我正在尝试创建一个图表,该图表从我们的一个数据库中获取数据,并计算过去 12 个月中每个月遇到的问题数量。查询运行正常,但即使数据库中有与查询匹配的记录,它仍然返回 0 结果。我正在使用的代码如下...
<?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'];
}
}
}
?>
我不明白为什么这没有返回任何结果,任何人都可以帮助我解决这个问题,我将不胜感激。
谢谢!
[示例数据] http://imgur.com/g0UzcpR
最佳答案
在您的查询中,您没有将 $value
作为字符串传递。尝试用撇号将 $value
括起来:
$sql = "SELECT `c_ID`, `department`, `departmentSub`, `todaysDate`, \n" . "SUM(CASE WHEN department='$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('$value')
并且 YEAR($value)
变为 YEAR(' $value')
此外,请确保在 foreach 中输出
循环,否则具有零值的日期将覆盖前一个。$loadingErrors
和 $custService
的值($months as $month => $value) {
关于php - PHP 中的 mysql 查询返回 0 个结果,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/44026791/