PHP表格無法更新

我目前正在爲客戶端數據庫管理製作一個系統。 mySQL中有四個表格用於這個系統，管理員，員工，客戶和項目。項目表中有一個來自客戶端表的外鍵，它是clientid。PHP表格無法更新

現在，我已經爲所有這些表格製作了表格，以便用戶可以將數據輸入到它們中。奇怪的是，唯一可以成功更新的形式是工作人員。客戶和項目表格都不能更新。它會成功返回，但數據不會被更改。

以下是員工更新代碼。

<?php 
    include 'database.php'; 

    $staffid = $_GET['staffid']; 
    $sql = "SELECT * FROM staff WHERE staffid='$staffid'"; 
    $result = mysqli_query($conn,$sql); 

    while ($row=mysqli_fetch_array($result)){ 
     $staffname = $row['staffname']; 
     $staffemail = $row['staffemail']; 
     $staffphone = $row['staffphone']; 
    } 

    if(isset($_POST['submit'])){ 
    $staffname = $_POST['staffname']; 
    $staffemail = $_POST['staffemail']; 
    $staffphone = $_POST['staffphone']; 

    $sql = "UPDATE staff SET 

    staffname='$staffname',staffemail='$staffemail',staffphone='$staffphone' WHERE staffid='$staffid'"; 

    $result = mysqli_query($conn,$sql); 

    if($result){ 
     echo "<table><td><tr><h4>Record has been updated successfully!<br></tr></td></h4></table>"; 
    } 
    else { 
     echo "<h4>Record has <b>NOT</b> been updated successfully<br></h4>"; 
    } 
} 
?> 


<form action="" method="post"> 
<table class ="table1"> 
<tr> 
<td>Staff Name:</td> <td><input type="text" name="staffname" size="50" value="<?php echo $staffname;?>"></td> 
</tr> 

<tr> 
<td>Staff Email:</td> <td><input type="text" name="staffemail" size="50" value="<?php echo $staffemail;?>"></td> 
</tr> 

<tr> 
<td>Staff Phone No:</td> <td><input type="text" name="staffphone" size="50" value="<?php echo $staffphone;?>"></td> 
</tr> 

<td><input type="submit" value="Update" name="submit"> <input type="button" value="View" name="view" onclick='location.href="viewstaff.php"'></td> 
</table> 
</form>

好了，現在是客戶端表的更新代碼。

<?php 
include 'database.php'; 

$clientid = $_GET['clientid']; 
$sql = "SELECT * FROM client WHERE clientid='$clientid'"; 
$result = mysqli_query($conn,$sql) or die ("Error in query: $query. ".mysqli_error()); 

while ($row=mysqli_fetch_array($result)){ 
    $clientid = $row['clientid']; 
    $clientname = $row['clientname']; 
    $clientno = $row['clientno']; 
    $clientemail = $row['clientemail']; 
    $clientadd = $row['clientadd']; 
} 

if(isset($_POST['submit'])){ 
    $clientid = $row['clientid']; 
    $clientname = $row['clientname']; 
    $clientno = $row['clientno']; 
    $clientemail = $row['clientemail']; 
    $clientadd = $row['clientadd']; 

    $sql = "UPDATE client SET clientid='$clientid',clientname='$clientname',clientno='$clientno',clientemail='$clientemail',clientadd='$clientadd' WHERE clientid='$clientid'"; 

    $result = mysqli_query($conn,$sql) or die ("Error in query: $query. ".mysqli_error()); 

    if($result){ 
     echo "<table><td><tr><h4>Record has been updated successfully!<br></tr></td></h4></table>"; 
    } 
    else { 
     echo "<h4>Record has <b>NOT</b> been updated successfully<br></h4>"; 
    } 
} 
?> 


<form action="" method="post"> 
<table class ="table1"> 
<tr> 
<td>Client ID:</td> <td><input type="text" name="clientid" size="50" value="<?php echo $clientid;?>"></td> 
</tr> 

<tr> 
<td>Client Name:</td> <td><input type="text" name="clientname" size="50" value="<?php echo $clientname;?>"></td> 
</tr> 

<tr> 
<td>Client Phone No.:</td> <td><input type="text" name="clientno" size="50" value="<?php echo $clientno;?>"></td> 
</tr> 

<tr> 
<td>Client Email:</td> <td><input type="text" name="clientemail" size="50" value="<?php echo $clientemail;?>"></td> 
</tr> 

<tr> 
<td>Client Address:</td> <td><input type="text" name="clientadd" size="50" value="<?php echo $clientadd;?>"></td> 
</tr> 

<td><input type="submit" value="Update" name="submit"> <input type="button" value="View" name="view" onclick='location.href="viewclient.php"'></td> 
</table> 
</form>

也許我是傻還是什麼，但我一直在試圖找出問題3小時，我這個接近哭了笑。一直在閱讀關於更新表單的所有主題，但仍然沒有答案。希望這裏的任何人都能幫助我。謝謝。

來源

2017-06-05 Taubah Jemali

**危險**：您很容易[SQL注入攻擊]（http://bobby-tables.com/）**，您需要[防禦]（http://stackoverflow.com/問題/ 60174/best-way-to-prevent-sql -injection-in-php）自己從。 – Quentin

借調上述，如果可能的話，我會建議使用[PDO Prepared Statements]（http://php.net/manual/en/pdo.prepared-statements.php）。 –

您還應該從更新查詢集字段中刪除'clientid'; – itzmukeshy7

您使用的客戶端表更新的代碼使用此代碼：

if(isset($_POST['submit'])){ 
    $clientid = $row['clientid'];  // $row should be $_POST 
    $clientname = $row['clientname']; // $row should be $_POST 
    $clientno = $row['clientno'];  // $row should be $_POST 
    $clientemail = $row['clientemail']; // $row should be $_POST 
    $clientadd = $row['clientadd'];  // $row should be $_POST

但這些$row S的關係是$_POST，否則更新後的數據將是相同的之前的數據（因爲$row是結果來自查詢SELECT * FROM client WHERE clientid='$clientid'）。您可以在工作人員表更新代碼做是正確的：

if(isset($_POST['submit'])){ 
    $staffname = $_POST['staffname']; 
    $staffemail = $_POST['staffemail']; 
    $staffphone = $_POST['staffphone'];

請注意，你的腳本是在SQL Injection Attack風險。看看Little Bobby Tables發生了什麼事。即使是if you are escaping inputs, its not safe!。改爲使用prepared parameterized statements。

來源

2017-06-05 09:11:47

PHP表格無法更新

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