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我在我的应用程序的一个类中工作正常,下面的代码。以下代码段从远程服务器收到一个字符串“Okay”。从sql server接收数据时出错
try {
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
nameValuePairs.add(new BasicNameValuePair("userid", et1.getText().toString()));
nameValuePairs.add(new BasicNameValuePair("pass", et2.getText().toString()));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
client.execute(httppost);
HttpResponse response = client.execute(httppost);
HttpEntity respEntity = response.getEntity();
if (respEntity != null) {
// EntityUtils to get the response content
content = EntityUtils.toString(respEntity);
Log.d("valueeeeeeeeeeee", content);
}
的PHP代码是在以下几点:
<?php
$host = "localhost";
$user = "admin";
$pass = "123";
$connection = mysql_connect($host,$user,$pass) or die (mysql_errno().": ".mysql_error()."<BR>");
$userid = mysql_real_escape_string($_POST['userid']);
$pass = mysql_real_escape_string($_POST['pass']);
$db_select=mysql_select_db("mydb");
if(!$db_select){
die(mysql_error());
echo "error";
}
$query = "select count(1) as count_users from wp_users where user_login = '".$userid."' and user_pass='".$pass."'";
$result = mysql_query($query);
$row = mysql_fetch_assoc($result);
if($row['count_users']>0)
{
echo "Okay";
}
else
{
echo "Not found";
}
if($medo=mysql_query($query)){
header("localhost/filename");
exit;
}else{
echo"<p> Error</p>";
die(mysql_error());
}
而当我试图在PHP运行下面的代码它给我同样采用Android代码中的错误。
<?php
$host = "localhost";
$user = "admin";
$pass = "123";
$connection = mysql_connect($host,$user,$pass) or die (mysql_errno().": ".mysql_error()."<BR>");
$userid = mysql_real_escape_string($_POST['userid']);
$pass = mysql_real_escape_string($_POST['pass']);
$db_select=mysql_select_db("mydb");
if(!$db_select){
die(mysql_error());
echo "error";
}
$query= "SELECT * FROM wp_users";
$result = mysql_query($query);
$num_rows = mysql_num_rows($result);
if("$num_rows" > 0)
{
echo "$num_rows";
}
else
{
echo "Not found";
}
if($medo=mysql_query($query)){
header("localhost/filename");
exit;
}else{
echo"<p> Error</p>";
die(mysql_error());
}
是否有不同的方式接收Android行末尾的行或列中的行数或数据?
它仍然给了我“NumberFormatException异常:无效的诠释“error – 2013-05-14 03:32:01
$ query =”SELECT COUNT(*)FROM wp_users“; <------并没有真正的帮助,因为它给了我不正确的行数。 – 2013-05-14 03:36:37
@MohammadSohaib您的查询是'$ query =“SELECT * FROM wp_users”;'因此'SELECT COUNT(*)FROM wp_users'给出了您在问题中提出的问题 - 结果集中的行数。要么改变你的查询,要么清楚地解释你试图得到的数字。 – peterm 2013-05-14 03:40:55