我的java函数
private void adddataintophp(String title, String date, String time, String channel){
ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
//http post
try{
nameValuePairs.add(new BasicNameValuePair("Title", title));
nameValuePairs.add(new BasicNameValuePair("Date", date));
nameValuePairs.add(new BasicNameValuePair("Time", time));
nameValuePairs.add(new BasicNameValuePair("Channel", channel));
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://10.0.2.2/insertprogram.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());
}
}我的php代码
<?php
$title = $_POST['Title'];
$date = $_POST['Date'];
$time = $_POST['Time'];
$channel = $_POST['Channel'];
mysql_connect("localhost","root","");
mysql_select_db("imammuda");
$sql=mysql_query("insert into Program (ID, Title, Date, Time, Channel) values ('NULL', '$title', '$date', '$time', '$channel')");
mysql_close();
?>当我执行它的时候,我会去数据库那里看看。
它已被添加,但值为空值。
因此,我的问题是如何将java中edittext获取的值传递到php中。
发布于 2011-05-19 22:12:26
您在Java中执行POST请求,但在PHP代码中读取GET请求。在发布数据时,它不会显示在URL中,数据应该根据GET动词显示在URL中。
发布于 2011-05-19 22:22:59
请记住转义SQL查询中的所有字符串以增强安全性:
$sql=mysql_query("insert into Program (ID, Title, Date, Time, Channel) values ('NULL', '" . mysql_real_escape_string($title) . "', '" . mysql_real_escape_string($date) . "', '" . mysql_real_escape_string($time) . "', '" . mysql_real_escape_string($channel) . "')");https://stackoverflow.com/questions/6059989
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