的foreach形式中选择

2017-07-02 62 views -2 likes

-2

我希望用户能够选择的行列1，我的代码的foreach形式中选择

$sqlUitlezenRanksUitlezen = mysqli_query($conn, "SELECT * FROM `Ranks` ORDER BY `RankID`"); 
//$sqlDataRanksUitlezen = mysqli_fetch_assoc($sqlUitlezenAccountBewerken); 
foreach($sqlUitlezenRanksUitlezen AS $sqlUitlezenRanksUitlezenEach) { 
    echo '<option value="'.$sqlUitlezenRankEach['RankID'].'">'.$sqlUitlezenRankEach['RankNaam'].'</option>'; 
}

但如果我这样做，它不会显示名次名称。有谁知道如何解决这个问题？

来源

2017-07-02 Lars J.

你永远不取结果。任何基本的mysql教程都会涵盖这一点。 –

我如何获取我的结果？ –

哦，我现在看到它，我的价值是不正确的 –

回答

试试这个： -

$sqlUitlezenRanksUitlezen = mysqli_query($conn, "SELECT * FROM `Ranks` ORDER BY `RankID`"); 
    $sqlDataRanksUitlezen = mysqli_fetch_assoc($sqlUitlezenAccountBewerken); 
    foreach($sqlDataRanksUitlezen AS $sqlUitlezenRanksUitlezenEach) { 
     echo '<option value="'.$sqlUitlezenRankEach['RankID'].'">'.$sqlUitlezenRankEach['RankNaam'].'</option>'; 
    }

来源

2017-07-02 14:01:35

的foreach形式中选择

回答

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