Question

这是我做SQL的第三天，我确信它很小。

我有这个代码。 $ con是我的SQL连接变量。

for ($i = 0; $i < mysqli_num_rows(mysqli_query($con, "SELECT * FROM offers WHERE active='y'")); $i++)
{
    echo '<div class="dottedBox">User: '.mysqli_fetch_row(mysqli_query($con, "SELECT poster FROM offers WHERE active='y'"))[$i]."<br>
    Has: ".mysqli_fetch_row(mysqli_query($con, "SELECT haveamt FROM offers WHERE active='y'"))[$i]." ".mysqli_fetch_row(mysqli_query($con, "SELECT have FROM offers WHERE active='y'"))[$i].'<br>
    Wants: '.mysqli_fetch_row(mysqli_query($con, "SELECT wantamt FROM offers WHERE active='y'"))[$i].' '.mysqli_fetch_row(mysqli_query($con, "SELECT want FROM offers WHERE active='y'"))[$i].'</div><br>';
}

第一项输出正常。但是，每个后续项都会导致以下错误：

Notice: Undefined offset: 1 in /var/www/civ/index.php on line 6

Notice: Undefined offset: 1 in /var/www/civ/index.php on line 7

Notice: Undefined offset: 1 in /var/www/civ/index.php on line 7

Notice: Undefined offset: 1 in /var/www/civ/index.php on line 8

Notice: Undefined offset: 1 in /var/www/civ/index.php on line 8

运行SELECT * FROM offers;在mysql中返回此内容（我正在使用Debian）：

mysql> SELECT * FROM offers;
+----------+-----------+---------+-----------+---------+---------+--------+
| poster   | have      | haveamt | want      | wantamt | offerid | active |
+----------+-----------+---------+-----------+---------+---------+--------+
| minicl55 | gfdsafdsa |       3 | fdsafdsa  |       2 |       1 | y      |
| minicl55 | jklgdsa   |     324 | jfklsdjkl |    4522 |       2 | y      |
+----------+-----------+---------+-----------+---------+---------+--------+

我有两个问题：

为什么输出不正确
我该如何解决这个问题？

Answer 1

您正在查询数据库中的循环中的每个值。您应该只需要查询一次数据库，然后遍历结果以显示它们。

$result = mysqli_query($con, "SELECT * FROM offers WHERE active='y'");

while ($row = mysqli_fetch_assoc($result))
{
    echo '<div class="dottedBox">User: '.$row['poster']."<br>
    Has: ".$row['haveamt']." ".$row['offers'].'<br>
    Wants: '.$row['wantamt'].' '.$row['offers'].'</div><br>';
}

mysqli_fetch_row仅适用于第一行

1 个答案: