Question

我有一个函数，它从我的数据库中请求一些信息，将它放在一个数组中并返回它。 sql语句是正确的。

如果在数据库中找不到当前日期，我想在我的网站上显示错误。错误消息也会添加到返回数组中。

首先我用COUNT检查数据库中是否存在相应的日期，如果count == 1，我用另一个语句从数据库中获取所有数据。如果计数！= 1，我将数组硬编码放在一起。

当在数据库中找到与给定参数相同的日期时，脚本就像一个魅力，但当我更改数据库中的日期时，我得到以下错误：

注意：未定义的变量：行第1257行/Applications/MAMP/htdocs/models/funcs.php

第1252-1257行：

while($stmt->fetch())
            {
                $row[] = array('id' => $id, 'datum' => $datum, 'mac' => $mac, 'ipad' => $ipad, 'iphone' => $iphone, 'imember' => $imember, 'applecare' => $applecare, 'verkoop_ochtend' => $verkoop_ochtend, 'verkoop_middag' => $verkoop_middag, 'verkoop_avond' => $verkoop_avond, 'vracht_ochtend' => $vracht_ochtend, 'vracht_middag' => $vracht_middag, 'vracht_avond' => $vracht_avond, 'service_ochtend' => $service_ochtend, 'service_middag' => $service_middag, 'service_avond' => $service_avond, 'werkzaamheden' => $werkzaamheden, 'error' => '');
            }
            $stmt->close();
            $data = $row;

有谁知道我在这里做错了什么？提前谢谢！

完整的功能：

function get_current_planning($date)
{
    GLOBAL $mysqli, $db_table_prefix;
    $stmt = $mysqli->prepare("SELECT COUNT(datum) FROM " . $db_table_prefix . "planning WHERE datum = '" . $date . "'");
    $result = $stmt->execute();
print_r($result);
var_dump($result);

    if($result == 1)
    {
        $stmt->prepare("SELECT
                                        *
                                        FROM " . $db_table_prefix . "planning
                                        WHERE datum = '" . $date . "'");
        $stmt->execute();
        $stmt->bind_result($id, $datum, $mac, $ipad, $iphone, $imember, $applecare, $verkoop_ochtend, $verkoop_middag, $verkoop_avond, $vracht_ochtend, $vracht_middag, $vracht_avond, $service_ochtend, $service_middag, $service_avond, $werkzaamheden);
        while($stmt->fetch())
            {
                $row[] = array('id' => $id, 'datum' => $datum, 'mac' => $mac, 'ipad' => $ipad, 'iphone' => $iphone, 'imember' => $imember, 'applecare' => $applecare, 'verkoop_ochtend' => $verkoop_ochtend, 'verkoop_middag' => $verkoop_middag, 'verkoop_avond' => $verkoop_avond, 'vracht_ochtend' => $vracht_ochtend, 'vracht_middag' => $vracht_middag, 'vracht_avond' => $vracht_avond, 'service_ochtend' => $service_ochtend, 'service_middag' => $service_middag, 'service_avond' => $service_avond, 'werkzaamheden' => $werkzaamheden, 'error' => '');
            }
            $stmt->close();
            $data = $row;
    }
    else
    {
        $row[] = array('id' => '', 'datum' => '', 'mac' => '', 'ipad' => '', 'iphone' => '', 'imember' => '', 'applecare' => '', 'verkoop_ochtend' => '', 'verkoop_middag' => '', 'verkoop_avond' => '', 'vracht_ochtend' => '', 'vracht_middag' => '', 'vracht_avond' => '', 'service_ochtend' => '', 'service_middag' => '', 'service_avond' => '', 'werkzaamheden' => '', 'error' => 'Er is geen planning gevonden voor de huidige datum!');
        $stmt->close();
        $data = $row;
    }
    return $data;
}

print_r（$ result）返回1

var_dump（$ result）返回bool（true）

Answer 1

您未fetch() SELECT COUNT()...查询的结果。在我看来，您的第二个查询中的while(...fetch())循环有时会运行零次。这将导致代码开始$row[] =无法运行，这反过来会导致在$row循环后while(...fetch())未定义。

无论如何，计算行然后获取它们有两个连续查询有点浪费。您可能希望完全跳过SELECT COUNT查询。相反，做这样的事情：

$row = Array();
while($stmt->fetch()) {
    $row[] = array('id' => $id, 'datum' => $datum, 'mac' => $mac, 'ipad' => $ipad, 'iphone' => $iphone, 'imember' => $imember, 'applecare' => $applecare, 'verkoop_ochtend' => $verkoop_ochtend, 'verkoop_middag' => $verkoop_middag, 'verkoop_avond' => $verkoop_avond, 'vracht_ochtend' => $vracht_ochtend, 'vracht_middag' => $vracht_middag, 'vracht_avond' => $vracht_avond, 'service_ochtend' => $service_ochtend, 'service_middag' => $service_middag, 'service_avond' => $service_avond, 'werkzaamheden' => $werkzaamheden, 'error' => '');
}
$stmt->close();
if (0 == count($row)) {
     /* deal with the no matching rows case */
}
$data = $row;

Answer 2

看起来您还没有定义变量$row。在开始将$row=Array();视为数组之前，您需要$row。

未定义的变量：行

2 个答案: