如何從PHP腳本同時寫入和讀取數據到Android設備

Question

我有一個PHP腳本，用於將數據寫入MYSQL數據庫。腳本從Android應用程序中檢索數據並將其寫入數據庫表。我想要的是：PHP腳本將根據輸入日期計算數據庫表中的數據計數，並將計數傳遞給Android設備，根據該設備我需要在Android應用中進行一些驗證。在同一個腳本中可以這樣做嗎？我的意思是可以通過單擊按鈕從Android應用程序向數據庫表寫入數據，也可以將相同腳本中的數據讀入App.Can任何人都可以幫我解決這個問題嗎？

我的PHP腳本是：

<?php 
require "conn.php"; 
require "SalesLogin.php"; 

$enquiry = $_POST["enquiry"]; 
$retail = $_POST["retail"]; 
$collection = $_POST["collection"]; 
$booking = $_POST["booking"]; 
$evaluation = $_POST["evaluation"]; 
$test_drive = $_POST["test_drive"]; 
$home_visit = $_POST["home_visit"]; 
$user_name = $_POST["user_name"]; 
$update_date = $_POST["date"]; 
$absent = $_POST["absent"]; 


$timezone = new DateTimeZone("Asia/Kolkata"); 
$date = new DateTime(); 
$date->setTimezone($timezone); 
$time = $date->format('H:i:s A'); 


$sql = "UPDATE employee_details SET 
enquiry_sum = (SELECT SUM(enquiry) +'$enquiry' FROM (SELECT * FROM employee_details WHERE date = CURDATE() AND name = '$user_name') AS x) 
WHERE date = CURDATE() AND name = '$user_name'"; 
$res = $conn->query($sql); 

$check = "UPDATE employee_details SET enquiry_target_status = (SELECT IF (MAX(enquiry_sum) = 52, 'ACHIEVED', 'NOT ACHIEVED') FROM (SELECT * from employee_details WHERE date = CURDATE() AND name = '$user_name') AS Y) WHERE date = CURDATE() AND name = '$user_name'"; 
$insert_status = $conn->query($check); 


$miss_count = "UPDATE employee_details SET enquiry_target_missed_by = (SELECT (50 - MAX(enquiry_sum)) FROM (SELECT * from employee_details WHERE date = CURDATE() AND name = '$user_name') AS Z) WHERE date = CURDATE() AND name = '$user_name'"; 
$insert_status = $conn->query($miss_count); 


$mysql_qry1 = "INSERT INTO employee_details(enquiry,retail, 
collection,booking, evaluation, test_drive, home_visit, name, date,time,absent) values ('$enquiry','$retail','$collection','$booking','$evaluation','$test_drive', 
'$home_visit','$user_name','$update_date','$time','$absent');"; 

if($conn->query($mysql_qry1) === TRUE) 
    echo "Your details has been successfully inserted."; 

else 
    echo "Error: " .$mysql_qry1. "<br>" . $conn->error; 


if($update_date != $date){ 
$mysql_qry2 = "UPDATE employee_data SET last_updated_date = (DATE_ADD('$update_date', INTERVAL 1 DAY)) WHERE name = '$user_name';"; 
$conn->query($mysql_qry2); 
echo "Date changed," .$mysql_qry2; 
} 


$mysql_qry3 = "SELECT COUNT(*) from employee_details WHERE date = '$update_date' and name LIKE '$user_name';"; 
$conn->query($mysql_qry3); 
     if($mysl_qry3 <= 2) 
       { 
        echo "You can login."; 
       } 
     else 
       { 
        echo "You cannot login anymore for today."; 
      } 


$conn->close(); 
?> 

我要檢查的條件，並通過結果爲以下部分和Android應用程序使驗證：

$mysql_qry3 = "SELECT COUNT(*) from employee_details WHERE date = '$update_date' and name LIKE '$user_name';"; 
$conn->query($mysql_qry3); 
     if($mysl_qry3 <= 2) 
       { 
        echo "You can login."; 
       } 
     else 
       { 
        echo "You cannot login anymore for today."; 
      } 

Answer 1

我正在的問題是：我沒有在PHP腳本中使用「AS」關鍵字。

更新後的PHP腳本是：

$mysql_qry3 = "SELECT COUNT(*) AS count from employee_details WHERE date = '$update_date' and name LIKE '$user_name';"; 
$result1 = mysqli_query($conn,$mysql_qry3); 

         $row = mysqli_fetch_assoc($result1); 
         $count = $row['count']; 

         echo "Count: " .$count; 
     if($count <= 2) 
       { 
        echo "You can login."; 
       } 
     else 
       { 
        echo "You cannot login anymore for today."; 
      } 

這給了期望的結果。