Question

我有一个表用户

user(id, username, email, profile_pic) // Id is primary key

现在我想在PHP代码中的每次迭代中一次获取两行结果，因为我需要从数据库中显示两行。我想以下面的方式显示我的结果。

示例

user1     user2
user3     user4
user5     user6

任何帮助将不胜感激。

感谢。

Answer 1

您不需要一次获取两行。您需要正确创建视图脚本。

$counter=0;
echo '<table><tr>';
while($row = $result->fetch()) {
    $counter++;
    echo '<td>'.$row['user'].'</td>';
    if ($counter%2 == 0) {
        echo '</tr><tr>';
    }
}

if ($counter%2 == 1) {
    echo '<td>&nbsp;</td>'
}

echo '</tr></table>'

Answer 2

最简单的方法是将所有用户加载到PHP中的数组中，并使用以下伪代码和PHP：

假设您的用户数组具有从0到用户数量的索引。您应在每个单元格中引用的值（x = 0行，y = 0-cols） CellValue（x + 1 + y）

祝你好运！

Answer 3

想法是将用户表加入到本身并根据你的意愿去做规则。所以这是代码。试试吧。

--- DATA ---

    create table user(id int, username varchar(100), email varchar(100), profile_pic varchar(100));

    insert into user
    (id , username , email , profile_pic )
    values
    (10, 'user1', 'x1@x.com', '111');


    insert into user
    (id , username , email , profile_pic )
    values
    (20, 'user2', 'x2@x.com', '222');

    insert into user
    (id , username , email , profile_pic )
    values
    (30, 'user3', 'x3@x.com', '333');

    insert into user
    (id , username , email , profile_pic )
    values
    (40, 'user4', 'x4@x.com', '444');

    insert into user
    (id , username , email , profile_pic )
    values
    (50, 'user5', 'x5@x.com', '555');

    insert into user
    (id , username , email , profile_pic )
    values
    (60, 'user6', 'x6@x.com', '666');


    insert into user
    (id , username , email , profile_pic )
    values
    (70, 'user7', 'x7@x.com', '777');

--- QUERY 1 ---

    SELECT v1.ID as id1, v1.email as email1, v2.ID as id2, v2.email as email2

    FROM

    (

    SELECT
        @i:=@i+1 AS num,
        u.*
    FROM
        (SELECT u1.* FROM USER u1 ORDER BY u1.id ASC) u,
        (SELECT @i:=0) AS tbl

    ORDER BY num ASC

    ) v1


    JOIN


    (

    SELECT
        @j:=@j+1 AS num,
        u.*
    FROM
        (SELECT u1.* FROM USER u1 ORDER BY u1.id ASC) u,
        (SELECT @j:=0) AS tbl

    ORDER BY num ASC

    ) v2 ON ( v1.num = v2.num - 1 AND mod(v1.num, 2) = 1);

如果你想获得最后一行（ID = 7的那一行），也可以这样做。

--- QUERY 2 ---

    SELECT v1.ID as id1, v1.email as email1, v2.ID as id2, v2.email as email2

    FROM

    (SELECT v0.* FROM 
            (

            SELECT
                @i:=@i+1 AS num,
                u.*
            FROM
                (SELECT u1.* FROM USER u1 ORDER BY u1.id ASC) u,
                (SELECT @i:=0) AS tbl

            ORDER BY num ASC

            ) v0 WHERE MOD(v0.num, 2) = 1 
    ) v1 

    LEFT JOIN


    (

    SELECT
        @j:=@j+1 AS num,
        u.*
    FROM
        (SELECT u1.* FROM USER u1 ORDER BY u1.id ASC) u,
        (SELECT @j:=0) AS tbl

    ORDER BY num ASC

    ) v2 ON ( v1.num = v2.num - 1 );

Answer 4

无需修改获取数据的方式。下面是一个明文，简短的解决方案，仅使用前端代码解决：

$rownum = 0;
while($row = $result->fetch()) {
    $rownum++;
    echo $row['username'] . ($rownum %2 == 0) ? "\n" : '     ');
}

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