如何将数组值转换为非对象到PHP

Question

我已经用邮递员表单数据提交了这些数据，那么如何从from-data数组转换为非对象JSON？

我必须尝试json_encode和json_decode，但不能将json或非对象结果输入控制器。

在CodeIgniter控制器中。

Array
(
    [------WebKitFormBoundaryv6rwIfAWUgyyzeZF
Content-Disposition:_form-data;_name] => "first_name"

Janak
------WebKitFormBoundaryv6rwIfAWUgyyzeZF
Content-Disposition: form-data; name="last_name"

Kumar
------WebKitFormBoundaryv6rwIfAWUgyyzeZF
Content-Disposition: form-data; name="mobile_no"

123456789
------WebKitFormBoundaryv6rwIfAWUgyyzeZF
Content-Disposition: form-data; name="country_code"

 91
------WebKitFormBoundaryv6rwIfAWUgyyzeZF
Content-Disposition: form-data; name="email_id"

[email protected]
------WebKitFormBoundaryv6rwIfAWUgyyzeZF
Content-Disposition: form-data; name="password"

123456
------WebKitFormBoundaryv6rwIfAWUgyyzeZF
Content-Disposition: form-data; name="device_token"

123HFDT3434
------WebKitFormBoundaryv6rwIfAWUgyyzeZF
Content-Disposition: form-data; name="device_type"

2
------WebKitFormBoundaryv6rwIfAWUgyyzeZF
Content-Disposition: form-data; name="agency_name"

google
------WebKitFormBoundaryv6rwIfAWUgyyzeZF

)

Answer 1

这就是你要追求的吗？

<?php
$content = <<<OUT
------WebKitFormBoundaryv6rwIfAWUgyyzeZF
Content-Disposition: form-data; name="last_name"

Kumar
------WebKitFormBoundaryv6rwIfAWUgyyzeZF
Content-Disposition: form-data; name="mobile_no"

123456789
------WebKitFormBoundaryv6rwIfAWUgyyzeZF
Content-Disposition: form-data; name="country_code"

 91
------WebKitFormBoundaryv6rwIfAWUgyyzeZF
Content-Disposition: form-data; name="email_id"

[email protected]'
OUT;

$clean = str_replace("\n",'',$content);
$parts = array_filter(explode('------',$clean));

$array = array();

foreach ($parts as $row){
    $parts2 = explode('name="',$row);
    $parts3 = explode('"',$parts2[1]);
    $array[$parts3[0]] = $parts3[1];
}

echo json_encode($array);
?>

我的数组演示有问题，因为"first_name"实际上是值，所以我不确定你是否正确输出了数据但让我知道。