curl在命令行上工作，但不在shell脚本中

Question

当我在命令行的curl命令中直接使用变量cookie的值时 - 它起作用;但它在脚本中不起作用。以下错误：

#!/bin/bash 

cookie=`tail -1000 cat.txt | grep -v "wm-ueug" | grep -v "JRECookie" | grep "JSESSIONID.*_ga=GA" | tail -1 | sed -r 's/.{26}//' | sed 's/.$//'` 

`curl "http://example.com/monitor?method=monitor&refresh=true&count=4&start=1&dateFrom=2017-02-21&dateTo=2017-02-28&runId=&hidden_status=&dojo.preventCache=1488290723103" -H "Host: example.com" -H "User-Agent: Mozilla/5.0 (Windows NT 6.1; WOW64; rv:50.0) Gecko/20400202 Firefox/50.0" -H "Accept: text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8" -H "Accept-Language: en-US,en;q=0.5" --compressed -H "Content-Type: application/x-www-form-urlencoded" -H "X-Requested-With: XMLHttpRequest" -H "Referer: http://example.com/monitor" -H "Cookie: $cookie" -H "Connection: keep-alive"` 

更新：删除返回刻度 - 现在看不到任何错误，但没有输出。

Answer 1

主要的变化使是运行curl命令subsitution之外：

# Not `curl ...` 
curl ... 

但是，您可能要分手的curl命令，使其更易于阅读和理解。

#!/bin/bash 

cookie=$(tail -1000 cat.txt | 
     grep -v "wm-ueug" | 
     grep -v "JRECookie" | 
     grep "JSESSIONID.*_ga=GA" | 
     tail -1 | sed -r 's/.{26}//' | sed 's/.$//') 

# Parameters can be passed to curl via the -d option, rather 
# than as a query string in the URL. 
url="http://example.com/monitor" 
parameters=(
    -d method=monitor 
    -d refresh=true 
    -d count=4 
    -d start=1 
    -d dateFrom=2017-02-21 
    -d dateTo=2017-02-28 
    -d runId=hidden_status 
    -d dojo.preventCache=1488290723103 
) 

headers=(
    -H "Host: example.com" 
    -H "User-Agent: Mozilla/5.0 (Windows NT 6.1; WOW64; rv:50.0) Gecko/20400202 Firefox/50.0" 
    -H "Accept: text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8" 
    -H "Accept-Language: en-US,en;q=0.5" 
    -H "Content-Type: application/x-www-form-urlencoded" 
    -H "X-Requested-With: XMLHttpRequest" 
    -H "Referer: http://example.com/monitor" 
    -H "Cookie: $cookie" 
    -H "Connection: keep-alive" 
) 
curl --compressed "${parameters[@]}" "${headers[@]}" "$url"