Question

我的php更新功能有问题，我有代码使功能工作如何才能更新最后一个功能。

<html>
<head>
<title>Update a Record in MySQL Database</title>
</head>
<body>

<?php
if(isset($_POST['update']))
{
$dbhost = '';
$dbuser = '';
$dbpass = '';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
  die('Could not connect: ' . mysql_error());
}

$stock_1 = $_POST['stock_1'];
$stock_2 = $_POST['stock_2'];
$stock_3 = $_POST['stock_3'];
$stock_4 = $_POST['stock_4'];
$stock_5 = $_POST['stock_5'];
$stock_6 = $_POST['stock_6'];

$sql = "UPDATE products ".
       "SET instock = $stock_1 ".
       "WHERE productid = 1" ;

$sql = "UPDATE products ".
       "SET instock = $stock_2 ".
       "WHERE productid = 2" ;
$sql = "UPDATE products ".
       "SET instock = $stock_3 ".
       "WHERE productid = 3" ;

$sql = "UPDATE products ".
       "SET instock = $stock_4 ".
       "WHERE productid = 4" ;

$sql = "UPDATE products ".
       "SET instock = $stock_5 ".
       "WHERE productid = 5" ;

$sql = "UPDATE products ".
       "SET instock = $stock_6 ".
       "WHERE productid = 6" ;


mysql_select_db('db_k0903037');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
  die('Could not update data: ' . mysql_error());
}
echo "Updated data successfully\n";
mysql_close($conn);
}
else
{
?>
<form method="post" action="<?php $_PHP_SELF ?>">
<table width="400" border="0" cellspacing="1" cellpadding="2">
<tr>
<td width="100">3.5" Seagate SATA 2TB</td>
<td><input name="stock_1" type="text" id="stock_1"></td>
</tr>
<tr>
<td width="100">Samsung 2.5" SATA Hard Drive</td>
<td><input name="stock_2" type="text" id="stock_2"></td>
</tr>
<tr>
<td width="100">8gb Kingston DDR3 RAM 1333mhz</td>
<td><input name="stock_3" type="text" id="stock_3"></td>
</tr>
<tr>
<td width="100">Apple MacBook Ram 8GB</td>
<td><input name="stock_4" type="text" id="stock_4"></td>
</tr>
<tr>
<td width="100">Gigabyte GA-970A-DS3</td>
<td><input name="stock_5" type="text" id="stock_5"></td>
</tr>
<tr>
<td width="100">Asus P8Z77-V PRO </td>
<td><input name="stock_6" type="text" id="stock_6"></td>
</tr>
<tr>
<td width="100"> </td>
<td> </td>
</tr>
<tr>
<td width="100"> </td>
<td>
<input name="update" type="submit" id="update" value="Update">
</td>
</tr>
</table>
</form>
<?php
}
?>
</body>
</html>

那是我的代码和我努力想知道它为什么不更新所有这些？显然我输入了正确的用户名和密码！

任何帮助将不胜感激。

Answer 1

您的mysql_select_db('db_k0903037');必须位于代码的顶部，然后您必须执行

$retval = mysql_query( $sql, $conn );
if(! $retval )
{
  die('Could not update data: ' . mysql_error());
}

每个$sql = "...";之后

。

您这样做的方法是，每次将$sql字符串设置为不同的字符串，但实际上并不执行该查询。 mysql_query执行实际执行：）

Answer 2

你为什么不在循环中这样做？

for($i=1;$i<=6;++$i){
  ${'stock_'.$i} = $_POST['stock_'.$i];
  $sql = "UPDATE products SET instock = ".${'stock_'.$i}." WHERE productid = ".$i ;
  $retval = mysql_query( $sql, $conn );
  // ...
}

或只是

for($i=1;$i<=6;++$i){
  $var = $_POST['stock_'.$i];
  $sql = "UPDATE products SET instock = ".$var." WHERE productid = ".$i ;
  $retval = mysql_query( $sql, $conn );
  // ...
}

多个UPDATE函数，php sql，只更新最后一个函数

2 个答案: