Php Codeigniter-post中json对象的数组

在FireBug中，我查看我的jquery post请求参数，如下所示

adults            1
applicants[]    [object Object]
attendees   1
children    0

({attendees:"2", 
  adults:"2", 
  children:"0", 
  grptype:"2", 
  'applicants[]':[{firstname:"John", lastname:"Doe", age:"33", allergies:"true", diabetic:"true",    lactose:"false", note:"nuts"}, {firstname:"Jane", lastname:"Doe", age:"34", allergies:"true", diabetic:"false", lactose:"false", note:"pollen"}]
})

adults  2
applicants[]    {firstname:"John", lastname:"Doe", age:"23", allergies:"true", diabetic:"true", lactose:"false", note:"nuts"}
applicants[]    {firstname:"Jane", lastname:"Doe", age:"23", allergies:"false", diabetic:"false", lactose:"false", note:""}
attendees   2
children    0

adults  1
applicants[]    {"firstname": "John", "lastname": "Doe", "age": "34", "allergies": "true", "diabetic": "true", "lactose": "false", "note": "nuts"}
attendees   1
children    0

在这篇post请求中，名为appenders的数组包含json对象，我希望迭代该对象并从codeigniter控制器中提取值。json字符串可能如下所示

adults            1
applicants[]    [object Object]
attendees   1
children    0

({attendees:"2", 
  adults:"2", 
  children:"0", 
  grptype:"2", 
  'applicants[]':[{firstname:"John", lastname:"Doe", age:"33", allergies:"true", diabetic:"true",    lactose:"false", note:"nuts"}, {firstname:"Jane", lastname:"Doe", age:"34", allergies:"true", diabetic:"false", lactose:"false", note:"pollen"}]
})

adults  2
applicants[]    {firstname:"John", lastname:"Doe", age:"23", allergies:"true", diabetic:"true", lactose:"false", note:"nuts"}
applicants[]    {firstname:"Jane", lastname:"Doe", age:"23", allergies:"false", diabetic:"false", lactose:"false", note:""}
attendees   2
children    0

adults  1
applicants[]    {"firstname": "John", "lastname": "Doe", "age": "34", "allergies": "true", "diabetic": "true", "lactose": "false", "note": "nuts"}
attendees   1
children    0

看看上面的申请者[]，我有两个人的信息作为json对象。我不确定如何访问控制器中的数据。看到这个了吗

$applicants = $this->input->post('applicants');
$this->output->append_output("<br/>Here: " . $applicants[0].firstname );

现在我仍然得到一个错误，说

**Message: json_decode() expects parameter 1 to be string, array given**

有什么想法吗

编辑2

好的，现在是这样的

adults            1
applicants[]    [object Object]
attendees   1
children    0

({attendees:"2", 
  adults:"2", 
  children:"0", 
  grptype:"2", 
  'applicants[]':[{firstname:"John", lastname:"Doe", age:"33", allergies:"true", diabetic:"true",    lactose:"false", note:"nuts"}, {firstname:"Jane", lastname:"Doe", age:"34", allergies:"true", diabetic:"false", lactose:"false", note:"pollen"}]
})

adults  2
applicants[]    {firstname:"John", lastname:"Doe", age:"23", allergies:"true", diabetic:"true", lactose:"false", note:"nuts"}
applicants[]    {firstname:"Jane", lastname:"Doe", age:"23", allergies:"false", diabetic:"false", lactose:"false", note:""}
attendees   2
children    0

adults  1
applicants[]    {"firstname": "John", "lastname": "Doe", "age": "34", "allergies": "true", "diabetic": "true", "lactose": "false", "note": "nuts"}
attendees   1
children    0

在控制器id中执行此操作

$applications = $this->input->post('applicants');
foreach ( $applications as $item)
{
  $item = json_decode($item, true);  
  $this->output->append_output(print_r($item));
}

这就是这种逻辑的结果

Array
(
    [firstname] => John
    [lastname] => Doe
    [age] => 34
    [allergies] => true
    [diabetic] => true
    [lactose] => false
    [note] => nuts
)

不确定我做错了什么，无论我如何访问dater，我都会收到一个错误，大意是我无法像那样访问它。如何提取值？

试试这个plz

$applicants = $this->input->post('applicants');
$json_output = json_decode($applicants );
foreach ( $json_output as $person)
{
  $this->output->append_output("<br/>Here: " . $person->firstname );
}

---

您必须在服务器上使用

$applications = json_decode($this->input->post('applicants'), true);

因此，它将成为一个关联数组，您可以像使用

数组一样使用它，而
json
中的第二个参数（true）将被转换为对象。在解码之前，它只是一个
字符串
（
json/java脚本对象表示法
string）

更新：因为它已经是一个对象数组，所以您不需要使用
json\u decode
，只需在
视图中循环数组即可

foreach($applicants as $item)
{
     echo $item->firstname . '<br />';
     echo $item->lastname . '<br />';
     // ...
}

foreach（申请人作为$item）
{
echo$item->firstname.“
”；
echo$item->lastname.“
”；
// ...
}

根据编辑2，它应该作为数组访问

echo $item['firstname'] . '<br />'

echo$item['firstname']
'
您在帖子中看到的
[object object]
意味着这正是要发送到服务器的内容-一个字符串
[object object]
。如何创建post数据？我在widard中有两个表单，在提交之前，我为每个表单创建json，并将json对象与jquery extend（）合并。然后我用jquert$.post（blablabla）发送。t=您是说将其作为字符串而不是对象发送吗？是的，您应该先将javascript对象转换为JSON字符串，然后再发送该字符串。然后，像@Sheikh Herra指出的那样，在服务器上解码JSON字符串。真不敢相信我这么笨，我用$item['firstname']来提取值。谢谢大家的指导。我一直收到一个错误，json_解码需要一个字符串，我给它一个数组。我做了一个打印，它刚刚输出了“object”。我正在发送一个json对象数组，该数组应该是字符串吗？@Binaryrespawn，那么你不需要使用
json\u decode
，它已经是一个对象数组，只需循环它。我按照上面的编辑2所述调整了我的json字符串。并使用php数组表示法而不是像$items['firstname']这样的对象表示法来提取值。这很有效。谢谢你的帮助。@Binaryrespawn，很高兴知道这一点。