Question

所以这是第一页上的代码：

<?php
    $db = mysql_connect(
  ':/Applications/MAMP/tmp/mysql/mysql.sock',
  'root',
  'root'
);
        if(!$db) die("Error connecting to MySQL database.");
        mysql_select_db('onlineform', $db);
$newQuery1 = mysql_query("SELECT newCampSessions FROM onlineformdata ORDER BY id DESC LIMIT 1") or die('Error ' . mysql_error());
$newFoo = mysql_fetch_array($newQuery1);
        $newQuery2 = mysql_query("SELECT pricePerWeek FROM onlineformdata ORDER BY id DESC LIMIT 1") or die('Error ' . mysql_error());
$newFoo1 = mysql_fetch_array($newQuery2);
$newOldString = $newFoo['newCampSessions'];
$newOldString2 = $newFoo1['pricePerWeek'];
$newChangedString = unserialize($newOldString);
$newChangedString2 = unserialize($newOldString2);
sql_close();
?>
<html>
<head> (all the tags in here) </head>
<body>
<form id="paymentform" action="amd8.php" method="post">
 <input type="hidden" name="pricePerWeek" value="<?php echo $newChangedString2 ?>"/>
 <input type="hidden" name="specificWeek" value="<?php echo $newChangedString ?>"/>

</form>
</body>
</html>

第一页还有更多内容，但我只是提供了相关的代码。在第一页上，一切都在PHP上工作。

当我尝试将其传递到第二页时，我正在尝试传递的特定数组的值为NULL。

amd8.php：

<?php

if ($_POST['formSubmit'] == "Submit")
{
    $newString = $_POST['pricePerWeek'];
    $newString2 = $_POST['specificWeek'];
}


$db = mysql_connect(
  ':/Applications/MAMP/tmp/mysql/mysql.sock',
  'root',
  'root'
);
        if(!$db) die("Error connecting to MySQL database.");
        mysql_select_db('onlineform', $db);

    $newQuery = mysql_query("SELECT newPrice,numberOfWeeks FROM onlineformdata ORDER BY id DESC LIMIT 1");
$newRow = mysql_fetch_row($newQuery);
$limit = $newRow[1];

    $totalPrice = 0;

    if (isset($_SESSION['campsessions']))
    {   

            for ($count == 0; $count < $limit; $count=$count+1)
        {

                foreach ($_SESSION['campsessions'] as $campsessions)
                {
            if ($campsessions == ($newString[$count]))
            {
                $totalPrice = $totalPrice + $newString2[$count];
            }
                } 
        }   
    }

?>

<html>
<head>(header info)</head>
<body>
<p>
<?php echo gettype($newString2);
      echo gettype($newString);
 ?>
</p>
</body>
</html>

正如您所看到的，我试图回应的gettypes() 是给我值NULL的

Answer 1

您需要将提交按钮命名为formSubmit，以使if语句返回true。像这样更改您的表单，它应该工作：

<form id="paymentform" action="amd8.php" method="post">
    <input type="hidden" name="pricePerWeek" value="<?php echo $newChangedString2 ?>"/>
    <input type="hidden" name="specificWeek" value="<?php echo $newChangedString ?>"/>
    <input type="submit" value="Submit" name="formSubmit">
</form>

Answer 2

只需更改第一个if：

if ($_POST['pricePerWeek'] || $_POST['specificWeek'])

另外，如果您愿意，可以在表单中添加隐藏的输入，并保留if as。

<input type="hidden" name="formSubmit" value="Submit" />

通过php在页面之间传递变量时遇到问题

2 个答案: