'

I want to send this to wget and then save the output in a file called temp.html. I tried this, but it doesn't work. Can someone explain why and/or give me a solution please?

#!/bin/bash
url = echo "http://maps.google.be/maps?saddr\=$1\&daddr\=$2" |  sed 's/ /%/g'
wget $url
  • For debugging something like this checking your variable values (by echo-ing them to the terminal) often gets you to the solution quickly. – kasterma Dec 13 '10 at 13:52
63

You can use backticks (`) to evaluate a command and substitute in the command's output, like:

echo "Number of files in this directory: `ls | wc -l`"

In your case:

wget `echo http://maps.google.be/maps?saddr\=$1\&daddr\=$2 | sed 's/ /%/g'`
| improve this answer | |
27

You could use "xargs". A trivial example:

ls -1 *.c | sort -n | xargs cat

You would have to take care that xargs doesn't split its stdin into two or more invocations of the comman ("cat" in the example above).

| improve this answer | |
  • 1
    I ended up here, because I couldn't remember this command name. My favorite use case was indeed: xargs -I % some_command --input=% -e -t -c (I put it here as a note for others that follow the same trace) – mpasko256 Dec 12 '17 at 21:22
  • 1
    Hi @Bruce Ediger! I randomly stumbled across you when googling this :D Hope you're doing well! – frederix May 18 at 20:49
  • 2
    Hey @frederix who let you in here? I see they will let just anyone in here these days! LOLOLOLOL – abgordon May 18 at 20:53
17

you're not actually executing your url line :

#!/bin/sh
url="$(echo http://maps.google.be/maps?saddr\=$1\&daddr\=$2 | sed 's/ /%/g')"
wget $url
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5

It seems you could use a combination of the answers here. I'm guessing you are wanting to replace space chars with their escaped ascii values in the url. To do this, you need to replace them with "%20", not just "%". Here's a solution that should give you a complete answer:

$ wget `echo http://maps.google.be/maps\?saddr\=$1\&daddr\=$2 | sed -e 's/\ /\%20/g'` -q -O temp.html

The backticks indicate that the enclosed command should be interpreted first, and the result sent to wget. Notice I escaped the space and % chars in the sed command to prevent them from being misinterpreted. The -q option for wget prevents processing output from the command being printed to the screen (handy for scripting when you don't care about the in-work status) and the -O option specifies the output file. FYI, if you don't want to save the output to a file, but just view it in the terminal, use "-" instead of a filename to indicate stdout.

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3

wget also accepts stdin with the - switch.

If you want to save the output in a file, use the -O switch.

echo http://maps.google.be/maps?saddr\=$1\&daddr\=$2 | sed 's/ /%/g' | wget -i- -O temp.html
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0

xargs is the best option to place output from a command into the argument of other commmand.

Suppose the output of command1 is 3 and you want your next command to take this 3 as an argument so you want something like

command2 3(which is output of 1st command) 4 5

For that you can go like

command1 | xrgs -I{} command2 {} 4 5

where 4 and 5 are other two arguments that may be needed for command2.

You can put those curly brackets on the argument place where you want the output from first command.

So, use

command1 | xrgs -I{} command2 {} 
| improve this answer | |

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