Convert JSON integers and floats to strings

Question

I want to pre-parse a JSON and convert integer and float value to string in the JSON. Also there are some string values in JSON.

For example:

{
    "first_name": "sample",
    "last_name": "lastname",
    "integer" : 100,
    "float" : 1555.20
}

I just use preg_replace() like this from here:

echo preg_replace('/\: *([0-9]+\.?[0-9e+\-]*)/', '"\\1"', $jsonString);

But its not working if I have a string value in my array, it only works if there are only integer and float values in array.

Can anyone help explain why this happens?

RomanPerekhrest · Accepted Answer · 2016-01-26T10:11:52.450

2

Here is solution:

$str = '{"first_name":"sample",
          "last_name": "lastname",
          "integer" : 100,
          "float" : 1555.20,
          "createddate":"2015-06-25 09:57:28"}';

$result = preg_replace("/(\"\w+\":\s*?)(\d+\.?[^,\}]*\b)/imu",'$1"$2"',$str);

var_dump($result);

// output:
string(121) "{"first_name":"sample",
"last_name": "lastname",
"integer" : 100,
"float" : 1555.20,
"createddate":"2015-06-25 09:57:28"}"

edited Jan 26 '16 at 10:11

answered Jan 26 '16 at 08:41

RomanPerekhrest

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this code works fine but when i add a new field datetime then this code gives output like this : "2015-06-25 08:"15":"14"" . – Nirali Kavar Jan 26 '16 at 08:58
it gives the same output. – Nirali Kavar Jan 26 '16 at 09:40
@NiraliKavar, strange ... I suppose that we have worked with different initial strings. What is your initial string? – RomanPerekhrest Jan 26 '16 at 09:43
my initial string is like that : {"first_name":"sample","last_name": "lastname","integer" : 100, "float" : 1555.20, "createddate":"2015-06-25 09:57:28"}. – Nirali Kavar Jan 26 '16 at 10:03
now json is generate as i want but it gives one more javascript error like that : TypeError: float.toFixed is not a function.. – Nirali Kavar Jan 26 '16 at 10:51
Your initial task was "convert integer and float value to string" and `toFixed` isn't a property of non-numeric variable types. Use this approach: `parseFloat(float).toFixed(2)` – RomanPerekhrest Jan 26 '16 at 11:13

Will · Answer 2 · 2016-01-26T07:34:53.127

1

This sounds like a job for lookaheads and lookbehinds!

$newJson = preg_replace('/("\s*:\s*)(?<!")[\d\.]+(?!")/', '\1"\2"', $json);

You can visualize this here.

(?<!") is a negative lookbehind. It is a "zero length expression" meaning that it doesn't actually match a character; it just prevents a match from occurring if a " was there. (?!") is a negative lookahead, and works the same way, but matches forward instead of backward.

edited Jan 26 '16 at 07:34

answered Jan 26 '16 at 07:04

Will

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I have one more field for date and its showing me like this, entry_date":"2"015"-"06"-"25" "10": how can i resolve this ? – Nirali Kavar Jan 26 '16 at 07:10
i used that new string from your edit but it makes all integer value blank,, Like this [ "integer"":" ] thanks Will :) – Nirali Kavar Jan 26 '16 at 08:38

Convert JSON integers and floats to strings

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