php - JSON 格式化

Question

使用发现的方法here ，我正在收集表单值并使用以下代码发布它们:

$.ajax({
    type: "POST",
    url: "http://"+document.domain+"/SimplaAdmin/includes/rpc.php",
    data: { data:postdata, method: 'addSite'},
    dataType: "json",
.......

发布的数据是:

data:{
    "textfield": ["",""],
    "dropdown": ["option1","option1"],
    "siteTitle":"this is the site title",
    "siteKey":"",
    "siteurl":"",
    "address1":"",
    "address2":"",
    "address3":"",
    "landline":"",
    "method":"addSite",
    "small-input":"",
    "medium-input":"",
    "large-input":""
}

然后我尝试使用以下方法获取值 siteTitle:

$data = $_POST['data'];
$obj=json_decode($data) ;
$title = $obj->{'siteTitle'};

但是不行，我的想法哪里出了问题？

Answer 1

您的语法不正确 - 您只需要 $title = $obj->siteTitle;

此外，我认为您出于本文的目的，在 JSON 字符串的开头添加了 data: ？那也不应该在那里。

举个例子:

<?php 

$string = '{"textfield":["",""],"dropdown":["option1","option1"],"siteTitle":"this is the site title","siteKey":"","siteurl":"","address1":"","address2":"","address3":"","landline":"","method":"addSite","small-input":"","medium-input":"","large-input":""}';

$obj = json_decode($string); 

print_r( $obj->siteTitle );

?>

哪些输出

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