我正在尝试从phpmyadmin数据库中检索特定值。我想从mytasksend表中删除to=$username所在的行。
 但是当我回显变量时,我没有任何输出。
我应该做些什么?

这是我的PHP文件

<?php
mysql_connect ("localhost","root","");
mysql_select_db("taskmanager");

$username= (isset($_POST['to'])) ? $_POST['to'] : '';
$q=mysql_query("SELECT `tasksentid` FROM `mytasksend` where `to` = $username")                or die(mysql_error());

$output=array();
   while($e=mysql_fetch_assoc($q))
   $output[]=$e;
print (json_encode($output));
mysql_close();

?>


这是我的java文件

try{
   HttpClient httpclient3 = new DefaultHttpClient();
   HttpPost httppost3 = new HttpPost("http://10.0.2.2:80/selection.php");
   ArrayList<NameValuePair>  nameValuePairs = new ArrayList<NameValuePair>();
   nameValuePairs.add(new BasicNameValuePair("to",id));
   Log.w("aaaaaaa",""+nameValuePairs);
   httppost3.setEntity(new UrlEncodedFormEntity(nameValuePairs));
   HttpResponse response = httpclient3.execute(httppost3);
   HttpEntity entity = response.getEntity();
   is = entity.getContent();

   Log.e("log_tag", "connection success ");
}

最佳答案

尝试将$username包装在查询中的单引号下。

"SELECT `tasksentid` FROM `mytasksend` where `to` = '$username'"

10-07 19:16
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