【发布时间】:2012-11-25 23:15:39
【问题描述】:
我正在尝试从 Android 应用程序进行简单插入。我可以通过连接?entry="Sample value from browser" 从浏览器运行我的php 脚本,但是当我从Android 运行应用程序时,我没有插入。
这里是我调用使用 JSON 并实现 AsyncTask 的插入类的地方:
package us.jtaylorok.android.sqlite.first;
import java.util.ArrayList;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
import org.apache.http.client.HttpClient;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.message.BasicNameValuePair;
import android.app.ProgressDialog;
import android.content.Context;
import android.os.AsyncTask;
import android.util.Log;
import android.widget.Toast;
public class RemoteInsert extends AsyncTask<Void, String,String >{
protected String TAG;
protected Context context;
protected String input;
protected ProgressDialog progressDialog;
public RemoteInsert(String i,Context c){
this.input = i;
this.context = c;
}
protected void onPreExecute() {
//ProgressDialog progressDialog; // = new ProgressDialog(context);
//progressDialog=ProgressDialog.show(,"Please Wait..","Sending data to database", false);
progressDialog=ProgressDialog.show(context,"Please Wait..","Sending data to database", false);
}
@Override
protected String doInBackground(Void... params) {
try {
HttpClient httpclient = new DefaultHttpClient();
//HttpPost httppost = new HttpPost("http://localhost/index.php");
//HttpPost httppost = new HttpPost("http://10.253.8.88/patient_data/patient_data.php");
HttpPost httppost = new HttpPost("http://10.100.205.72/patient_data/patient_data.php");
ArrayList<NameValuePair> postParameters = new ArrayList<NameValuePair>();
postParameters.add(new BasicNameValuePair("entry", "Input from Android"));
httppost.setEntity(new UrlEncodedFormEntity(postParameters));
HttpResponse response = httpclient.execute(httppost);
Log.i("postData", response.getStatusLine().toString());
} catch(Exception e) {
Log.e(TAG, "Error: "+e.toString());
}
return "";
}
protected void onPostExecute(String result) {
progressDialog.dismiss();
Toast.makeText(context, "Finished", Toast.LENGTH_LONG).show();
}
}
这是我的 PHP 脚本:
<?php
// mysql_connect("host","username","password");
mysql_connect("localhost","user1","mypassword");
mysql_select_db("test");
$entry_value = $_REQUEST["entry"];
$query = "INSERT INTO patientdata (entry) values (".$entry_value.");";
if( !mysql_query($query) ) {
/*insert failed*/
}
mysql_close();
?>
同样,如果我从浏览器中调用它,它会完美运行,但它会在实现 AsyncTask 之前引发异常。
我确实让 AVD 显示添加和删除,但是当我这样做时,我的 apache2 access_log 或 error_log 中没有请求。有什么建议吗?
【问题讨论】:
-
也许当我向 user1 授予权限时,它需要来自 localhost 以外的其他地方。我能够从 $_REQUEST 获取值并将它们写入文本文件,因此 mysql_query 函数是不稳定的,我认为我已经读过它已被弃用,或者当来自时 user1@localhost 权限问题存在问题安卓。我什至可以通过在 android 的浏览器中运行 php 文件来实现这一点。因此,欢迎提出任何建议。
-
解决了我的 php.ini 中的配置问题
标签: java php android android-asynctask android-dialog