無法通過 XMLHttpRequest 將數據從 php 頁面發送到另一個 php 頁面

Question

我有一個表，每一行都有唯一的 ID，所以我試圖將此 ID 發送到另一個 PHP 頁面以在 SQL 語句中使用此 ID。 我嘗試做一些回顯語句來檢查請求是否成功但回顯語句沒有執行

發送數據的函數

<script>
    function promote(id) {
        var xhttp = new XMLHttpRequest();
        xhttp.onreadystatechange = function () {
            if (this.readyState == 4 && this.status == 200) {
                alert("Promoted to QA successfully");
            }  
        };
        xhttp.open("GET", "promoteAccount.php?id="+id, true);
        xhttp.send();
    }
</script>

促進帳戶.php

<?php
    $id = $_REQUEST['id'];
    require_once 'dbconnection.php';
    $sql = "SELECT * FROM staf WHERE StaffID = ".$id;
    $result = mysqli_query($conn, $sql);
    if ($result) {
        if ($row = mysqli_fetch_array($result)) {
            if ($row['Role'] == "Receptionist") {
                $updateSQL = "UPDATE staff SET Role = 'QA' WHERE StaffID = ".$id;
                $updateResult = mysqli_query($conn, $updateSQL);
            }
        }
    }
?>

Answer 1

根據上面關於 GET 與 POST 對於這種性質的操作的評論，我建議您將 PHP 更改為僅接受 POST 請求，並在您的 Javascript 中使用更新且功能更強大的fetch api 。 似乎可以簡化 SQL，以便您使用where子句中的列role發出單個查詢來驗證更新是否適用，而不是使用單獨的查詢並遍歷記錄集。

<?php
    
    if( $_SERVER['REQUEST_METHOD']=='POST' && !empty( $_POST['id'] ) ){
    
        require_once 'dbconnection.php';
        
        $code='QA';
        $role='Receptionist';
        # error: missing ` from after staffid!!
        $sql='update `staff` set `role`=? where `staffid`=? and `role`=?';
        $stmt=$conn->prepare( $sql );
        $stmt->bind_param('sss', $code, $_POST['id'], $role );
        $stmt->execute();
        $result=$stmt->affected_rows;
        $stmt->close();
        $conn->close();
        
        http_response_code( 200 );
        exit( $result==1 ? 'Success' : 'Failed' );
    }
    
    http_response_code( 404 );
    
    
?>

以及使用Fetch修改的 Javascript 函數

const promote=( id=false )=>{
    if( id ){
        let fd=new FormData();
            fd.set('id',id);
            
        fetch( 'promoteAccount.php',{ method:'post',body:fd } )
            .then(r=>r.text())
            .then(text=>{
                alert('Promoted to QA successfully');
                console.log( text );
            })
    }
}

我創建的單頁演示來測試這個：

<?php
    
    if( $_SERVER['REQUEST_METHOD']=='POST' && !empty( $_POST['id'] ) ){
    
        #require_once 'dbconnection.php';
        
        chdir('../../dbo');
        require 'db-conn-details.php';
        require 'mysqli-conn.php';
        
        
        
        $code='QA';
        $role='Receptionist';
        
        $sql='update `staff` set `role`=? where `staffid`=? and `role` = ?';
        
        $stmt=$conn->prepare( $sql );
        $stmt->bind_param('sss', $code, $_POST['id'], $role );
        
        $stmt->execute();
        $result=$stmt->affected_rows;
        
        $stmt->close();
        $conn->close();
        
        http_response_code( 200 );
        exit( $result==1 ? 'Success' : 'Failed' );
    }
    
?>
<!DOCTYPE html>
<html lang='en'>
    <head>
        <meta charset='utf-8' />
        <title></title>
    </head>
    <body>
        <input type='button' data-id=1 value='Promote 1' />
        <input type='button' data-id=2 value='Promote 2' />
        <input type='button' data-id=3 value='Promote 3' />
        <input type='button' data-id=4 value='Promote 4' />
        <input type='button' data-id=5 value='Promote 5' />
        <input type='button' data-id=6 value='Promote 6' />
        <input type='button' data-id=7 value='Promote 7' />
        <script>
            const promote=( id=false )=>{
                if( id ){
                    let fd=new FormData();
                        fd.set('id',id);
                        
                    fetch( 'promoteAccount.php',{ method:'post',body:fd } )
                        .then(r=>r.text())
                        .then(text=>{
                            alert('Promoted to QA successfully');
                            console.log( text );
                        })
                }
            }
            document.querySelectorAll('input[type="button"]').forEach( bttn=>bttn.addEventListener('click',(e)=>{
                promote( e.target.dataset.id );
            }))
        </script>
    </body>
</html>

使用下表架構和數據：

+---------+------------------+------+-----+---------+----------------+
| Field   | Type             | Null | Key | Default | Extra          |
+---------+------------------+------+-----+---------+----------------+
| staffid | int(10) unsigned | NO   | PRI | NULL    | auto_increment |
| name    | varchar(50)      | NO   |     | 0       |                |
| role    | varchar(50)      | NO   |     | 0       |                |
+---------+------------------+------+-----+---------+----------------+



+---------+-----------+--------------+
| staffid | name      | role         |
+---------+-----------+--------------+
|       1 | Paula     | QA           |
|       2 | Daniela   | PA           |
|       3 | Susan     | Cleaner      |
|       4 | Isobel    | Receptionist |
|       5 | Carrie    | Team Leader  |
|       6 | Chantelle | IT support   |
|       7 | Helen     | Receptionist |
+---------+-----------+--------------+

無法通過 XMLHttpRequest 將數據從 php 頁面發送到另一個 php 頁面

問題描述

1 個解決方案

解決方案1
0 2022-06-03 13:46:21

無法通過 XMLHttpRequest 將數據從 php 頁面發送到另一個 php 頁面

問題描述

1 個解決方案

解決方案1 0 2022-06-03 13:46:21

解決方案1
0 2022-06-03 13:46:21