Question

目前，我从数据库中检索数据就像那样

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();
    }
}

用这个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();
?>

现在我想将数据插入数据库。怎么做？

我找到了sql命令，它是“插入表（column1，column2）的值（'value1'，'value2'）”。

这是带有常量值的插入，这是php中的类型。

我想要的是从java获得用户输入然后将此输入复制到php'value1'之后运行php来更新数据库。

Answer 1

取决于您使用的是Get还是Post

我将假设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')");

?>

详细了解如何使用db here

<强>更新

知道正确的请求方法，你可以使用它。

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

现在您可以使用$ req作为请求变量：

$value = $req['value'];

如何使用用户输入将数据插入服务器数据库？

1 个答案: