- How to Check If String is a Json in PHP?
- You might also like.
- PHP RFC: is_json
- Proposal
- Description
- Parameters
- Return values
- Extra behavior
- Examples
- Fundaments/Reasons
- Disadvantages of using json_decode
- Disadvantages of using regex
- Needs from major projects and developers
- Complexity added in the core
- Backward Incompatible Changes
- Proposed PHP Version(s)
- RFC Impact
- Check if String is a Valid Json in PHP
- Method 1: Using json_decode()
- Frequently Asked:
- Method 2: Using json_decode() with error suppression
- Summary
- Related posts:
- Share your love
- Leave a Comment Cancel Reply
- Terms of Use
- Disclaimer
How to Check If String is a Json in PHP?
Here, I will show you how to works How to check if string is a json in PHP. you can see Validate json in PHP example. we will help you to give example of Check if string is a json or not. you can understand a concept of Json functions and operators in PHP.
json_decode($data, true) will return an object converted into an associative array, and then the is_array() function checks whether the object is an array or not. The json_last_error() returns the last error if there were any errors when json_decode() was invoked.
function json_validator($data) <
if (!empty($data)) <
return is_string($data) &&
is_array(json_decode($data, true)) ? true : false;
>
return false;
>
$data1 = »;
$data2 = »;
echo «Result for data1: «;
echo (json_validator($data1) ? «JSON is Valid\n»
: «JSON is Not Valid\n»);
echo «Result for data2: «;
echo (json_validator($data2) ? «JSON is Valid»
: «JSON is Not Valid»);
?>
Result for data1: JSON is Valid
Result for data2: JSON is Not Valid
function json_validator($data) <
if (!empty($data)) <
@json_decode($data);
return (json_last_error() === JSON_ERROR_NONE);
>
return false;
>
$data1 = »;
$data2 = »;
echo «Result for data1: «;
echo (json_validator($data1) ? «JSON is Valid\n»
: «JSON is Not Valid\n»);
echo «Result for data2: «;
echo (json_validator($data2) ? «JSON is Valid»
: «JSON is Not Valid»);
?>
Result for data1: JSON is Valid
Result for data2: JSON is Not Valid
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PHP RFC: is_json
This RFC introduces a new function called is_json() to validate if an string contains a valid json.
Most userland implementations to achieve this are done using json_decode() which by design generates an object/array while parsing the string, ergo using memory and processing, that could be save.
Proposal
Description
is_json(string $json, int $depth = 512, int $flags = 0): bool
Parameters
The json string being analyzed.
This function only works with UTF-8 encoded strings.
PHP implements a superset of JSON as specified in the original » RFC 7159.
Maximum nesting depth of the structure being decoded.
Bitmask of JSON_INVALID_UTF8_IGNORE, JSON_THROW_ON_ERROR. The behavior of these constants is described on the JSON constants page.
Function description, examples, technical strategy, current JSON parser, etc.
Return values
Returns true if the string passed contains a valid json, otherwise returns false.
Extra behavior
Examples
Fundaments/Reasons
Disadvantages of using json_decode
By design, json_decode() generates an object/array while parsing the string, ergo using memory and processing for it, that is not needed if the only thing to discover is if a string contains a valid json.
Disadvantages of using regex
Using a regex for this task forces different, error-prone, hard to maintain, implementations.
Needs from major projects and developers
In the “References” section, there is a list of major open-source php projects needing this feature; also in th mntioned section can find a link to one of the most popular StackOverflow questions, which somehow reflects the need from our developers to have a feature like this included.
Complexity added in the core
At the moment, there is a JSON parser in the core, used by json_decode to do its job, so there is no need to write a new JSON parser for this RFC ; the proposed function will use the existing JSON parser exclusively to parse an string without generating any object/array in memory for it.
Backward Incompatible Changes
None, as this is a new function only.
is_json will no longer be available as a function name, could break potential userland implementations.
Proposed PHP Version(s)
RFC Impact
This RFC has no impact on SAPIs, existing extensions, Opcache, etc.
Check if String is a Valid Json in PHP
This tutorial will discuss about unique ways to check if string is a valid json in php.
Table Of Contents
Method 1: Using json_decode()
The json_decode() accepts a JSON encoded string as parameter, and if it is a valid json string then it converts the string into a PHP value. Where if it is not a valid JSON then the last error will not be JSON_ERROR_NONE.
We have create a function to validate JSON string i.e.
It accepts a string as argument and returns true if given string is a valid JSON string.
Frequently Asked:
Let’s see the complete example,
$strValue = '< "name": "Ritika", "age" : 32, "city": "London" >'; if (isStringJson($strValue)) < echo "The string represents valid JSON."; >else < echo "The string does not represent valid JSON."; >?>
The string represents valid JSON.
Method 2: Using json_decode() with error suppression
If error suppression is enabled, then we can rely on the return valud of json_decode() .
The json_decode() accepts a JSON encoded string as parameter, and if it is a valid json string then it converts the string into a PHP value. If it is not a valid JSON string then it returns null.
We have create a function to validate JSON string i.e.
It accepts a string as argument and returns true if given string is a valid JSON string.
Let’s see the complete example,
$strValue = ''; if (isValidJSON($strValue)) < echo "The string represents valid JSON."; >else < echo "The string does not represent valid JSON."; >?>
The string represents valid JSON.
Summary
We learned about two ways to check if a string is a valid JSON string or not in PHP.
Related posts:
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