dongzhong1891 2014-04-19 20:27
浏览 19

无法使这个if语句工作

while ( $org = $orgid->fetch_array()) {
    $getorg = "SELECT Name FROM organisations WHERE ID='" . $org['orgid'] . "'";
    $orgname = $db->query($getorg) or die ($db->error());
    if ( $org['orgid'] !== false ) {
            echo("<td>" . $orgname->fetch_object()->Name . "</td>" );
        $hasOrg = true;
    }
    else {
        echo("<td>Player not in an org.</td>");
    }
}

Link To Image

Here is my code, and above is the result (the link).

CG Staff is the organization, for players with it.

The query only has two results, so that is why it is returning true all the time. How, though, can I make it say something in the for rows in the table that don't have a result?

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1条回答 默认 最新

  • doudao8283 2014-04-19 20:46
    关注

    In your if statement

    if ( $org['orgid'] !== false )
    

    $org['orgid'] has to be identical to FALSE. A zero(0) or NULL will evaluate as true, make sure that column contains a FALSE.

    You could re-write this like this to evaluate the contents of $org['orgid']

    if ( $org['orgid'] ) {
            echo("<td>" . $orgname->fetch_object()->Name . "</td>" );
            $hasOrg = true;
    } else {
        echo("<td>Player not in an org.</td>");
    }
    
    评论

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