String format a JSON string gives KeyError

Question

Why does this code give a KeyError?

output_format = """
{ 
    "File": "{filename}", 
    "Success": {success}, 
    "ErrorMessage": "{error_msg}", 
    "LogIdentifier": "{log_identifier}" 
}
"""

print output_format.format(filename='My_file_name',
                           success=True,
                           error_msg='',
                           log_identifier='123')

Error message:

KeyError: ' "File"'

Martijn Pieters · Accepted Answer · 2019-11-19T23:50:24.277

75

You need to double the outer braces; otherwise Python thinks { "File".. is a reference too:

output_format = '{{ "File": "{filename}", "Success": {success}, "ErrorMessage": "{error_msg}", "LogIdentifier": "{log_identifier}" }}'

Result:

>>> print output_format.format(filename='My_file_name',
...                            success=True,
...                            error_msg='',
...                            log_identifier='123')
{ "File": "My_file_name", "Success": True, "ErrorMessage": "", "LogIdentifier": "123" }

If, indicentally, you are producing JSON output, you'd be better off using the json module:

>>> import json
>>> print json.dumps({'File': 'My_file_name',
...                   'Success': True,
...                   'ErrorMessage': '',
...                   'LogIdentifier': '123'})
{"LogIdentifier": "123", "ErrorMessage": "", "Success": true, "File": "My_file_name"}

Note the lowercase true in the output, as required by the JSON standard.

edited Nov 19 '19 at 23:50

answered May 03 '13 at 10:31

Martijn Pieters

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1

If I could give this answer 10 upvotes I would. Thanks – Brian Moeskau Jun 16 '17 at 20:44
1

I'll add this comment as an answer for clarity. Note that for nested dictionaries you need to double up on any { and } of the nested dictionaries. – RaamEE Jul 26 '18 at 12:24
3

@RaamEE: you need to escape **all** `{..}` pairs that are not placeholders. – Martijn Pieters Jul 26 '18 at 14:15
There's a typo on the `output_format ` assignment: isn't it `"Success": "{success}"` ? – andreagalle Aug 23 '23 at 17:26
1

@andreagalle: no, because the `success` value is not a string. Note that the _question_ set the format here. – Martijn Pieters Aug 23 '23 at 21:34

score 4 · Answer 2 · answered Oct 20 '20 at 17:43

As mentioned by Tudor in a comment to another answer, the Template class was the solution that worked best for me. I'm dealing with nested dictionaries or list of dictionaries and handling those were not as straightforward.

Using Template though the solution is quite simple.

I start with a dictionary that is converted into a string. I then replace all instances of { with ${ which is the Template identifier to substitute a placeholder.

The key point of getting this to work is using the Template method safe_substitute. It will replace all valid placeholders like ${user_id} but ignore any invalid ones that are part of the dictionary structure, like ${'name': 'John', ....

After the substitution is done I remove any leftovers $ and convert the string back to a dictionary.

In the code bellow, resolve_placeholders returns a dictionary where each key matches a placeholder in the payload string and the value is substituted by the Template class.

from string import Template
.
.
.
payload = json.dumps(payload)
payload = payload.replace('{', '${')
replace_values = self.resolve_placeholders(payload)
if replace_values:
    string_template = Template(payload)
    payload = string_template.safe_substitute(replace_values)
payload = payload.replace('${', '{')
payload = json.loads(payload)

I used this approach because it's quite elegant after utilizing RegEx and a negative lookbehind to allow my JSON templates to still use `${` to clearly indicate template tags. So your `payload.replace( ... )` becomes `re.sub(r"(?<!\$)({)", '${', payload)` to only replacing `{` and ignore all occurrences of `${`. The final step is okay for `.replace( ...)` but for completeness I used `re.sub(r"(\$\{)", '{', payload)` — theVinchi, Dec 09 '20 at 18:39

score 2 · Answer 3 · edited May 26 '20 at 11:29

To extend on Martijn Pieters answer and comment:

According to MArtijn' comment, escaping the {..} pairs that are not placeholders is they way to go with nested dictionaries. I haven't succeded in doing that, so I suggest the following method.
For nested dictionaries I tried doubling up on any { and } of the nested dictionaries.

a='{{"names":{{"a":"{name}"}}}}'

a.format(name=123) output:

output: '{"names":{"a":"123"}}'

But this makes using format to change values inside a json string, a over-complex method, so I use a twist on the format command. I replace ${param_name} in a json string. For example:

My predefined JSON looks like this:

my_json_dict = {
     'parameter': [
         {
             'name': 'product',
             'value': '${product}'
         },
         {
             'name': 'suites',
             'value': '${suites}'
         },
         {
             'name': 'markers',
             'value': '${markers}'
         }
     ]
 }

I provide this dictionary as values to replace instead of the parameters

parameters = {
    'product': 'spam',
    'suites': 'ham',
    'markers': 'eggs'
}

And use this code to do the replacment

json_str = json.dumps(my_json_dict)
for parameter_name, parameter_value in parameters.iteritems():
        parameter_name = '${'+parameter_name+'}'
        json_str = json_str.replace(parameter_name, parameter_value)

json_dict = json.loads(json_str)

Maybe [string.Template](https://docs.python.org/3/library/string.html#template-strings) might help. — Tudor, Feb 07 '19 at 15:02
You'd be better off just using Python dictionaries and then using `json.dumps()` to output the JSON format. — Martijn Pieters, Aug 23 '23 at 21:35

String format a JSON string gives KeyError

3 Answers3

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