Resolve PHP json decode error: Difference between revisions
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* [https://php.net/manual/en/function.json-last-error.php PHP: json_last_error - Manual] | * [https://php.net/manual/en/function.json-last-error.php PHP: json_last_error - Manual] | ||
* [https://github.com/Seldaek/jsonlint Seldaek/jsonlint: JSON Lint for PHP] | * [https://github.com/Seldaek/jsonlint Seldaek/jsonlint: JSON Lint for PHP] | ||
* [https://www.php.net/manual/en/class.jsonexception.php PHP: JsonException - Manual] (PHP 7.3.0) | |||
[[Category:Data Science]] [[Category:PHP]] [[Category:Json]] [[Category: | [[Category:Data Science]] [[Category:PHP]] [[Category:Json]] [[Category:String manipulation]] | ||
Latest revision as of 19:32, 9 March 2021
Syntax error, malformed JSON[edit]
Possible causes
- the input is empty entirely or combination of multiple white spaces
- the input contains Byte order mark (BOM)[1]
- One file was combined with multiple JSON which were separated by return symbol
- Use ' instead of " for quotation
Solution:
- Not allowed empty input
- Remove Byte order mark (BOM)
- Use " for quotation
- Use JSONLint - The JSON Validator to validate the JSON format
Unexpected control character found[edit]
Possible causes
- the input string contains control characters
Solution:
- Remove control character. Using PHP:
$replacement = '';
preg_replace('/[\x00-\x1F]/', $replacement, $input);