按行更新行

Question

單擊按鈕編輯時，我試圖更新/插入我的數據庫（col：note）。但是，如果我更改一行中的值，所有行都會更新相同的值。我嘗試添加一個WHERE條件（基於編號< - 唯一ID）插入和更新SQL，但無法讓它工作。請幫忙嗎？

$conn = new PDO('mysql:host=localhost;dbname=jr', 'root', 'Js'); 
    $conn->exec("set names utf8"); 
    $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); 
    $num_rows = $conn->query('SELECT COUNT(*) FROM employees')->fetchColumn(); 
    $pages = new Paginator($num_rows,9,array(15,3,6,9,12,25,50,100,250,'All')); 
    echo $pages->display_pages(); 
    echo "<span class=\"\">".$pages->display_jump_menu().$pages->display_items_per_page()."</span>"; 
    $stmt = $conn->prepare("SELECT employees.afnumber,employees.name,employees.dateofemployment,employees.actualpost,employees.department FROM employees WHERE employees.status='Employed' AND (employees.afnumber LIKE '%$search%' OR employees.name LIKE '%$search%') ORDER BY employees.afnumber DESC LIMIT :start,:end"); 
    $stmt->bindParam(':start', $pages->limit_start, PDO::PARAM_INT); 
    $stmt->bindParam(':end', $pages->limit_end, PDO::PARAM_INT); 
    $stmt->execute(); 
    $result = $stmt->fetchAll(); 

    $ewhtable = "<table class='sortable'><tr><th>AFNumber</th><th>Employee Name</th><th>Years of Service</th><th>Actual Post</th><th>Department</th><th>Note</th><th>Deducted Hours</th></tr>\n"; 
    foreach($result as $row) { 
     $years=explode("/", $row[2]); 
     $years[2]=intval(date ('Y')) - $years[2]; 

$sql="SELECT note,deductedwh FROM editedworkhours WHERE afnumber='$row[0]'"; 

$var = ""; 
$varr = ""; 

$stmt = $conn->prepare($sql); 
$stmt->execute(); 
$result2 = $stmt->fetchAll(); 

foreach ($result2 AS $row2) { 
    $var .= $row2['deductedwh'] . "\n"; 
    $varr .= $row2['note'] . "\n"; 
} 
     $Id = $row[0]; 
     $ewhtable .= "<tr><td>$row[0]</td><td>$row[1]</td><td>$years[2]</td><td>$row[3]</td><td>$row[4]</td><td><form method='post'><input type='text' name='Note' value='$varr' style=' padding: 10px;border: solid 2px #c9c9c9; width:200px; height:2px;'><input type='submit' id='search' name='edit' alt='search' value=''></form></td><td>$var</td></tr>\n"; 
    } 



    $ewhtable .= "</table>\n"; 
    echo $ewhtable; 

    exportTable(str_replace("&","",$ewhtable),"EmployeeDeductedWorkHoursTable"); 
    echo $pages->display_pages(); 
    echo "<p class=\"paginate\">Page: $pages->current_page of $pages->num_pages</p>\n"; 

      if(isset($_POST['edit'])) 
    { 
     $note = $_POST['Note']; 

    $sql1="SELECT Note FROM editedworkhours"; 

if ($result=mysqli_query($con,$sql1)) 
    { 

    $rowcount=mysqli_num_rows($result); 
    } 
    if($rowcount==0) 
    { 
$sql="INSERT INTO editedworkhours (Note) VALUES ('$note')"; 
$result = mysqli_query($con,$sql); 
    } 
    else 
    { 
$sql2 = "UPDATE editedworkhours SET Note= '$note'"; 
$result2 = mysqli_query($con,$sql2); 
    } 

    } 
    echo "</div>"; 

Answer 1

有太多的錯誤，在此代碼，從缺乏錯誤報告，SQL注入的直接危險的開始，混合不同數據庫的API，缺乏正確的SQL等等 - 太多使它在一個交代帖子。

這個問題必須關閉太寬泛，你必須坐在你的課本幾個小時。