php 将某些字段更新到数据库

Question

我正在尝试开发一个库存系统，用户可以在该系统中查看他们的库存并使用仅由用户输入的值更新数量，其余部分从数据库中保持不变。但它不起作用，请帮我找出我做错的地方。它将回显成功消息，但数据库未更新。

<form name="form" method="post">
<table width="70%" border="5" align="center"><tr>
<th scope="row">SKU</th>
<th scope="row">Item Description</th>
<th scope="row">Current Qunatity</th>
<th scope="row">Update Quantity</th>
<th scope="row">Unit Price</th>
</tr>
<tr>
<th scope="row">
<?php
include('connect.php');
$result = mysqli_query("SELECT * FROM products") 
        or die(mysqli_error());

while($row = mysqli_fetch_array( $result )) {
    echo "<tr>";
    echo '<td><a name="sku[]">'.$row['sku_id'].'</a></td>';
    echo '<td>'.$row['description'].'</td>';
    echo '<td>'.$row['quantity'].'</td>';
    echo '<td><input name="qty[]" /></td>';
    echo '<td>'.$row['unit_price'].'</td>';
    echo "</tr>"; 
    }
?>
</table>
<input style="float:right" name="update" type="submit" id="update" value="Update"/>
</form>
<?php

if(isset($_POST['update']))
{
    $qty = $_POST['qty'];
    $sku = $_POST['sku'];
    foreach($qty as $key => $value) 
    {
        if(empty($value))
        {
            continue;
        }
        else
        {
            $sql = "UPDATE products SET quantity ='".$value."' WHERE sku_id = '".$sku[$key]."'";
            mysql_query($sql);
        }
    }   
    $retval = mysqli_query($sql);
    if(! $retval)
    {
        die('Could not update data: '. mysql_error());
    }
    echo 'Update data successfully!';
}
?>

Answer 1

你在这里使用mysql_query：

$sql = "UPDATE products SET quantity ='".$value."' WHERE sku_id = '".$sku[$key]."'";
mysql_query($sql);

而不是mysqli_query：

$sql = "UPDATE products SET quantity ='".$value."' WHERE sku_id = '".$sku[$key]."'";
mysqli_query($sql);

此外，您在这里也使用了mysql_error：

die('Could not update data: '. mysql_error());

附：不要忘记转义您在数据库查询中使用的任何用户输入！虽然理想情况下你应该使用类似 PDO 或 MySQLi 的预处理语句

Answer 2

这应该是你的完整答案（使用 mysqli 更新）：

<form name="form" method="post">
<table width="70%" border="5" align="center">
<tr>
<th scope="row">SKU</th>
<th scope="row">Item Description</th>
<th scope="row">Quantity</th>
<th scope="row">Unit Price</th>
</tr>
<?php
include('connect.php');
$result = mysqli_query("SELECT * FROM products") 
        or die(mysqli_error());

while($row = mysqli_fetch_array( $result )) {
    echo "<tr>";
    echo '<td>'.htmlspecialchars($row['sku_id']).'</td>';
    echo '<td>'.htmlspecialchars($row['description']).'</td>';
    echo '<td><input name="qty['.htmlspecialchars($row['sku_id']).']" value="'.htmlspecialchars($row['quantity']).'"/></td>';
    echo '<td><input name="price['.htmlspecialchars($row['sku_id']).']" value="'.htmlspecialchars($row['unit_price']).'"/></td>';
    echo "</tr>"; 
    }
?>
</table>
<input style="float:right" name="update" type="submit" id="update" value="Update"/>
</form>
<?php

if(isset($_POST['update']))
{
    $qty = $_POST['qty'];
    $price = $_POST['price'];
    $stmt =  $mysqli->stmt_init(); // <- mysqli class way of doing this
    $stmt->prepare("UPDATE products SET quantity = ?, unit_price = ? WHERE sku_id = ?");
    foreach($qty as $key => $value) 
    {
        $data = array($qty[$key], $price[$key], $key);
        $stmt->execute($sql, $data);
    }
    echo 'Update data successfully!';
}
?>

出于测试目的，post 的处理可以更改为：

if(isset($_POST['update']))
{
    $qty = $_POST['qty'];
    $price = $_POST['price'];
    //$stmt =  $mysqli->stmt_init(); // <- mysqli class way of doing this
    //$stmt->prepare("UPDATE products SET quantity = ?, unit_price = ? WHERE sku_id = ?");
    foreach($qty as $key => $value) 
    {
        echo "UPDATE products SET quantity = ".$qty[$key].", unit_price = ".$price[$key]." WHERE sku_id = " . $key . "<br/>\n";
        //$data = array($qty[$key], $price[$key], $key);
        //$stmt->execute($sql, $data);
    }
    echo 'Update data successfully!';
}

Answer 3

如果您执行 var_dump($_POST);你会看到你的输入没有值。

您需要在表单上指定值。

不过，我更愿意这样做：

echo '<input name="sku['.$row['sku_id'].']" value="'.$row['quantity'].'" />';

然后就可以循环$_POST['sku']，将Key作为sku_id，Value作为新的值（数量），进行更新