Shell script: output text after cURL variable

Question

I am trying to do something simple but it's not working. I am trying to echo a stock price from a variable and follow it with text. The problem is that the text does not follow the variable and goes in front of it.

#!/bin/sh
stock=$(curl -s 'http://download.finance.yahoo.com/d/quotes.csv?s=AAPL&f=l1')
echo "$stock"

echo ".. $stock .."

the second output doesn't show correctly, what am I doing wrong?

Thx

linuxfan · Accepted Answer · 2014-10-22 20:43:47Z

1

The variable $stock has \r\n at its end.

$ echo $stock | hexdump -c 
0000000   1   0   2   .   9   9  \r  \n                                
0000008

You can modify your call to curl as

stock=$(curl -s 'http://download.finance.yahoo.com/d/quotes.csv?s=AAPL&f=l1'| tr -d '\r\n')

$ echo ".. $stock .."
.. 102.99 ..

answered Oct 22, 2014 at 20:43

linuxfan

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