php 字符是否在json中,检查字符串是否是PHP中的JSON的最快方法？
回答问题
该函数json_last_error返回JSON编码和解码期间发生的最后一个错误。因此，检查有效JSON的最快方法是
// decode the JSON data
// set second parameter boolean TRUE for associative array output.
$result = json_decode($json);
if (json_last_error() === JSON_ERROR_NONE) {
// JSON is valid
}
// OR this is equivalent
if (json_last_error() === 0) {
// JSON is valid
}
请注意，json_last_errorPHP> = 5.3.0仅支持。
完整的程序来检查确切的错误
在开发期间知道确切的错误总是好的。这是完整的程序，以检查基于PHP文档的确切错误。
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;
}
使用有效的JSON INPUT进行测试
$json = '[{"user_id":13,"username":"stack"},{"user_id":14,"username":"over"}]';
$output = json_validate($json);
print_r($output);
有效的输出
Array
(
[0] => stdClass Object
(
[user_id] => 13
[username] => stack
)
[1] => stdClass Object
(
[user_id] => 14
[username] => over
)
)
使用无效的JSON进行测试
$json = '{background-color:yellow;color:#000;padding:10px;width:650px;}';
$output = json_validate($json);
print_r($output);
无效的输出
Syntax error, malformed JSON.
额外注意事项(PHP> = 5.2 && PHP <5.3.0)
由于json_last_errorPHP 5.2不支持，您可以检查编码或解码是否返回布尔值FALSE。这是一个例子
// decode the JSON data
$result = json_decode($json);
if ($result === FALSE) {
// JSON is invalid
}
希望这是有帮助的。快乐的编码！