Question

我刚刚开始学习PHP，我很难将mysql_result转换为使用sqlsrv的东西。

我想要转换的代码是：（编辑为包含完整代码）

 function database($querydb) {

global $global;
global $field;

if (isset($global['queries'])) {
    $global['queries']++;
} else {
    $global['queries'] = "1";
}
$field['queries'] = $global['queries'];
if (isset($global['query_log'])) {
    $global['query_log'] .= "\n<br>$querydb";
} else {
    $global['query_log'] = "$querydb";
}

$serverName = "XXX";
$uid = "XXX";
$pwd = "XXX";
$dbName = "XXX";


$connectionInfo = array("UID" => $uid, "PWD" => $pwd, "Database" => $dbName, "ReturnDatesAsStrings"=>true);

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

$stmt = sqlsrv_query($conn, $querydb) or return_error("Query Error: $querydb");

    while( $row = sqlsrv_fetch_array( $stmt, SQLSRV_FETCH_NUMERIC) ) 
        {
        $global['dbresult'] = $row;
        }


if ((substr($querydb,0,6)!="INSERT") && (substr($querydb,0,6)!="UPDATE") && (substr($querydb,0,6)!="DELETE")) {

while ($row1 = sqlsrv_fetch_array($stmt))
{
$global['dbnumber'] = mysql_numrows($global['dbresult']);  // original $dbnumber
}

return;

}

function return_error($error) {
print $error;
exit;
}

function date_status($date, $username) {

global $global;
global $field;
global $input;
global $text;

$status = "0";

if ($username!="") {

    $query = "SELECT countedrow.total, id, start_date, end_date FROM calendar JOIN (SELECT total = COUNT(*) FROM oc_calendar) AS countedrow ON 1=1";

    database($query);


    for ($i = 0; $i < $global['dbnumber']; $i++) {

        $status = "1";

        $event_id = sqlsrv_fetch_array($global['dbresult'],$i,"id"); 


    }
}

return $status;

}

我试过sqlsrv_get_field，sqlsrv_fetch，sqlsrv_fetch_array，但我显然没有正确的语法，并且因为我收到错误而误解了这个：

sqlsrv_fetch_array（）期望参数1是资源，在...中给出的数组

......无论我做什么。

如何从数组中提取id以设置$ event_id？我哪里错了？

Answer 1

$querydb = "SELECT countedrow.total, id, start_date, end_date FROM oc_calendar JOIN (SELECT total = COUNT(*) FROM oc_calendar) AS countedrow ON 1=1";

$stmt = sqlsrv_query($conn, $querydb);

if( $stmt === false) {
    die( print_r( sqlsrv_errors(), true) );
}

while($row = sqlsrv_fetch_array($stmt, SQLSRV_FETCH_ASSOC)) {
    $event_id = $row["id"];
}

这有用吗？

为什么你在一段时间内将所有行分别映射到globalvar，然后再次循环遍历全局变量！如果你需要将它从函数返回全局，然后将其作为返回数组传递而不是全局

，则直接将它全部循环

Answer 2

您说您不知道如何为sqlsrv_fetch_array()提供有效的第一个参数，但您的代码已正确执行：

$stmt = sqlsrv_query($conn, $querydb) or return_error("Query Error: $querydb");
while( $row = sqlsrv_fetch_array( $stmt, SQLSRV_FETCH_NUMERIC) ) {
}

然而你迷失在你自己的意大利面中：

while( $row = sqlsrv_fetch_array( $stmt, SQLSRV_FETCH_NUMERIC) ) 
{
    $global['dbresult'] = $row;
}
$event_id = sqlsrv_fetch_array($global['dbresult'],$i,"id");

老实说，我建议您删除此代码并从头开始。

mysql_result转换为sqlsrv

2 个答案: