Question

我试图将我的应用程序列入其中，以便批准或拒绝。我之前已经选择了很多并且显示但是由于某种原因这个没有用。它显示表格......例如：用户名，你是谁？但它没有显示$ applicationRow ['Username']等。我已经通过mysql编辑器运行查询，它工作正常，所以我知道它不是那样的。它进入了while循环，因为它显示了表...但是为什么它不显示结果???

$applicationQuery = "SELECT tblMembers.Username, tblMembers.Bio, tblApplications.WhyJoin, tblApplications.Games, tblApplications.FoundBy, tblApplications.Joke
                                FROM tblApplications
                                INNER JOIN tblMembers ON tblMembers.ID = tblApplications.MemberID
                                WHERE Approved=0";
        echo $applicationQuery;
        $applicationResults = mysql_query($applicationQuery) or die(mysql_error());

        while($applicationRow = mysql_fetch_row($applicationResults))
        {
            echo "<table>";
            echo "</tr><td>Username:</td><td></td>".$applicationRow['Username']."</tr>";
            echo "</tr><td>Who are you?</td><td></td>".$applicationRow['Bio']."</tr>";
            echo "</tr><td>What games do you play?</td><td></td>".$applicationRow['Games']."</tr>";
            echo "</tr><td>How did you find us?</td><td></td>".$applicationRow['FoundBy']."</tr>";
            echo "</tr><td>Why do you want to join?</td><td></td>".$applicationRow['WhyJoin']."</tr>";
            echo "</tr><td>Tell us a joke:</td><td></td>".$applicationRow['Joke']."</tr>";
            echo "</table><hr/>";
        }

Answer 1

mysql_fetch_row()返回一个数值数组。

请尝试使用mysql_fetch_assoc()。这将返回一个关联数组，它出现在你正在寻找的位置。

Answer 2

更改此

$applicationQuery = "SELECT tblMembers.Username, tblMembers.Bio, tblApplications.WhyJoin, tblApplications.Games, tblApplications.FoundBy, tblApplications.Joke
                                FROM tblApplications
                                INNER JOIN tblMembers ON tblMembers.ID = tblApplications.MemberID
                                WHERE Approved=0";

到

$applicationQuery = "SELECT tblMembers.Username as Username, tblMembers.Bio as Bio, tblApplications.WhyJoin as WhyJoin, tblApplications.Games as Games, tblApplications.FoundBy as FoundBy, tblApplications.Joke ad Joke
                                FROM tblApplications
                                INNER JOIN tblMembers ON tblMembers.ID = tblApplications.MemberID
                                WHERE Approved=0";

因为当您在while循环中获取时，列别名不匹配

并使用mysql_fetch_Array代替mysql_fetch_row

MySql结果即将进入但未显示

2 个答案: