PHP-MySQL:在下面的代码中连接查询和代码操作 - php-mysql : concat query in below code and manipulation of code

php-mysql : concat query in below code and manipulation of code

>这是我的代码

<?php
$page = isset($_POST['page']) ? intval($_POST['page']) : 1;
$rows = isset($_POST['rows']) ? intval($_POST['rows']) : 10;
$offset = ($page-1)*$rows;
$result = array();
include 'conn.php';
//$rs1 = "select CONCAT(id,' ',firstname) AS password FROM users";
$rs = mysql_query("select count(*) from users");
$row = mysql_fetch_row($rs);

$result["total"] = $row[0];
$rs = mysql_query("select * from users limit $offset,$rows");

$items = array();
while($row = mysql_fetch_object($rs)){
    array_push($items, $row);
}
$result["rows"] = $items;
echo json_encode($result);
?>

我的输出是

ID   firstname     lastname  password*
 1    john          abc 
 2    don           def
 3    man           ghi

我希望密码与id+名字一起坎卡特..请帮助我在上面的代码...

更正的代码：

<?php
$page = isset($_POST['page']) ? intval($_POST['page']) : 1;
$rows = isset($_POST['rows']) ? intval($_POST['rows']) : 10;
$offset = ($page-1)*$rows;
$result = array();
include 'conn.php';
//$rs1 = "select CONCAT(id,' ',firstname) AS password FROM users";
$rs = mysql_query("select count(*) from users");
$row = mysql_fetch_row($rs);

$result["total"] = $row[0];
$rs = mysql_query("select *, CONCAT(id, firstname, password) AS pwd  from users limit $offset,$rows");

$items = array();
while($row = mysql_fetch_object($rs)){
    array_push($items, $row);
}
$result["rows"] = $items;
echo json_encode($result);
?>