Question

我正在尝试创建一个图表，该图表从我们的某个数据库获取数据，并计算过去12个月中每个问题的数量。查询运行正常，但即使数据库中存在与查询匹配的记录，它仍会返回0结果。我使用的代码如下......

 <?php
if (isset($_POST['dept'])) {
    $dept = $_POST['dept'];

} else {

    $dept = '';

}


for ($i = 1; $i <= 12; $i++) {
    $months[] = date("Y-m-d", strtotime(date('Y-m-01') . " -$i months"));

}

foreach ($months as $month => $value) {

    $theMonth = date("M", strtotime($value));
    $theYear  = date("Y", strtotime($value));

    $sql = "SELECT `c_ID`, `department`, `departmentSub`,  `todaysDate`, \n" . "SUM(CASE WHEN department='$dept' AND  departmentSub LIKE 'Loading Error' AND MONTH(todaysDate) = MONTH($value) AND YEAR(todaysDate) = YEAR($value) THEN 1 ELSE 0 END) AS loadingErrors,\n" . "SUM(CASE WHEN department='$dept' AND  departmentSub LIKE 'Bad Customer Service' AND MONTH(todaysDate) = MONTH($value) AND YEAR(todaysDate) = YEAR($value) THEN 1 ELSE 0 END) AS custService\n" . "FROM `nonConformance`";


    if ($result = mysqli_query($conn, $sql)) {
        // Return the number of rows in result set
        while ($row = $result->fetch_assoc()) {
            $loadingErrors = $row['loadingErrors'];
            $custService   = $row['custService'];
        }

    }
}
?>

我无法弄清楚为什么这不会返回任何结果，任何人都可以帮助我，我会非常感激。

谢谢！

[示例数据] http://imgur.com/g0UzcpR

Answer 1

在您的查询中，您不会将$value作为字符串传递。尝试使用撇号围绕$value：

 $sql = "SELECT `c_ID`, `department`, `departmentSub`,  `todaysDate`, \n" . "SUM(CASE WHEN department='$dept' AND  departmentSub LIKE 'Loading Error' AND MONTH(todaysDate) = MONTH('$value') AND YEAR(todaysDate) = YEAR('$value') THEN 1 ELSE 0 END) AS loadingErrors,\n" . "SUM(CASE WHEN department='$dept' AND  departmentSub LIKE 'Bad Customer Service' AND MONTH(todaysDate) = MONTH('$value') AND YEAR(todaysDate) = YEAR('$value') THEN 1 ELSE 0 END) AS custService\n" . "FROM `nonConformance`";

MONTH($value)变为MONTH('$value')，YEAR($value)变为YEAR('$value')

另外，请确保您在$loadingErrors循环中输出$custService和foreach ($months as $month => $value) {的值，否则具有零值的日期将覆盖之前的值。

Answer 2

您是否在查询中尝试了此操作？

替换

    departmentSub LIKE 'Loading Error'

与

    departmentSub = 'Loading Error'

或

    departmentSub LIKE '%Loading Error%'

如果您要查找精确的字符串值，为什么要使用LIKE？

在PHP中使用mysql查询返回0结果

2 个答案: