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'"