Java 我所需要做的就是发布3个值!Android到PHP服务器脚本
android设备上的代码是:Java 我所需要做的就是发布3个值!Android到PHP服务器脚本,java,php,android,mysql,Java,Php,Android,Mysql,android设备上的代码是: public JSONObject getJSONFromUrl(String url, List<NameValuePair> params) { // Making HTTP request try { // defaultHttpClient DefaultHttpClient httpClient = new DefaultHttpClient(); HttpPost http
public JSONObject getJSONFromUrl(String url, List<NameValuePair> params) {
// Making HTTP request
try {
// defaultHttpClient
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(url);
httpPost.setEntity(new UrlEncodedFormEntity(params));
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
try {
BufferedReader reader = new BufferedReader(new InputStreamReader(
is, "iso-8859-1"), 8);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "n");
}
is.close();
json = sb.toString();
Log.e("JSON", json);
} catch (Exception e) {
Log.e("Buffer Error", "Error converting result " + e.toString());
}
// try parse the string to a JSON object
try {
jObj = new JSONObject(json);
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
// return JSON String
return jObj;
}
其中storeUser是:
public function storeUser($name, $email, $password) {
$result = mysql_query("INSERT INTO users(unique_id, name, email, encrypted_password, salt, created_at) VALUES('$uuid', '$name', '$email','$encrypted_password', '$salt', NOW())");
}
这个过程是有效的,会向android设备发送一条消息,但细节不清楚
插入MySQL数据库
我只想从android设备发布详细信息,并将其插入我服务器上的MySQL数据库
谁知道为什么会这样,我做错了什么
$name = $_POST['name'];
$name = $_POST['email'];
$name = $_POST['password'];
这显然是错误的。将变量更改为$email
和$password
此外,停止使用mysql_查询,因为它正在变得越来越强大。使用或代替。我从上面找出了我的代码的错误 java/android代码中传递给httppost的url不能作为我使用的api的简单地址,我尝试了www.domain.com/index.php?用“?”号。添加此代码使代码正常工作 我只是回到了最基本的部分,看看是什么让一篇文章如此精彩
这里是对维基百科上的文章的参考,它帮助我找出了http POST的组成部分,警告您的代码容易受到sql注入攻击。如何防止这种攻击的教程?
$name = $_POST['name'];
$name = $_POST['email'];
$name = $_POST['password'];