Question

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我遇到了这个困境：

users  ID  username  firstname  middlename lastname
        1  bilbodog    Casper               Thomsen
        2  bilbodog2   Smith      John      Andersen


flirts    flirter_id  flirted_id  date  time
              1          2          X    X

我想获得flirter_id'1'所做的所有flirted_id，并将flirted_id与用户进行比较并收集名字，中间名和＆amp;来自flirted_id的姓氏。另外，我想从调情表中收集日期和时间。然后我想在HTML表格中回显结果。

所以现在我这样做：

<?php
$ID=$_SESSION['ID'];
// Create connection
$conn = new mysqli($mysql_hostname, $mysql_user, $mysql_password, $mysql_database);
// Check connection
if ($conn->connect_error) {
  die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT flirted_id, date, time FROM flirts WHERE flirter_id='$ID'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
  ----> ECHO TABLE <----
} else {
  echo 'No flirts found.';
}
$conn->close();
?>

但这不起作用。

我在这一点上很困惑。我不知道如何从1个表中收集3个结果并使用其中一个结果在表2中进行比较，以收集该ID的人的名字，中间名和姓氏。

提前致谢！

Answer 1

如果我理解正确，您想要联系以获取table1中的信息以及表2中列为{1}}的人员在表1中的信息

added_id

如果您还想了解select t1.date, t1.time, t2.first, t2.middle, t2.last from table1 t1 join table2 t2 on (t1.added_id = t2.id) where t1.adder_id = 1;的信息（假设这些人也存储在表2中）：

adder_id

在这种情况下，你应该给cols别名，以便更清楚地识别它们，例如select t1.date, t1.time, t2.first, t2.middle. t2.last, t3.first, t3.middle, t3.last from table1 t1 join table2 t2 on (t1.added_id = t2.id) join table2 t3 on (t1.added_id = t3.id) where t1.adder_id = 1;等等

注意我假设t2.first as added_id_first, t3.first as adder_id_first和adder_id字段中没有NULL值。如果他们可以在声明中使用added_id而不是LEFT JOIN。

Answer 2

试试吧

select concat (table2.first," ",table2.middle ) as adder_id_Name,AddedName.* from table2,(SELECT concat (tb2.first," ",tb2.middle ) as added_id_Name,  tb1.adder_id from table2 tb2,table1 tb1 where tb1.added_id = tb2.ID) as AddedName where table2.ID = AddedName.adder_id

Answer 3

我自己发现了。我扼杀了我的大脑并得到了这个答案：

                              <?php
                                    // Create connection
                                    $conn = new mysqli($mysql_hostname, $mysql_user, $mysql_password, $mysql_database);
                                    // Check connection
                                    if ($conn->connect_error) {
                                        die("Connection failed: " . $conn->connect_error);
                                    }
                                        $sql = "SELECT from_id, message, date, time FROM messages WHERE to_id='$ID'";
                                        $result = $conn->query($sql);
                                    if ($result->num_rows > 0) {
                                        // output data of each row
                                        while($row = $result->fetch_assoc()) {
                                            $from_id=$row["from_id"];
                                            $message=$row["message"];
                                            $date=$row["date"];
                                            $time=$row["time"];
                                            $sql10 = "SELECT firstname, middlename, lastname FROM users WHERE ID='$from_id'"; 
                                            $sql11 = mysql_query($sql10); 
                                            $sql12 = mysql_fetch_row($sql11);
                                            $firstname=$sql12[0];
                                            $middlename=$sql12[1];
                                            $lastname=$sql12[2];
                                            ?>
                                            <a href="#" class="list-group-item">
                                                <span class="label label-default"><?php echo $date, ' ', $time; ?></span>
                                                <span class="fa fa-star"></span>
                                                <span class="name" style="min-width: 120px; display: inline-block;">Casper Thomsen</span>
                                                <span class="text-muted" style="font-size: 11px;">- <?php echo $message; ?></span>
                                                <span class="pull-right">
                                                    <button type="submit" id="view_message" name="view_message" class="btn btn-default btn-xs pull-right">L&aelig;s  Besked <i class="fa fa-angle-right"></i></button>
                                                </span>
                                            </a>
                                            <?php
                                        }
                                    } else {
                                        echo '
                                        <div class="block block-team_member_block span12 first cf">
                                            <div class="member">
                                                <div class="member-social">
                                                </div>
                                                <h3>No members or could not connect to database..</h3>
                                                <div class="bline"></div>
                                            </div>
                                        </div>';
                                    }
                                    $conn->close();
                                ?>

从table1中选择3个值并使用table2中table1中的一个值来获取结果？

3 个答案: