How to define a string as a variable in python

Question

I'm trying to get a string as a variable to represent my column in DB.

l_id =1
level=7
test = "level"+level
device_level_mapping = DeviceLevelMapping.objects.filter(test = l_id)

It says keyword cant be an expression.

I need to do something like this. In PHP we do like this.

$a = "test"+"1";
$$a = 10;
echo $test1; // now test1 has become a variable.

How do I make this happen in python?

Is this even legal code? `'level'+7` should fail. Also, I see no exceptions that you are trying to raise. I suspect you have posted the wrong section of code — inspectorG4dget, Oct 30 '13 at 06:57
Still not legal code. You have one extra `)` in the third line. Also, what is the `= l_id` doing there. Also, what is the value of `l_id`? — inspectorG4dget, Oct 30 '13 at 07:02
In my opinion the real problem is: The DB layout is wrong and now you're going to work around it by even uglier code. — Matthias, Oct 30 '13 at 07:03
@inspectorG4dget I gave both of them in the answer, I dont know why I am still downvoted. :( — thefourtheye, Oct 30 '13 at 07:04
@thefourtheye: I didn't downvote you. IMHO the question is far too underspecified for me to be able to provide/comprehend an answer. Therefore, I cannot vote on any answer. I daresay you didn't deserve the downvote, though. Also, it seems that OP wants to troll on us (see previous comment at me). I wonder if this is the `mod_wsgi` troll we saw a while back — inspectorG4dget, Oct 30 '13 at 07:07
@user1162512 Be nice to people. inspectorG4dget points out what is wrong with your question. Please fix your question first. — thefourtheye, Oct 30 '13 at 07:08
@user1162512: it is up to you, as the OP, to provide valid code, from which we can reproduce your errors, so that we can help you (see [SSCCE](http://sscce.org/) for more. — inspectorG4dget, Oct 30 '13 at 07:10
@tripleee - I have already gone through this thread and I didnt got the expected o/p. Any more suggestions please with a demo code. Thanks — PythonEnthusiast, Oct 30 '13 at 07:21

score 1 · Accepted Answer · answered Oct 30 '13 at 07:35

1

test = "level"+level
d = {test: l_id}
device_level_mapping = DeviceLevelMapping.objects.filter(**d)

answered Oct 30 '13 at 07:35

Naster

344
1
7

How to define a string as a variable in python

1 Answers1