我已经使用在互联网上发布的方法成功从远程数据库中获取数据。但是我无法将数据推送(插入)到同一个表中。 我添加了一个静态计数器,只是为了检查代码是否到达给定的URL,但正如预期的那样,它失败了。下面是我在远程服务器文件管理器中保存的php文件。
<?php
$json=$_GET [ 'json']; $json=f ile_get_contents( 'php://input');
$obj=json_decode($json);
$conn=mysql_connect( "mydatabasename", "myusername", "mypassword") or die( "error connecting");
mysql_select_db( "mydatabasename",$conn)or die("database couldnot connect");
error_reporting(E_ALL);
$tid=$_POST[ 'tid'];
$name=$_POST[ 'name'];
mysql_query( "insert into mytablename(tid,name) values($tid,$name)");
?>
我在android布局中输入了两个输入,tid,name并尝试发送到远程数据库。 注意:出于安全目的,已隐藏数据库名称和其他详细信息。
答案 0 :(得分:0)
如果你正在寻找一个干净的方式..i建议你做这样的事情: register.php
<?php
include('connect.php');
$response = array();
if (isset($_POST['Nom']) && isset($_POST['Prenom']) && isset($_POST['Email'])&& isset($_POST['Mdp'])) { //checking if the required fields are set
$Nom = $_POST['Nom'];//the family name
$Prenom = $_POST['Prenom']; //last name
$Email = $db->real_escape_string($_POST['Email']);
$Mdp = $db->real_escape_string($_POST['Mdp']); //the password
if ($res = $db->query("SELECT * FROM `patient` WHERE `Email_p`='$Email' ")) {
$row_cnt = $res->num_rows; }
if($row_cnt>0) {
$response["success"] = 0;
$response["message"] = "Email exists"; }
if ($row_cnt <1){
$result = mysqli_query($db,"INSERT INTO `patient`(`Id_p`, `Nom`, `Prenom`, `Email_p`, `Mdp`) VALUES ('','$Nom','$Prenom','$Email','$Mdp')");
if ($result ) {
$response["success"] = 1; // if account created we set success value to 1
$response["message"] = "account created";
} else {
$response["success"] = 0;
$response["message"] = "Oops Error";
}}
}else {
$response["success"] = 0;
$response["message"] = "Fields messing";
}
echo json_encode($response);
?>
在android ... yourActivity.java
class CreerNouveauCompte extends AsyncTask<String, String, String> {
@Override
protected void onPreExecute() {
super.onPreExecute();
pDialog = new ProgressDialog(inscription.this);
pDialog.setMessage(getString(R.string.inscriEnCours));
pDialog.setIndeterminate(false);
pDialog.setCancelable(true);
pDialog.show();
}
protected String doInBackground(String... args) {
String nom = edtNom.getText().toString();
String prenom = edtPrenom.getText().toString();
String email = edtEmail.getText().toString();
String mdp = edtMdp.getText().toString();
JSONObject json;
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("Nom", nom));
params.add(new BasicNameValuePair("Prenom", prenom));
params.add(new BasicNameValuePair("Email", email));
params.add(new BasicNameValuePair("Mdp", mdp));
json= jsonParser.makeHttpRequest(url_register (your url),
"POST", params);
}
try {if(json != null && !json.isNull("success")){
int success = json.getInt("success");
s2=json.getString("message");
if (success == 1) { //the acount created
Intent intent;
SharedPreferences settings = getSharedPreferences("compte", 0);
SharedPreferences.Editor editor = settings.edit();
editor.putString("nom",edtNom.getText().toString() );
editor.putString("prenom",edtPrenom.getText().toString() );
editor.putString("email",edtEmail.getText().toString());
editor.putString("mdp",edtMdp.getText().toString());
editor.apply();
intent = new Intent(inscription.this,MainActivity.class);
}
startActivity(intent);
finish();
} else {
}}
} catch (JSONException e) {
e.printStackTrace();
}
return null;
}
protected void onPostExecute(String file_url) {
pDialog.dismiss();
}
}
call it like this
new CreerNouveauCompte().execute();
connect.php
<?php
$db = new mysqli('localhost', 'root', '', 'rechmed');
mysqli_set_charset($db,'utf8');
?>
//note i use mysqli ..in your case use mysql
答案 1 :(得分:0)
如果您尝试使用以下格式解码JSONObject {“tid”:“myTidValue”,“name”:“myNameValue”}
,请使用下面的代码<?php
$json=stripslashes($_GET['json']);
$obj=json_decode($json, true);
$conn=mysql_connect( "mydatabasename", "myusername", "mypassword") or die( "error connecting");
mysql_select_db( "mydatabasename",$conn)or die("database couldnot connect");
error_reporting(E_ALL);
$tid=$json['tid'];
$name=$json['name'];
mysql_query("insert into mytablename(tid,name) values('$tid','$name')");
?>
并使用以下格式解码JSONArray [{“tid”:“myTidValue1”,“name”:“myNameValue1”},{“tid”:“myTidValue2”,“name”:“myNameValue2”}]你可以使用下面的代码
<?php
$json=stripslashes($_GET['json']);
$obj=json_decode($json, true);
$conn=mysql_connect( "mydatabasename", "myusername", "mypassword") or die( "error connecting");
mysql_select_db( "mydatabasename",$conn)or die("database couldnot connect");
error_reporting(E_ALL);
foreach ($json as $data){
$tid=$data['tid'];
$name=$data['name'];
mysql_query("insert into mytablename(tid,name) values('$tid','$name')");
}
?>
如果两者都不起作用,请发布您的Android代码