初學者形式與「更新」和「刪除」按鈕

Question

我是一個PHP的初學者。我看了一個教程來創建一個修改我的wamp創建的mysql數據庫表的表單。首先複製視頻，但後來從頭開始創建自己的桌面，並試圖升級它。

我的添加行工作正常，但更新和刪除沒有。我認爲WHERE條款不正確，參考reg_id。

我創建了一個唯一的主鍵，它自動遞增並且不能修改;這是我想在更改時引用（因爲它不能更改）。

if (isset($_POST['update'])){ 
$UpdateQuery = "UPDATE register SET First_Name='$_POST[first_name]', Last_Name='$_POST[last_name]', Breed='$_POST[breed]', Weight='$_POST[weight]', Age='$_POST[age]', Sex='$_POST[sex]' WHERE '$_POST[reg_id]'='$_POST[reg_id]'"; 
mysqli_query($con,$UpdateQuery);}; 

if (isset($_POST['delete'])){ 
$DeleteQuery = "DELETE FROM register WHERE reg_id='$_POST[reg_id]'"; 
mysqli_query($con,$DeleteQuery);}; 

這裏是它的其餘部分在窗體位於：

while($record=mysqli_fetch_array($myData)){ 
    echo "<form action=register.php method=post>"; 
    echo "<tr>"; 
    echo "<td>" . $record['reg_id'] . " </td>"; 
    echo "<td>" . "<input type=text name=first_name value=" . $record['First_Name'] . " </td>"; 
    echo "<td>" . "<input type=text name=last_name value=" . $record['Last_Name'] . " </td>"; 
    echo "<td>" . "<input type=text name=breed value=" . $record['Breed'] . " </td>"; 
    echo "<td>" . "<input type=int name=weight value=" . $record['Weight'] . " </td>"; 
    echo "<td>" . "<input type=int name=age value=" . $record['Age'] . " </td>"; 
    echo "<td>" . "<input type=text name=sex value=" . $record['Sex'] . " </td>"; 
    echo "<td>" . "<input type=submit name=update value=update" . " </td>"; 
    echo "<td>" . "<input type=submit name=delete value=delete" . " </td>"; 
    echo "</tr>"; 
    echo "</form>"; 
} 

請幫我解決這個問題。

Answer 1

if (isset($_POST['update'])){ 
$UpdateQuery = "UPDATE register SET First_Name='".$_POST['first_name']."',Last_Name='".$_POST['last_name']."', Breed='".$_POST['breed']."', Weight='".$_POST['weight']."', Age='".$_POST['age']."', Sex='".$_POST['sex']."' WHERE reg_id ='".$_POST['reg_id']."'"; 
mysqli_query($con,$UpdateQuery); 
}; 

if (isset($_POST['delete'])){ 
$DeleteQuery = "DELETE FROM register WHERE reg_id='".$_POST['reg_id']."'"; 
mysqli_query($con,$DeleteQuery); 
}; 

應「封閉，即optional.add隱藏會更加美好

echo "<td><input type='hidden' name='reg_id' value='".$record['reg_id']."'></td>"; 
echo "<td><input type='submit' name='update' value='update'></td>"; 
echo "<td><input type='submit' name='delete' value='delete'></td>"; 

Answer 2

您正在使用$_POST沒有'。

試試這個：{$_POST['first_name']}並根據這個替換全部$_POST。

所以，你的更新查詢會是這樣：

"UPDATE register SET First_Name='{$_POST['first_name']}', Last_Name='{$_POST['last_name']}', Breed='{$_POST['breed']}', Weight='{$_POST['weight']}', Age='{$_POST['age']}', Sex='{$_POST['sex']}' WHERE reg_id='{$_POST['reg_id']}'"; 

沒有與名稱reg_id沒有外地，所以你$_POST['reg_id']不會work.Also請改變你的WHERE條件。你在哪裏條件匹配相同的值。

而且你刪除查詢將是：

"DELETE FROM register WHERE reg_id='{$_POST['reg_id']}'"; 

您所查詢的是開放的SQL注入攻擊。請參閱此：How can I prevent SQL injection in PHP?

Answer 3

顯示頁面

while($record = mysqli_fetch_array($myData)) { 
      echo "<table>"; 
      echo "<tr>"; 
      echo "<td>".$record['reg_id']."</td>"; 
      echo "<td>".$record['First_Name']."</td>"; 
      echo "<td>".$record['Last_Name']."</td>"; 
      echo "<td>".$record['Breed']."</td>"; 
      echo "<td>".$record['Weight']."</td>"; 
      echo "<td>".$record['Age']."</td>"; 
      echo "<td>".$record['Sex']."</td>"; 
      echo "<td><a href='edit.php?reg_id=".$record['reg_id']."'>EDIT</a></td>"; 
      echo "<td><a href='delete.php?reg_id=".$record['reg_id']."'>DELETE</a></td>"; 
      echo "</tr>"; 
      echo "</table>"; 
     } 

delete.php

<?php 
    if (isset($_POST['delete'])){ 
    $DeleteQuery = "DELETE FROM `register` WHERE `reg_id`={$_GET['reg_id']}'"; 
    mysqli_query($con,$DeleteQuery); 
    header("Location: your display page"); 
}; 
    ?> 

編輯表單

while($record = mysqli_fetch_array($myData)) { 
    echo '<form action="edit.php" method="Post"> 
      <input type="text" name="First_Name" value="'.$record['reg_id'].'"/> 
      <input type="text" name="First_Name" value="'.$record['First_Name'].'"/> 
      <input type="text" name="Last_Name" value="'.$record['Last_Name'].'"/> 
      <input type="text" name="Breed" value="'.$record['Breed'].'"/> 
      <input type="text" name="Weight" value="'.$record['Weight'].'"/> 
      <input type="text" name="Age" value="'.$record['Age'].'"/> 
      <input type="text" name="Sex" value="'.$record['Sex'].'"/> 
      <imput type="submit" value="save" name="submit" /> 
     </form>'; 
    } 

edit.php

 if (isset($_POST['update'])){ 
     $UpdateQuery = "UPDATE `register` SET `First_Name`='{$_POST['first_name']}', `Last_Name`='{$_POST['last_name']}', `Breed`='{$_POST['breed']}', Weight='{$_POST['weight']}', `Age`={$_POST['age']}, Sex='{$_POST['sex']}' WHERE `reg_id`={$_GET['reg_id']}"; 
     mysqli_query($con,$UpdateQuery); 
     header("Location: your display page"); 
};