1
我有一個MySQL查詢,按月將我的數據庫中的所有ClientCostToDate
行(類型爲DECIMAL
)相加,然後將數據作爲JSON返回。在MySQL數據庫的列中計數值的查詢
我的PHP腳本是:
//the sql query to be executed
$estimates_query = "SELECT DATE_FORMAT(CreatedDate, '%M %Y') AS CreatedMonth,
SUM(ClientCostToDate) AS ClientCostsTotal,
EXTRACT(YEAR_MONTH FROM CreatedDate) AS CreatedYearMonth
FROM Estimates
WHERE CreatedDate IS NOT NULL
AND EXTRACT(YEAR_MONTH FROM CreatedDate) >= EXTRACT(YEAR_MONTH FROM CURDATE())-100
GROUP BY DATE_FORMAT(CreatedDate, '%M')
ORDER BY CreatedYearMonth";
//storing the result of the executed query
$result = $conn->query($estimates_query);
//initialize the array to store the processed data
$estimatesJsonArray = array();
//check if there is any data returned by the sql query
if ($result->num_rows > 0) {
//converting the results into an associative array
while ($row = $result->fetch_assoc()) {
$jsonArrayItem = array();
$jsonArrayItem['date'] = $row['CreatedMonth'];
$jsonArrayItem['clientCostsTotal'] = $row['ClientCostsTotal'];
//append the above created object into the main array
array_push($estimatesJsonArray, $jsonArrayItem);
}
}
//close the connection to the database
$conn->close();
//set the response content type as json
header('Content-type: application/json');
//output the return value of json encode using the echo function
echo json_encode($estimatesJsonArray, JSON_PRETTY_PRINT);
這裏是JSON:
[
{
"date": "February 2016",
"clientCostsTotal": "21211.25"
},
{
"date": "March 2016",
"clientCostsTotal": "206996.25"
},
{
"date": "April 2016",
"clientCostsTotal": "74667.50"
},
{
"date": "May 2016",
"clientCostsTotal": "61128.75"
},
{
"date": "June 2016",
"clientCostsTotal": "267740.50"
},
{
"date": "July 2016",
"clientCostsTotal": "200946.75"
},
{
"date": "August 2016",
"clientCostsTotal": "0.00"
}
]
有一個在MySQL數據庫Status
(VARCHAR
型)的另一列。它包含以下值:新估計,已批准,已開票,正在等待結算,已取消,有客戶,已提交成本。
我需要編寫一個MySQL查詢,該查詢給出了組成SUM(ClientCostToDate) AS ClientCostsTotal
的行的所有狀態。然後,我需要計算每種狀態的數量(新估計,已批准,已開票,等待結算,已取消,有客戶,已提交成本)。什麼是完成這個最好的方法?
謝謝。這完全符合我喜歡的方式。爲了獲得所有'Status'值的總數,你會運行一個SUM(num_NewEstimate,num_Approved,...)嗎?這是完成這一難題的最佳方式嗎? – Liz
@LizBanach:MySQL不允許你在同一個查詢的SELECT列表中引用列別名。您可以使此查詢成爲內聯視圖(具有必需的性能損失),並在外部查詢中引用列名稱。或者,1)重複相同的表達式(即得到單獨的計數),但將它們與加法運算符(+)組合,而不是使用逗號分隔符。或2)做另一個表達式,測試所有值SUM(狀態IN('新估計','批准',...))AS num_in_status_list'。 – spencer7593
謝謝你的指導@ spencer7593 – Liz