PHP接收http post请求为$ _POST

Question

如果我试图发布像＃34; abc＆＃34;我收到了回复，但当我尝试发布大量数据（如108 KB）时，它没有显示响应可能是我需要增加一些限制，但我无法访问php.ini文件。

还有其他方法吗？

我正在发布android的数据，所以在android中有任何字符串缩短编码并在php中解码可用。我已经使用base64来获取图像数据。

我的PHP代码回应了回报。

<?php
header('Access-Control-Allow-Origin: *');
header('Access-Control-Allow-Methods: PUT, GET, POST, DELETE, OPTIONS');
header('Access-Control-Allow-Headers: Content-Type');


if(isset($_POST['incidentDetails']))
{
    echo 'Responce from server:  ' . $_POST['incidentDetails'];
}
else
{
    echo 'Request without paramenter';  
}
?>

网址 - http://bumba27.byethost16.com/incidentManagments/api/adding_incident_details.php

我需要发布参数名称incidentDetails- http://www.filedropper.com/log_2

Android代码

// Create a new HttpClient and Post Header
        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost("http://bumba27.byethost16.com/incidentManagments/api/adding_incident_details.php");

        try 
        {
            List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(1);
            nameValuePairs.add(new BasicNameValuePair("incidentDetails", headerData));
            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

            HttpResponse response = httpclient.execute(httppost);

            responseCode = response.getStatusLine().getStatusCode();
            responseBody = EntityUtils.toString(response.getEntity());
        } 
        catch (Throwable t ) 
        {
            Log.d("Error Time of Login",t+"");
        } 

Answer 1

您是否确保在发送图像数据的HTML表单上的表单标记中包含enctype =“multipart / form-data”？

<form name="someform" id="someform" enctype="multipart/form-data" method="post" action="somefile.php">

修改 根据PHP.NET网站，你必须做一个保存文件的过程 - 不只是打印出$ _POST数组：

$name= $_FILES["myfile"]["name"];
$type= $_FILES["myfile"]["type"];
$size= $_FILES["myfile"]["size"];
$temp= $_FILES["myfile"]["temp_name"];
$error= $_FILES["myfile"]["error"];

if ($error > 0)
    die("Error uploading file! code $error.");
else
   {
    if($type=="image/png" || $size > 2000000)//condition for the file
    {
    die("Format  not allowed or file size too big!");
    }
    else
    {
     move_uploaded_file($temp, "uploaded/" .$name);
     echo "Upload complete!"; 
     }
}