<?php
$serverName = "192.168.0.98"; //数据库服务器地址

$uid = "sa";     //数据库用户名

$pwd = "123456"; //数据库密码

$connectionInfo = array("UID"=>$uid, "PWD"=>$pwd, "Database"=>"AIS20160314104736");

$conn = sqlsrv_connect($serverName, $connectionInfo);

if( $conn == false)

{

    echo "连接失败！";
    exit;
}


$sql = "selsect ICStockBill.FBillNo,t_ICItem.FName form ICStockBill left join t_ICItem on ICStockBill.FItemID = t_ICItem.FItemID";
$query = sqlsrv_query( $conn, $sql);
while ($row = sqlsrv_fetch_array($query)) {
  echo $row['FBillNo'] ;
  
  }

?>


提示出错：
Warning: sqlsrv_fetch_array() expects parameter 1 to be resource, bool given in D:\WWW\index.php on line 23

sqlsrv_query

Returns a statement resource on success and FALSE if an error occurred.

"selsect ICStockBill.FBillNo,t_ICItem.FName form

这里应该用from还是应该用form？
如果不知道哪里错了，那么可以 echo $sql;exit;,把SQL语句输出出来，贴到mysql里直接运行一下看看是不是报错。