分配变量时发生Python语法错误

Question

I am new to python I am trying to assign variable it says systax error

protocol = input.readLine()
connectUrlHttp='http'
connectUrlHttps='https'


if protocol== "t3s":
 connectUrl=connectUrlHttps
elif protocol== "iiops":
  connectUrl=connectUrlHttps
else:
connectUrl=connectUrlHttp


  sca_deployComposite(connectUrl"://"+host+":"+port,emdroot+"/"+compositeLoc,owrite,user, password,default)

I am facing below error

(no code object) at line 0
  File "/scratch/agentHome/sdappaji2/core/12.1.0.3.0/EMStage/PAF/DeployCompositesDP1367835748253/deploycompositesscripts/deployComposites.py", line 36
    connectUrl=connectUrlHttp

Please guide me and give some pointers

Answer 1

In Python, indentation levels are significant:

Use 4 spaces per indentation level.

-- PEP 8 -- Style Guide for Python Code

Your code should probably look like this:

protocol = input.readLine()
connectUrlHttp = 'http'
connectUrlHttps = 'https'

if protocol == "t3s":
    connectUrl = connectUrlHttps
elif protocol == "iiops":
    connectUrl = connectUrlHttps
else:
    connectUrl = connectUrlHttp

# Note the `+` after `connectUrl` on the next line:
sca_deployComposite(connectUrl + "://" + host + ":" + port, emdroot + "/" +
                    compositeLoc, owrite, user, password, default)

Answer 2

The actual syntax error is here:

sca_deployComposite(connectUrl"://"+host+":"+port,emdroot+"/"+compositeLoc,owrite,user, password,default)
                            ^^^

You probably missed a + there.

The indentation is also wrong after the else , as Ashwini points out.

Answer 3

Your code is not properly indented after else:

if protocol== "t3s":
 connectUrl=connectUrlHttps
elif protocol== "iiops":
  connectUrl=connectUrlHttps
else:
    connectUrl=connectUrlHttp

sca_deployComposite(connectUrl+"://"+host+":"+port,emdroot+"/"+compositeLoc,owrite,user, password,default)

Plus you were missing a + after sca_deployComposite(connectUrl .