[英]join more than tables in json
我想按json文件顯示數據時遇到問題。 如果我在一個表中顯示數據也可以,但是當我要連接多個表時,沒有數據顯示
<?php
mysql_connect($hostname,$username, $password) or die ("<html><script language='JavaScript'>alert('Unable to connect to database! Please try again later.'),history.go(-1)</script></html>");
mysql_select_db($dbname);
$query = "SELECT Product.Product_Name, Product.Price, Product.Image, Gender.Description, Age.Description, Status.Availability from Product join Age on Age.Age_ID join Gender on Gender.Gender_ID join Status on Status.ID";
$result = mysql_query($query);
//Create an array
$json_response = array();
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$row_array['Product_Name'] = $row['Product_Name'];
$row_array['Price'] = $row['Price'];
$row_array['Image'] = base64_encode($row["Image"]);
$row_array['Description'] = $row['Description'];
$row_array['Description'] = $row['Description'];
$row_array['Availability'] = $row['Availability'];
//push the values in the array
array_push($json_response,$row_array);
}
echo json_encode($json_response);
//Close the database connection
fclose($db)
?>
您最后一次加入on Status.ID
是問題。 Status
表沒有ID
列。 根據您的圖表,您具有Status.Status_ID
(沒有Status.ID
)。此外,您的數據還必須具有相關數據,其中每個表具有相同的值,否則,您將獲得空結果
您的圖表:
更改查詢
SELECT
Product.Product_Name,
Product.Price,
Product.Image,
Gender.Description,
Age.Description,
Status.Availability
FROM
Product
JOIN
Age ON Age.Age_ID
JOIN
Gender ON Gender.Gender_ID
JOIN
Status ON Status.ID
至
SELECT
Product.Product_Name,
Product.Price,
Product.Image,
Gender.Description,
Age.Description,
`Status`.Availability
FROM
Product
JOIN
Age ON Product.Age_ID = Age.Age_ID
JOIN
Gender ON Product.Gender_ID = Gender.Gender_ID
JOIN
`Status` ON Product.Status_ID = `Status`.Status_ID
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