我想輸出我的控制器在我的視圖中輸出的內容作爲json,但我認爲我輸出的很奇怪。CakePHP json輸出
在網絡上搜索我的JSON,並將其與輸出看起來像這樣出現:
{"menu": {
"id": "file",
"value": "File",
"popup": {
"menuitem": [
{"value": "New", "onclick": "CreateNewDoc()"},
{"value": "Open", "onclick": "OpenDoc()"},
{"value": "Close", "onclick": "CloseDoc()"}
]
}
}}
但是我的僅僅是未格式化,看起來像這樣。
[{"Customer":{"id":"1","first_name":"Ian","last_name":"Smith","address_1":"10 High Streets","address_2":"","town_city":"Plymouth","county":"Devon","postcode":"PL1 2JD"}},{"Customer":{"id":"2","first_name":"David","last_name":"Smith","address_1":"52 Low Avenue","address_2":"","town_city":"Exeter","county":"Devon","postcode":"EX2 1KO"}}]
我怎樣才能輸出它看起來像第一個?
EDIT
控制器
$user = $this->Customer->find('all');
$this->set('users', $user);
查看
<?php echo json_encode($user); ?>
好的......看編輯。 – user667430 2013-03-08 16:37:11
你爲什麼在意它看起來像什麼?這是正確格式的正確數據 - 在這種情況下代碼的視覺外觀應該不重要,並且您希望它儘可能精簡。 – Dave 2013-03-08 16:55:39