Can’t output json_encode with PHP7.3 [duplicate]? Don’t Panic, We’ve Got You Covered!
Image by Leonard - hkhazo.biz.id

Can’t output json_encode with PHP7.3 [duplicate]? Don’t Panic, We’ve Got You Covered!

By Posted on

Are you stuck with the infamous “can’t output json_encode with PHP7.3” error? Fear not, dear developer, for we’re about to dive into the depths of this frustrating issue and emerge victorious with a solution that’ll make your JSON encoding woes disappear like magic!

Table of Contents
  1. The Problem: Understanding the Error
  2. Solution 1: Validate Your Data Before Encoding
  3. Solution 2: Use the Correct Data Types
  4. Solution 3: Use the JSON_FORCE_OBJECT Option
  5. Solution 4: Update Your PHP Version (If Possible)
  6. Solution 5: Use a Third-Party Library (If All Else Fails)
  7. Conclusion
    1. Bonus Tip: JSON Encoding Best Practices

The Problem: Understanding the Error

Before we tackle the solution, let’s quickly grasp the essence of the problem. When you encounter the “can’t output json_encode with PHP7.3” error, it’s often due to one of the following reasons:

  • Incorrect data types being passed to json_encode()
  • Invalid or corrupted data in the array/object being encoded
  • PHP7.3’s new JSON encoding settings and restrictions
  • Compatibility issues with older PHP versions

Solution 1: Validate Your Data Before Encoding

Ensure that the data you’re trying to encode is valid and correctly formatted. A single misplaced or invalid character can cause the entire encoding process to fail.


 'John Doe',
    'age' => 30,
    ' occupation' => 'Developer' // Note the space in the key
  );

  // This will fail because of the space in the key
  echo json_encode($data);
?>

In the above example, the space in the key ” occupation” will cause json_encode() to throw an error. Remove the space, and voilà! The encoding process will succeed.

Solution 2: Use the Correct Data Types

PHP7.3 has become more stringent when it comes to data types. Make sure you’re passing the correct data types to json_encode().


 'John Doe',
    'age' => '30' // Incorrect data type, should be an integer
  );

  // This will fail because of the incorrect data type
  echo json_encode($data);
?>

In this example, the ‘age’ key has a string value instead of an integer. Correct the data type to integer, and the encoding will work as expected.

Solution 3: Use the JSON_FORCE_OBJECT Option

If you’re encountering issues with json_encode() returning an empty string or boolean values, try using the JSON_FORCE_OBJECT option.

Solution 4: Update Your PHP Version (If Possible)

If you’re still stuck with an older PHP version, it’s worth considering an upgrade to PHP7.3 or later. The newer versions have resolved many JSON encoding issues and introduced more robust error handling.

PHP Version JSON Encoding Support
PHP 5.x Poor support, prone to errors
PHP 7.0-7.2 Better support, but still some issues
PHP 7.3+ Robust support, improved error handling

Solution 5: Use a Third-Party Library (If All Else Fails)

In rare cases where none of the above solutions work, you might need to resort to using a third-party library like json-picker or json-serialize.


 'bar');

  $json = JsonPicker::encode($data);

  echo $json;
?>

Conclusion

By following these solutions, you should be able to resolve the “can’t output json_encode with PHP7.3” error and successfully encode your data into JSON format. Remember to validate your data, use the correct data types, and take advantage of PHP7.3’s improved JSON encoding features.

If you’re still encountering issues, don’t hesitate to seek help from online communities, forums, or consult with a seasoned developer. Happy coding, and may your JSON encoding woes become a thing of the past!

Bonus Tip: JSON Encoding Best Practices

To avoid common JSON encoding pitfalls, follow these best practices:

  1. Use a consistent data structure and formatting
  2. Validate and sanitize your data before encoding
  3. Use the correct data types and encoding options
  4. Test your JSON output regularly to catch errors early
  5. Consider using a JSON schema validator for added security

By following these best practices, you’ll be well on your way to becoming a JSON encoding master and avoiding the “can’t output json_encode with PHP7.3” error for good!

Frequently Asked Question

Stuck with json_encode not working in PHP 7.3? Don’t worry, we’ve got you covered! Here are some frequently asked questions and answers to help you troubleshoot and solve the issue:

Q1: Why is json_encode not working in PHP 7.3?

This is likely due to the fact that PHP 7.3 has more strict type checking, which can cause issues with json_encode. Specifically, if you’re passing an object with properties that are not serializable (e.g., resources, closures), json_encode will fail.

Q2: How do I debug json_encode errors in PHP 7.3?

You can use the json_last_error() function to get the last error message. Additionally, you can use the json_last_error_msg() function to get a human-readable error message.

Q3: Can I use json_encode with objects that have private or protected properties?

No, json_encode will not serialize private or protected properties by default. You can use the JsonSerializable interface or implement a custom serialize method to include these properties.

Q4: Why is json_encode returning an empty string?

This can happen if the input data is not serializable or if there’s an error during the encoding process. Check the input data and ensure it’s valid and serializable.

Q5: Are there any alternatives to json_encode in PHP 7.3?

Yes, you can use the Zend Json component or other third-party libraries like JsonSerializable or Symfony’s JsonSerializer.