$ _GET变量 - 为什么之前的数据没有隐藏？

时间：2015-01-21 23:17:03

标签： php variables get hide

我正在创建一个包含类别，子类别和列表的网站。在点击类别时，即计算页面改变以显示来自数据库中处于计算类别中的所有子类别。然后，当单击子类别时，将显示该子类别的列表，并将子类别选择发布到URL。然后，当点击列表时，会发生同样的事情。这一切都还可以，但是当我尝试显示列表细节时，它应该隐藏列表，但这不会发生。有谁知道为什么？我正在使用的代码如下：

<?php
include 'core/init.php';
include 'includes/overall/header.php';
?>

<h1> Computing </h1>

<?php
$subcategory = 'Computing';

if (isset($_GET['subcategory']) && trim($_GET['subcategory']) != '') {
    $subcategory = trim($_GET['subcategory']) != '';
}

$sql = "SELECT * FROM categories WHERE category = '$subcategory' ORDER BY subcategory ASC";
$result = mysql_query($sql) or die(mysql_error() . "<br>" . $sql);
$catNo = 1;
echo "<table class='categorytable'> <tr class='categoryrow'>";
while ($row = mysql_fetch_array($result)) {
    echo '<td class="categorydata"><a href="computing.php?subcategory=' . strtolower($row['subcategory']) . '"><img class="catimg" src="' . $row['subcategory_img_path'] . '" border="0" /></br>' . $row['subcategory'] . '</a></td>';
    if ($catNo % 3 == 0) {

        echo "</tr><tr>";
    }
    $catNo++;
}
echo "</tr> </table>";
?>
<?php
if (isset($_GET['subcategory'])) {
    $subcat = $_GET['subcategory'];

    $sql2 = "SELECT * FROM listings WHERE subcategory = '$subcat' ORDER BY title ASC";
    $result2 = mysql_query($sql2) or die(mysql_error() . "<br>" . $sql2);
    $catNo = 1;
    echo "<table class='categorytable'> <tr class='categoryrow'>";
    while ($row2 = mysql_fetch_array($result2)) {
        echo '<td class="categorydata"><a href="computing.php?subcategory=' . $subcat . '&' . 'listingid=' . ($row2['listing_id']) . '"><img class="catimg" src="' . $row2['main_image'] . '" border="0" /></br>' . $row2['title'] . '&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;' . '<i class="fa fa-cog"></i>' . ' ' . $row2['cogs'] . ' per week' . '</a></td>';
        if ($catNo % 3 == 0) {

            echo "</tr><tr>";
        }
        $catNo++;
    }
    var_dump($_GET);


    echo "</tr> </table>";
?>
<?php
    if (isset($_GET['listingid'])) {
        $listingid = $_GET['listingid'];

        $sql3 = "SELECT * FROM listings WHERE listing_id='$listingid'";
        $result3 = mysql_query($sql3) or die(mysql_error() . "<br>" . $sql3);
        while ($row3 = mysql_fetch_assoc($result3)) {
            //whatever your values are
            echo $row3['description'] . "<br />";
        }
    }
}
?>

</br>

<?php
include 'includes/overall/footer.php';
?>

我知道不推荐使用php_函数，并且还没有考虑安全性，但是当我完成所有工作时会担心这个问题。

6 个答案:

答案 0 :(得分：2)

我认为您的代码很好，只需修复此行：

if (isset($_GET['subcategory']) && trim($_GET['subcategory']) != '') {
    $subcategory = trim($_GET['subcategory']) != '';
}

为什么要为$subcategory var指定一个布尔值，然后在其上触发查询。声明：

$subcategory = trim($_GET['subcategory']) != '';

如果TRUE为非空或非空，则会给1（$_GET['subcategory']）。 您应该将行更改为：

if (isset($_GET['subcategory']) && trim($_GET['subcategory']) != '') {
    $subcategory = trim($_GET['subcategory']);
}

希望这会对你有所帮助。

答案 1 :(得分：1)

将while部分放入if。

if(!isset($_GET['subcategory'])){
$sql = "SELECT * FROM categories WHERE category = '$subcategory' ORDER BY subcategory ASC";
$result = mysql_query($sql) or die(mysql_error() . "<br>" . $sql);
$catNo = 1;
echo "<table class='categorytable'> <tr class='categoryrow'>";
while ($row = mysql_fetch_array($result)) {
    echo '<td class="categorydata"><a href="computing.php?subcategory=' . strtolower($row['subcategory']) . '"><img class="catimg" src="' . $row['subcategory_img_path'] . '" border="0" /></br>' . $row['subcategory'] . '</a></td>';
    if ($catNo % 3 == 0) {

        echo "</tr><tr>";
    }
    $catNo++;
}
echo "</tr> </table>";

}

答案 2 :(得分：1)

我认为我有更好的解决方案。检查此代码：

<?php
if (isset($_GET['subcategory'])) {
    $subcat = $_GET['subcategory'];

    $sql2 = "SELECT * FROM listings WHERE subcategory = '$subcat' ORDER BY title ASC";
    $result2 = mysql_query($sql2) or die(mysql_error() . "<br>" . $sql2);
    $catNo = 1;
    echo "<table class='categorytable'> <tr class='categoryrow'>";
    while ($row2 = mysql_fetch_array($result2)) {
        echo '<td class="categorydata"><a href="computing.php?listingid=' . ($row2['listing_id']) . '"><img class="catimg" src="' . $row2['main_image'] . '" border="0" /></br>' . $row2['title'] . '&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;' . '<i class="fa fa-cog"></i>' . ' ' . $row2['cogs'] . ' per week' . '</a></td>';
        if ($catNo % 3 == 0) {

            echo "</tr><tr>";
        }
        $catNo++;
    }
    var_dump($_GET);


    echo "</tr> </table>";
}
?>
<?php
if (isset($_GET['listingid'])) {
    $listingid = $_GET['listingid'];

    $sql3 = "SELECT * FROM listings WHERE listing_id='$listingid'";
    $result3 = mysql_query($sql3) or die(mysql_error() . "<br>" . $sql3);
    while ($row3 = mysql_fetch_assoc($result3)) {
        //whatever your values are
        echo $row3['description'] . "<br />";
    }
}
?>

在这里，我更改了代码以匹配一种视图，在我看来，这是最接近修复代码的方式。由于未根据子类别名称选择列表数据库表，我们可以做的是将列表块与子类别块分隔为if(isset($_GET['subcategory'])) if块在if (isset($_GET['listingid']))之前结束并删除子类别当您调用列表时，从URL中获取GET变量，这样您就不会在阻止时运行子类别。 当您在URL中使用子类别变量提供URL并且在尝试显示您正在显示的列表并隐藏页面的这些阶段时，单独未将子类别提供给URL时，会发生这样的情况。 您不需要子类别值来显示列表，只显示列表清单，然后您应该只关心listingid GET变量。

答案 3 :(得分：1)

使用以下代码：

<?php
$category = 'Computing';

if (isset($_GET['category']) && trim($_GET['category']) != '') {
    $category = trim($_GET['category']);
}

$sql = "SELECT * FROM categories WHERE category = '$category' ORDER BY category ASC";
$result = mysql_query($sql) or die(mysql_error() . "<br>" . $sql);
$catNo = 1;
echo "<table class='categorytable'> <tr class='categoryrow'>";
while ($row = mysql_fetch_array($result)) {
    echo '<td class="categorydata"><a href="computing.php?category='. $category.'&subcategory=' . strtolower($row['subcategory']) . '"><img class="catimg" src="' . $row['subcategory_img_path'] . '" border="0" /></br>' . $row['subcategory'] . '</a></td>';
    if ($catNo % 3 == 0) {

        echo "</tr><tr>";
    }
    $catNo++;
}
echo "</tr> </table>";
?>
<?php
if (isset($_GET['subcategory'])) {
    $subcat = $_GET['subcategory'];

    $sql2 = "SELECT * FROM subcategory WHERE subcategory = '$subcat' ORDER BY title ASC";
    $result2 = mysql_query($sql2) or die(mysql_error() . "<br>" . $sql2);
    $catNo = 1;
    echo "<table class='categorytable'> <tr class='categoryrow'>";
    while ($row2 = mysql_fetch_array($result2)) {
        echo '<td class="categorydata"><a href="computing.php?category='. $category.'&subcategory=' . $subcat . '&' . 'listingid=' . ($row2['listing_id']) . '"><img class="catimg" src="' . $row2['main_image'] . '" border="0" /></br>' . $row2['title'] . '&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;' . '<i class="fa fa-cog"></i>' . ' ' . $row2['cogs'] . ' per week' . '</a></td>';
        if ($catNo % 3 == 0) {

            echo "</tr><tr>";
        }
        $catNo++;
    }
    var_dump($_GET);


    echo "</tr> </table>";
?>
<?php
    if (isset($_GET['listingid'])) {
        $listingid = $_GET['listingid'];

        $sql3 = "SELECT * FROM listings WHERE listing_id='$listingid'";
        $result3 = mysql_query($sql3) or die(mysql_error() . "<br>" . $sql3);
        while ($row3 = mysql_fetch_assoc($result3)) {
            //whatever your values are
            echo $row3['description'] . "<br />";
        }
    }
}
?>

</br>

<?php
include 'includes/overall/footer.php';
?>

答案 4 :(得分：0)

我同意zairwolf，基本上Steven你需要一种标记何时应该显示列表的方法，对页面加载结构的一些规划可能会有所帮助。以下是您可能实施的流程，如果我了解您的需求，可以改进您的实施。

//Proposed layout

function show_categories () {
   //draw the categories down the left side of screen with "/?category={$record->CATEGORYID}" in anchor href
}

function show_subcategories ($category) {
  //select the subcategories based on the passed category and output also with href as you have already
}

function show_listings ($subcategory) {
  //select all the listings based on the sub category sent here
}

//always show the categories ?
show_categories ();

//other logic when to display sub categories or listings
if ($_REQUEST["subcategory"]) { 
 show_listings ($_REQUEST["subcategory"]);
} else {
 show_subcategories ($_REQUEST["category"])
}

答案 5 :(得分：0)

尝试使用$ _POST方法，它会起作用

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