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Php sqlsrv_查询在while循环中缺少第一个结果_Php_Php4 - Fatal编程技术网

Php sqlsrv_查询在while循环中缺少第一个结果

Php sqlsrv_查询在while循环中缺少第一个结果,php,php4,Php,Php4,嗨,我有以下代码 function dropDownMenuMobile($parentId,$siteId,$root=false) { $menu = ""; $sql = "EXEC spGenerateDropdownMenuMobile ".$parentId.",".$siteId; $stmt = sqlsrv_query($GLOBALS['conn'],$sql); sqlsrv_fetch($stmt); if($stmt &&

嗨,我有以下代码

function dropDownMenuMobile($parentId,$siteId,$root=false)
{
    $menu = "";
    $sql = "EXEC spGenerateDropdownMenuMobile ".$parentId.",".$siteId;
    $stmt = sqlsrv_query($GLOBALS['conn'],$sql);
    sqlsrv_fetch($stmt);
    if($stmt && sqlsrv_has_rows($stmt) === true)
    {
        $menu .= "<ul>";
            while($rs = sqlsrv_fetch_array($stmt, SQLSRV_FETCH_NUMERIC))
            {
                $urlLink = str_replace("<br />"," ",$rs[2]);
                $urlLink = replaceIllegalChars($urlLink);
                $actualLink = (strlen($rs[7]) > 2) ? $rs[7] : $rs[0]."/".makeSEurl($urlLink);
                $liclass = ($rs["5"] > 0)?"expand":"";
                $menu .= "<li class=\"".$liclass."\">";
                    $menu .= "<a href=\"/".$actualLink."\">".$rs[2]."</a>";
                    $menu .= ($rs[5] > 0) ? dropDownMenuMobile($rs[0],$siteId,"") : "";
                $menu .= "</li>";
            }
        $menu .= "</ul>";
    }
    return $menu;
}
函数dropDownMenuMobile($parentId,$siteId,$root=false)
{
$menu=“”;
$sql=“EXEC spGenerateDropdownMenuMobile”$parentId.,“$siteId;
$stmt=sqlsrv_查询($GLOBALS['conn',$sql);
sqlsrv_fetch($stmt);
如果($stmt&&sqlsrv_有_行($stmt)==true)
{
$menu.=“
    ”; 而($rs=sqlsrv\u fetch\u数组($stmt,sqlsrv\u fetch\u NUMERIC)) { $urlink=str_replace(“
    ,”,$rs[2]); $urlink=replaceIllegalChars($urlink); $actualLink=(strlen($rs[7])>2)$rs[7]:$rs[0]。“/”。makeSEurl($urlink); $liclass=($rs[“5”]>0)?“扩展”:“; $menu.=“”; $菜单=“”; $menu.=($rs[5]>0)?dropDownMenuMobile($rs[0],$siteId,”):; $menu.=“”; } $menu.=“
”; } 返回$menu; }

它在许多其他地方运行,但由于某些原因,在while循环中,每次调用它时似乎都会丢失第一个结果

我想是因为你跑了

sqlsrv_fetch($stmt);
它将在运行后获取第一个结果

sqlsrv_fetch_array($stmt, SQLSRV_FETCH_NUMERIC)
继续获取第二个结果

只需删除第一个fetch