Answers for "php is json"

PHP
2

php check if json

function isJson($string) {
 json_decode($string);
 return (json_last_error() == JSON_ERROR_NONE);
}
Posted by: Guest on April-09-2020
15

json php

//Json Encode

$person = array( 
    "name" => "KINGASV", 
    "title" => "CTO"
); 
$personJSON=json_encode($person);//returns JSON string

//Json Decode

$personJSON = '{"name":"KINGASV","title":"CTO"}';

$person = json_decode($personJSON);

echo $person->name; // KINGASV
Posted by: Guest on July-16-2020
7

php is json string

function IsJsonString(str) {
    try {
        JSON.parse(str);
    } catch (e) {
        return false;
    }
    return true;
}
Posted by: Guest on April-10-2020
1

php if is json object

//Simple
if (is_object(json_decode($var))) { 
  ....
}

//Else
var $x = json_decode($var);
var $y = is_object($x)?$x:....;

//Better
function json_validate($string) {
    // decode the JSON data
    $result = json_decode($string);

    // switch and check possible JSON errors
    switch (json_last_error()) {
        case JSON_ERROR_NONE:
            $error = ''; // JSON is valid // No error has occurred
            break;
        case JSON_ERROR_DEPTH:
            $error = 'The maximum stack depth has been exceeded.';
            break;
        case JSON_ERROR_STATE_MISMATCH:
            $error = 'Invalid or malformed JSON.';
            break;
        case JSON_ERROR_CTRL_CHAR:
            $error = 'Control character error, possibly incorrectly encoded.';
            break;
        case JSON_ERROR_SYNTAX:
            $error = 'Syntax error, malformed JSON.';
            break;
        // PHP >= 5.3.3
        case JSON_ERROR_UTF8:
            $error = 'Malformed UTF-8 characters, possibly incorrectly encoded.';
            break;
        // PHP >= 5.5.0
        case JSON_ERROR_RECURSION:
            $error = 'One or more recursive references in the value to be encoded.';
            break;
        // PHP >= 5.5.0
        case JSON_ERROR_INF_OR_NAN:
            $error = 'One or more NAN or INF values in the value to be encoded.';
            break;
        case JSON_ERROR_UNSUPPORTED_TYPE:
            $error = 'A value of a type that cannot be encoded was given.';
            break;
        default:
            $error = 'Unknown JSON error occured.';
            break;
    }

    if ($error !== '') {
        // throw the Exception or exit // or whatever :)
        exit($error);
    }
    // everything is OK
    return $result;
}
$output = json_validate($json);
Posted by: Guest on March-30-2020

Browse Popular Code Answers by Language