将变量传递给curl命令以调用rest api url n python

Question

我正在尝试使用变量triggername传递报告标题名称。当我执行代码时，我在api url中没有捕获triggername的值。

有人可以帮我解决这个问题吗？

**#!/usr/bin/python2.7
import requests
import os
import commands
triggername="GSD_Trigger_Active"
print("**********",triggername, type(triggername))
cmd="""curl -v -u $(ops_unv_cred_user_35ab1e1285be4e659f9ab15fca0a35fc):$(ops_unv_cred_pwd_35ab1e1285be4e659f9ab15fca0a35fc)  -H "Accept: application/json" -X GET 'https://xxxxx.cloud/resources/report/run?reporttitle=triggername'
"""
os.system(cmd)

我收到以下错误消息：

Could not find report with name "triggername"

Answer 1

认为您需要对变量进行插值。最简单的方法是这样。

triggername = "GSD_Trigger_Active"
cmd= f'curl -v -u $(ops_unv_cred_user_35ab1e1285be4e659f9ab15fca0a35fc):$(ops_unv_cred_pwd_35ab1e1285be4e659f9ab15fca0a35fc)  -H "Accept: application/json" -X GET "https://xxxxx.cloud/resources/report/run?reporttitle={triggername}"'

这应该给你的cmd为'curl -v -u $(ops_unv_cred_user_35ab1e1285be4e659f9ab15fca0a35fc):$(ops_unv_cred_pwd_35ab1e1285be4e659f9ab15fca0a35fc) -H "Accept: application/json" -X GET "https://xxxxx.cloud/resources/report/run?reporttitle=GSD_Trigger_Active"'

添加屏幕截图：

python3.6之前使用此：

cmd = 'curl -v -u $(ops_unv_cred_user_35ab1e1285be4e659f9ab15fca0a35fc):$(ops_unv_cred_pwd_35ab1e1285be4e659f9ab15fca0a35fc)  -H "Accept: application/json" -X GET "https://xxxxx.cloud/resources/report/run?reporttitle={}"'.format(triggername)