[英]sending data from android to php using json
嘿,我正在尝试将数据从我的 android 应用程序发送到 Web 服务器并使用 mysql 将其发送到数据库,尽管它似乎没有任何错误,但我的代码不起作用
安卓Java代码:
String categorys=category.getText().toString();
String authors = author.getText().toString();
String quess = question.getText().toString();
String anss = answer.getText().toString();
try {
JSONObject json = new JSONObject();
json.put("category",categorys);
json.put("ques",quess);
json.put("ans",anss);
json.put("authors",authors);
postData(json);
} catch (JSONException e) {
e.printStackTrace();
}
}
});
}
public void postData(JSONObject json) throws JSONException {
HttpClient httpclient = new DefaultHttpClient();
try {
HttpPost httppost = new HttpPost("http://shlomo.webuda.com/androidtomy.php");
List<NameValuePair> nvp = new ArrayList<NameValuePair>(2);
nvp.add(new BasicNameValuePair("json", json.toString()));
//httppost.setHeader("Content-type", "application/json");
httppost.setEntity(new UrlEncodedFormEntity(nvp));
HttpResponse response = httpclient.execute(httppost);
if(response != null) {
InputStream is = response.getEntity().getContent();
//input stream is response that can be shown back on android
}
}
catch (Exception e)
{
e.printStackTrace();
}
}`
代码
mysql_connect("something","something","something");
mysql_select_db("something");
$json = $_SERVER['HTTP_JSON'];
echo "JSON: \n";
var_dump($json);
echo "\n\n";
$data = json_decode($json,true);
var_dump($data);
$category=$data['category'];
$author=$data['authors'];
$question=$data['ques'];
$answer=$data['ans'];
$sql = 'INSERT INTO Ques(Ques_Author, Ques_Question,Ques_Answer,Ques_Category,Ques_Approve) values("category","author","question","answer","0")';
mysql_query($sql);
}
?>
正如评论中所指出的,PHP 文件有几个问题。 请尝试以下方法:
mysql_connect("something","something","something");
mysql_select_db("something");
$json = $_REQUEST['json'];
echo "JSON: \n";
var_dump($json);
echo "\n\n";
$data = json_decode($json,true);
var_dump($data);
$category=$data['category'];
$author=$data['authors'];
$question=$data['ques'];
$answer=$data['ans'];
$sql = "INSERT INTO Ques(Ques_Author, Ques_Question,Ques_Answer,Ques_Category,Ques_Approve) values($category,$author,$question,$answer,0)";
mysql_query($sql);
此外,您不应使用 mysql_query 函数。 您应该使用 mysqli_ * *-functions 或 PDO ( http://php.net/manual/en/book.pdo.php )。 mysql_query 从 PHP5.5 开始被弃用是有充分理由的。 特别是当您将似乎正确的 url 发布到上述 php 文件时。
如果您仍然遇到问题,请提供 $json 变量的内容,以及来自您的 Java 代码的请求副本
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