0

这是我的 java 代码，它发出一个 http 请求并将 json 对象发送到 php 脚本。

登录.java

    String username = jTextField1.getText();
    String password = jPasswordField1.getText();

    JSONObject obj = new JSONObject();

    obj.put("username", username);
    obj.put("password", password);

    //JSONArray list = new JSONArray();
    //list.add(username);
    //list.add(password);


    //obj.put("logindata", list);
    try {
        HttpClient httpclient = new DefaultHttpClient();
        HttpResponse response;
        HttpPost httppost = new HttpPost("http://localhost/kolitha/json_test/index.php");
        StringEntity se = new StringEntity("myjson" + obj.toString());
        httppost.setEntity(se);
        System.out.print(se);
        httppost.setHeader("Accept", "application/json");
        httppost.setHeader("Content-type", "application/json");
        System.out.println(obj.toString());

        //response = httpclient.execute(httppost);

        HttpGet httpget = new HttpGet("http://localhost/kolitha/json_test/index.php");
        response = httpclient.execute(httpget);
        HttpEntity entity = response.getEntity();
        System.out.println(EntityUtils.toString(entity));



    } catch (Exception e) {
        e.printStackTrace();
        System.out.print("Cannot establish connection!");

    }

这是我的 index.php 脚本，我无法从 json 对象中提取用户名和密码。

索引.php

$obj = json_decode(file_get_contents('php://input'));

$username=$obj->{'username'};
$password=$obj->{'password'};


$connect=mysql_connect('localhost', 'root', '');
IF (!$connect){
die ('Failed Connecting to Database: ' . mysql_error());}

$d = mysql_select_db("kolitha_json_test");
if(!$d){ echo "db not selected";}

$sql="SELECT * FROM login WHERE username='$username' AND password='$password' ";
$result=mysql_query($sql) or die (mysql_error());
$count=mysql_num_rows($result);

 // If result matched $myusername and $mypassword, table row must be 1 row
if($count==1)
{
echo "true";
}
else
 {
echo "false";
}
?>
4

2 回答 2

0

好的，我发表了评论，但似乎没有人注意评论。错误在 Java 代码中

StringEntity se = new StringEntity("myjson" + obj.toString());

它为名为的 json 字符串添加了一个前缀myjson

于 2013-07-16T10:00:02.527 回答
0

为什么同时使用HttpPost 和HttpGet。您正在向HttpPost请求添加字符串实体而不使用它..

而您使用的是尚未设置用户名和密码的Httpget（因此显然没有与 httpget 请求关联的 json 数据）

在此处进行更改

 response = httpclient.execute(httppost);//don comment this line

如果您使用的是 httppost，请在此处添加评论

//HttpGet httpget = new HttpGet("http://localhost/kolitha/json_test/index.php");
//response = httpclient.execute(httpget);
于 2013-07-16T11:05:43.230 回答