[英]Retrieve data from mysql using android
我有一个问题,例如:我们有2个用户,因此用户A将数据插入数据库,可以毫无问题地检索到它。 用户B将数据插入数据库中,并且当用户B要检索数据时,用户B获取用户A而不是用户B的数据。
是什么会导致该问题?
这是我的用于从mysql数据库获取数据的php代码:
<?php
// array for JSON response
$response = array();
// include db connect class
require_once __DIR__ . '/db_connect.php';
// connecting to db
$db = new DB_CONNECT();
// check for post data
if (isset($_GET["subject_id"])) {
$subject_id = $_GET['subject_id'];
// get a product from products table
$result = mysql_query("SELECT * FROM subject_offered WHERE subject_id = $subject_id");
if (!empty($result)) {
// check for empty result
if (mysql_num_rows($result) > 0) {
$result = mysql_fetch_array($result);
$product = array();
$product["subject_id"] = $result["subject_id"];
$product["lecturer_name"] = $result["lecturer_name"];
$product["time_offered"] = $result["time_offered"];
$product["subject_details"] = $result["subject_details"];
//$product["updated_at"] = $result["updated_at"];
// success
$response["success"] = 1;
// user node
$response["product"] = array();
array_push($response["product"], $product);
// echoing JSON response
echo json_encode($response);
} else {
// no product found
$response["success"] = 0;
$response["message"] = "No subject found";
// echo no users JSON
echo json_encode($response);
}
} else {
// no product found
$response["success"] = 0;
$response["message"] = "No subject found";
// echo no users JSON
echo json_encode($response);
}
} else {
// required field is missing
$response["success"] = 0;
$response["message"] = "Required field(s) is missing";
// echoing JSON response
echo json_encode($response);
}
?>
// get a product from products table
$result = mysql_query("SELECT * FROM subject_offered WHERE subject_id = $subject_id");
应该
// get a product from products table
$result = mysql_query("SELECT * FROM subject_offered WHERE subject_id = '$subject_id' ");
//some times when you miss '' this single commas then it also create a problem
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.