如何通過用戶輸入將數據插入服務器數據庫？

Question

目前，我檢索數據從數據庫就是這樣

private void getdatafromphp(){ 
    ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(); 
    //http post 
    try{ 
     HttpClient httpclient = new DefaultHttpClient(); 
     HttpPost httppost = new HttpPost("http://10.0.2.2/video.php"); 
     httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs)); 
     HttpResponse response = httpclient.execute(httppost); 
     HttpEntity entity = response.getEntity(); 
     is = entity.getContent(); 
    }catch(Exception e){ 
     Log.e("log_tag", "Error in http connection"+e.toString()); 
    } 
     //convert response to string 
    try{ 
     BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8); 
     sb = new StringBuilder(); 
     sb.append(reader.readLine() + "\n"); 
     String line="0"; 
     while ((line = reader.readLine()) != null) { 
      sb.append(line + "\n"); 
     } 
     is.close(); 
     result=sb.toString(); 
    }catch(Exception e){ 
     Log.e("log_tag", "Error converting result "+e.toString()); 
    } 
       //paring data 
    try{ 
     jArray = new JSONArray(result); 
     JSONObject json_data=null; 
     json_data = jArray.getJSONObject(jArray.length()-1); 
     url=json_data.getString("VideoUrl"); 
    }catch(JSONException e1){ 
    }catch(ParseException e1) { 
     e1.printStackTrace(); 
    } 
} 

這個PHP

<?php 
mysql_connect("localhost","root",""); 
mysql_select_db("imammuda"); 
$sql=mysql_query("select * from Video"); 
while($row=mysql_fetch_assoc($sql)) 
$output[]=$row; 
print(json_encode($output)); 
mysql_close(); 
?> 

現在我想將數據插入到數據庫中。怎麼做？

我找到了「insert into table（column1，column2）values（'value1'，'value2'）」的sql命令。

這是插入常量值，這是在php中鍵入。

我想從java那裏得到用戶的輸入，然後將該輸入複製到php'value1'後運行php來更新數據庫。

Answer 1

取決於您是否使用Get或Post

我將承擔GET

$value = $_GET['value']; // this will retrieve the value from the url and save it in a variable 

mysql_connect("localhost","root",""); 

// escape the value first 
$value = mysql_real_escape_string($value); 


mysql_select_db("imammuda"); 
$result = mysql_query("insert into Video (value) values ('$value')"); 

?> 

瞭解與db here

UPDATE工作

知道正確請求方法你可以使用這個。

$req; 
if ($_SERVER['REQUEST_METHOD'] == 'GET') { 
    $req = $_GET; 
}else { 
    $req = $_POST; 
} 

現在你可以使用$ REQ爲您的請求變量：

$value = $req['value'];