blocks|key|3095055|text|在JSON中，数组[]只有In有数字键，而对象{}有string属性。在第二个示例中包含数组键，必然会迫使整个外部结构成为一个对象。因为包含了字符串键a,b,c,d，所以这两个示例的内部对象都作为对象。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3095056|如果在第一个示例中使用JSON_FORCE_OBJECT选项，应该会得到与第二个类似的结构，外部结构是对象而不是数组。如果没有将其指定为对象，则外部数组中缺少字符串键会导致PHP假定它将被编码为JSON中的等效数组结构。|3095057|$arrytest+=+array(array('a'=>1,+'b'=>2),array('c'=>3),array('d'=>4));

//+Force+the+outer+structure+into+an+object+rather+than+array
echo+json_encode($arrytest+,+JSON_FORCE_OBJECT);

//+{"0":{"a":1,"b":2},"1":{"c":3},"2":{"d":4}}|code-block|syntax|javascript|3095058|entityMap^0|9|2|N|2|23|7|0|B|H|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|N|$]]

In JSON, arrays <code>[]</code> only every have numeric keys, whereas objects <code>{}</code> have string properties. The inclusion of a array key in your second example forces the entire outer structure to be an object by necessity. The inner objects of both examples are made as objects because of the inclusion of string keys <code>a,b,c,d</code>.

If you were to use the <code>JSON_FORCE_OBJECT</code> option on the first example, you should get back a similar structure to the second, with the outer structure an object rather than an array. Without specifying that you want it as an object, the absence of string keys in the outer array causes PHP to assume it is to be encoded as the equivalent array structure in JSON.

<pre><code>$arrytest = array(array('a'=&gt;1, 'b'=&gt;2),array('c'=&gt;3),array('d'=&gt;4));

// Force the outer structure into an object rather than array
echo json_encode($arrytest , JSON_FORCE_OBJECT);

// {"0":{"a":1,"b":2},"1":{"c":3},"2":{"d":4}}
</code></pre>

blocks|key|226512|text|具有连续数字键的数组被编码为JSON数组。事情就是这样。为什么？因为这是有道理的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|226513|由于键可以通过数组编码隐式表示，因此没有理由显式地将它们编码为对象键。|226514|请参阅json_encode+documentation中的所有示例。|offset|length|style|CODE|226515|entityMap|0|LINK|mutability|MUTABLE|url|http://php.net/manual/en/function.json-encode.php^0|0|0|3|B|3|P|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|T|8|@$F|U|G|V|H|I]]|9|@$F|W|G|X|1|Y]]|A|$]]|$1|J|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|K|$L|$5|M|N|O|A|$P|Q]]]]

Arrays with continuous numerical keys are encoded as JSON arrays. That's just how it is. Why? Because it makes sense.

Since the keys can be expressed implicitly through the array encoding, there is no reason to explicitly encoded them as object keys.

See all the examples in the <a href="http://php.net/manual/en/function.json-encode.php" rel="nofollow"><code>json_encode</code> documentation</a>.

blocks|key|228741|text|在第一个选项中，您只有数字索引(0、1和2)。尽管它们没有显式声明，但php会自动创建它们。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|228742|在第二种选择中，您将其中一个索引声明为字符串，这将使PHP在内部将所有索引转换为字符串。|228743|当您对第一个数组进行json编码时，没有必要在生成的json字符串中显示整数，因为任何解码器都应该能够“猜测”它们是0、1和2。|228744|但在第二个数组中，这是必要的，因为解码器必须知道数组中的密钥值。|228745|这相当简单。是否未在数组中声明索引？它们是0，1，2，3等等。|228746|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|N|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|O|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|P|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|K|$]]

At the first option you only have numeric indexes (0, 1 and 2). Although they are not explicitly declared, php automatically creates them.

At the second option, you declare one of the indexes as an string and this makes PHP internally transform all indexes to string.

When you json encode the first array, it's not necessary to show the integers in the generated json string because any decoder should be able to "guess" that they are 0, 1 and 2.

But in the second array, this is necessary, as the decoder must know the key value in your array.

It's pretty simple. No indexes declared in array? Them they are 0, 1, 2, 3 and so on.

blocks|key|227078|text|json格式的输出是year1{a，b}，year2{c}，year3{d}+**a的值为1，b=2，c=3，d=4分别存储在年份1的a，b年份2的c和年份3的d的数组中|type|unstyled|depth|inlineStyleRanges|entityRanges|data|227079|$array1+=+array('a'=>1,+'b'=>2);
++++$array2+=+array('c'=>3);
++++$array3+=+array('d'=>4)
++++$form+=+array("year1"+=>$array1,
++++++++++++++++++"year2"+=>$array2,
++++++++++++++++++"year3"+=>$array3,
++++++++++++);

++++$data+=+json_encode($form);|code-block|syntax|javascript|227080|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

output of this as in json form is year1{a,b},year2{c}, year3{d}
**a has value 1 ,b=2,c=3,d=4 stored in array of year1's a,b years2's c and years3's d respectivily 

<pre><code>$array1 = array('a'=&gt;1, 'b'=&gt;2);
 $array2 = array('c'=&gt;3);
 $array3 = array('d'=&gt;4)
 $form = array("year1" =&gt;$array1,
 "year2" =&gt;$array2,
 "year3" =&gt;$array3,
 );

 $data = json_encode($form);
</code></pre>

When doing a json_encode a multidimensional array in PHP, I'm noticing a different output simply by naming one of the arrays, as opposed to not naming them. For Example:

<pre><code>$arrytest = array(array('a'=&gt;1, 'b'=&gt;2),array('c'=&gt;3),array('d'=&gt;4));
json_encode($arrytest)
</code></pre>

gives a single array of multiple json objects

<pre><code>[{"a":1,"b":2},{"c":3},{"d":4}];
</code></pre>

<hr>

whereas simply assigning a name to the middle array

<pre><code>$arrytest = array(array('a'=&gt;1, 'b'=&gt;2),"secondarray"=&gt;array('c'=&gt;3),array('d'=&gt;4));
json_encode($arrytest)
</code></pre>

creates a single json object with multiple json objects inside

<pre><code>{"0":{"a":1,"b":2},"secondarray":{"c":3},"1":{"d":4}};
</code></pre>

why would the 1st option not return the same reasults as the 2nd execpt with "1" in place of "secondarray"

JSON_ENCODE of multidimensional array giving different results

PHP语言

JSON

当在PHP语言中做一个多维数组的json_encode时，我注意到一个不同的输出，只需命名其中一个数组，而不是不命名它们。例如：$arrytest = array(array('a'=>1, 'b'=>2),array('c'=>3),array('d'=>4));json_encode($arrytest)提供由多个...

问多维数组的JSON_ENCODE给出了不同的结果
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