我正在MYSQL数据库中上传数据,同时我想检索我已插入的属性之一,以满足我的成功上传。当我第一次按下按钮时,它只将数据上传到服务器,而不返回任何内容。同样,当我点击按钮时,它会同时执行两个过程(插入和检索数据),因此我无法第一次以 json 对象的形式返回值。
这是我的 php 代码 engrdatainsert.php
<?php
$sqlCon=mysql_connect("localhost","root","");
mysql_select_db("PeopleData");
//Retrieve the data from the Android Post done by and Engr...
$adp_no = $_REQUEST['adp_no'];
$building_no = $_POST['building_no'];
$contractor_name = $_POST['contractor_name'];
$officer_name = $_POST['officer_name'];
$area = $_POST['area'];
--------------------插入从Android接收到的值----------||
$sql = "INSERT INTO engrdata (adp_no, building_no,area,contractor_name,officer_name) VALUES('$adp_no', '$building_no', '$are', '$contractor_name', '$officer_name')";
//--------现在查看插入数据的事务状态---------||
$q=mysql_query("SELECT adp_no FROM engrdata WHERE adp_no='$adp_no'");
while($e=mysql_fetch_assoc($q))
$output[]=$e;
print(json_encode($output));//conveting into json array
mysql_close();
?>
我的 Android 代码
public void insertdata()
{
InputStream is=null;
String result=null;
ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(5);
nameValuePairs.add(new BasicNameValuePair("adp_no",adp));//"34"));
nameValuePairs.add(new BasicNameValuePair("building_no",bldng));//"72"));
nameValuePairs.add(new BasicNameValuePair("area",myarea));//"72"));
nameValuePairs.add(new BasicNameValuePair("contractor_name",cntrct));//"72"));
nameValuePairs.add(new BasicNameValuePair("officer_name",ofcr));//"72"));
//http post
try{
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://10.0.2.2/androidconnection/engrdatainsert.php");
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
is = entity.getContent();
Log.i("postData", response.getStatusLine().toString());
}
catch(Exception e)
{
Log.e("log_tag", "Error in http connection "+e.toString());
}
//convert the input strem into a string value
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();
result=sb.toString();
}
catch(Exception e)
{ Log.e("log_tag", "Error converting result "+e.toString()); }
try
{
JSONArray jArray = new JSONArray(result);
for(int i=0;i<jArray.length();i++)
{
JSONObject json_data = jArray.getJSONObject(i);
Toast.makeText(this, "data is "+json_data.getString("adp_no")+"\n", Toast.LENGTH_LONG).show();
String return_val = json_data.getString("adp_no");
if(return_val!=null)
{
Intent offff=new Intent(this,MainActivity.class);
offff.addFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP);
offff.addFlags(Intent.FLAG_ACTIVITY_NEW_TASK);
//startActivity(offff);
}
}
}
//}
catch(JSONException e)
{ Log.e("log_tag", "Error parsing data "+e.toString()); }
// return returnString;//*/
}
最佳答案
在您的 PHP 代码中,您没有执行 INSERT 查询。你需要做这样的事情:
--------------------插入从Android接收到的值----------||
$sql = "INSERT INTO engrdata (adp_no, building_no,area,contractor_name,officer_name) VALUES('$adp_no', '$building_no', '$are', '$contractor_name', '$officer_name')";
mysql_query($sql) or die(mysql_error());
//--------现在查看插入数据的事务状态---------||
注意我添加的行,它实际执行查询。
关于php - 来自 php 服务器的 Json 数据不起作用。,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/13973355/