PHP检索数据,但返回资源ID#4

时间:2013-12-02 05:35:20

标签: php sql-server sqlsrv

我尝试使用sqlsrv_query从microsoft数据库中获取数据。这是我的代码:

$serverName = "mssql.somee.com\sqlexpress"; //serverName\instanceName


$connectionInfo = array("UID"=>"id","PWD"=>"password","Database"=>"db");

$conn = sqlsrv_connect( "db.mssql.somee.com", $connectionInfo);

if( $conn ) {
 echo "Connection established.<br />";
}else{
 echo "Connection could not be established.<br />";
 die( print_r( sqlsrv_errors(), true));
}



$sql = "select UserName,Email from HTS_User where UserID = 7";

$stmt = sqlsrv_query($conn,$sql);//query is working
if($stmt === false){
die(print_r(sqlsrv_errors(),true));
}else{
echo "sqlsrv_query is working";
echo "<br />";
}

if(sqlsrv_fetch($stmt) ===false){
echo "fetch error";
die(print_r(sqlsrv_errors(),true));
}else{
echo "sqlsrv_fetch is working";
echo "<br />";
}

$n = sqlsrv_get_field($stmt,0);
echo "$n: ";
echo "<br />"

$e = sqlsrv_get_field($stmt,1);
echo "$e: ";

但输出是这而不是真实数据:

资源ID#4: 资源ID#5:

有谁知道这个问题?非常感谢 杰森

1 个答案:

答案 0 :(得分:1)

尝试

if(sqlsrv_fetch_array($stmt) ===false){

而不是

if(sqlsrv_fetch($stmt) ===false){

参考:

http://php.net/manual/en/function.sqlsrv-fetch-array.php

http://www.php.net/manual/en/function.sqlsrv-fetch.php