为什么json_decode对于空数组返回null？

Question

为什么会输出“NULL”？在我的理解中它应该被解码为空数组。

我是否漏掉了某些显而易见的东西？

<?php

$json = json_encode(array());
$json_decoded = json_decode($json, true);
// same with json_decode($json);

if ($json_decoded == null){
    echo "NULL";
} else
{
    echo "NOT NULL";
}

?>

- netdigger

---

如果你尝试 $json_decoded === null 会怎样？ - mgraph

@Aelios：在PHP中，数组没有new关键字。 - prodigitalson

你期望返回不同的东西吗？ - Sena

Answer 1

这是因为array()==NULL。在这种情况下，它不会检查对象类型。

gettype(null)返回null，而

gettype(array())返回array。希望你明白了它们之间的区别。

可能你需要的是

if ($json_decoded === null) {
   echo "NULL";
} else
{
   echo "NOT NULL";
}

- gopi1410

Answer 2

打印 $json_decoded 值，它会返回一个空数组。 :)

$json = json_encode(array());
$json_decoded = json_decode($json, true);


if ($json_decoded == null){
    print_r($json_decoded);
} else
{
    echo "NOT NULL";
}

输出：数组（）
这是因为使用 == 运算符时，空数组会被类型混淆为 null

- Mithun Satheesh

Answer 3

如果您的数据中包含一些\n，json_decode 也可能会默默地失败。

- Hafenkranich

Answer 4

这一定能解决问题

 if ($json_decoded === null)

来自手册的示例：

<?php
$a = array('<foo>',"'bar'",'"baz"','&blong&', "\xc3\xa9");

echo "Normal: ",  json_encode($a), "\n";
echo "Tags: ",    json_encode($a, JSON_HEX_TAG), "\n";
echo "Apos: ",    json_encode($a, JSON_HEX_APOS), "\n";
echo "Quot: ",    json_encode($a, JSON_HEX_QUOT), "\n";
echo "Amp: ",     json_encode($a, JSON_HEX_AMP), "\n";
echo "Unicode: ", json_encode($a, JSON_UNESCAPED_UNICODE), "\n";
echo "All: ",     json_encode($a, JSON_HEX_TAG | JSON_HEX_APOS | JSON_HEX_QUOT | JSON_HEX_AMP | JSON_UNESCAPED_UNICODE), "\n\n";

$b = array();

echo "Empty array output as array: ", json_encode($b), "\n";
echo "Empty array output as object: ", json_encode($b, JSON_FORCE_OBJECT), "\n\n";

$c = array(array(1,2,3));

echo "Non-associative array output as array: ", json_encode($c), "\n";
echo "Non-associative array output as object: ", json_encode($c, JSON_FORCE_OBJECT), "\n\n";

$d = array('foo' => 'bar', 'baz' => 'long');

echo "Associative array always output as object: ", json_encode($d), "\n";
echo "Associative array always output as object: ", json_encode($d, JSON_FORCE_OBJECT), "\n\n";
?>

输出：

Normal: ["<foo>","'bar'","\"baz\"","&blong&","\u00e9"]
Tags: ["\u003Cfoo\u003E","'bar'","\"baz\"","&blong&","\u00e9"]
Apos: ["<foo>","\u0027bar\u0027","\"baz\"","&blong&","\u00e9"]
Quot: ["<foo>","'bar'","\u0022baz\u0022","&blong&","\u00e9"]
Amp: ["<foo>","'bar'","\"baz\"","\u0026blong\u0026","\u00e9"]
Unicode: ["<foo>","'bar'","\"baz\"","&blong&","é"]
All: ["\u003Cfoo\u003E","\u0027bar\u0027","\u0022baz\u0022","\u0026blong\u0026","é"]

Empty array output as array: []
Empty array output as object: {}

Non-associative array output as array: [[1,2,3]]
Non-associative array output as object: {"0":{"0":1,"1":2,"2":3}}

Associative array always output as object: {"foo":"bar","baz":"long"}
Associative array always output as object: {"foo":"bar","baz":"long"}

- greenLizard

Answer 5

你需要使用严格相等运算符===，自己观察：

$json = json_encode(array());
var_dump($json); // string(2) "[]"
$json_decoded = json_decode($json, true);
var_dump($json_decoded); // array(0) { }

所以这样做：

$json = json_encode(array());
$json_decoded = json_decode($json, true);

if ($json_decoded === null) 
{
   echo "NULL";
} else
{
   echo "NOT NULL";
}

应该放在else条件中打印NOT NULL

- Sarfraz