Question

我试图允许用户根据名称，价格等对mysql查询进行排序。这种下拉使他们有能力做到这一点，它会根据用户选择的内容更改“ORDER BY”子句。现在是凌晨3点，我无法发现错误：

    <?php
        $sortBy = $_POST['sortBy'];
        if ($sortBy) {

            $priceLowToHigh = $_POST['price-low-to-high'];
            $priceHighToLow = $_POST['price-high-to-low'];
            $dateMostRecent = $_POST['date-most-recent'];
            $dateOldest = $_POST['date-oldest'];
            $alphabeticalOrder = $_POST['alphabetical-order'];

            if ($priceLowToHigh) {
                $sortOrder = "price ASC";
            } elseif ($priceHighToLow) {
                $sortOrder = "price DESC";
            } elseif ($dateMostRecent) {
                $sortOrder = "date DESC";
            } elseif ($dateOldest) {
                $sortOrder = "date ASC";
            } elseif ($alphabeticalOrder) {
                $sortOrder = "name ASC";
            }

        } else {
            $sortOrder = "date DESC";
        }
    ?>
    <form action="" method="post">
        <select name="sortBy" onchange="this.form.submit()">
            <option>Sort By</option>
            <option value="price-low-to-high">Price (low to high)</option>
            <option value="price-high-to-low">Price (high to low)</option>
            <option value="date-most-recent">Date (most recent)</option>
            <option value="date-oldest">Date (oldest)</option>
            <option value="alphabetical-order">Alphabetical Order</option>
        </select>
    </form>

稍后在ORDER BY子句中使用如下：

    $query = mysql_query("SELECT * FROM products WHERE category = $categoryId ORDER BY $sortOrder");

Answer 1

您正在错误地处理菜单。您的$ sortBy变量已经是用户选择的变量，而不是单独的$ _POST字段。

这是你应该做的 -

<?php
    $sortBy = $_POST['sortBy'];
    if($sortBy == 'price-low-to-high') {
        $sortOrder = "price ASC";
    } else if($sortBy == 'price-high-to-low') {
        $sortOrder = "price DESC";
    } else if($sortBy == 'date-most-recent') {
        $sortOrder = "date DESC";
    } else if($sortBy == 'date-oldest') {
        $sortOrder = "date ASC";
    } else if($sortBy == 'alphabetical-order') {
        $sortOrder = "name ASC";
    } else {
        $sortOrder = "date DESC";
    }
?>

Answer 2

我无法评论你的回答拉斐尔，因为我不够高，但我只是想提一下可能使用switch语句来提高可读性。但是你就在那里。给予好评

Answer 3

看起来你的代码是双引号，当它应该是单一的，让php替换你的代码。您应该将该sql打印到屏幕上，您将看到它可能没有替换该值。

我指的是你调用mysql_query的行。

Answer 4

你的基本代码有点不对劲。它应该如下： -

<?php
$sortBy = $_POST['sortBy'];

if($sortBy == 'price-low-to-high')
  $sortOrder = "price ASC";
else if($sortBy == 'price-high-to-low')
  $sortOrder = "price DESC";
else if($sortBy == 'date-most-recent')
  $sortOrder = "date DESC";
else if($sortBy == 'date-oldest')
  $sortOrder = "date ASC";
else if($sortBy == 'alphabetical-order')
  $sortOrder = "name ASC";
else
  $sortOrder = "date DESC";
?>

<form action="" method="post">
        <select name="sortBy" onchange="this.form.submit()">
            <option>Sort By</option>
            <option value="price-low-to-high">Price (low to high)</option>
            <option value="price-high-to-low">Price (high to low)</option>
            <option value="date-most-recent">Date (most recent)</option>
            <option value="date-oldest">Date (oldest)</option>
            <option value="alphabetical-order">Alphabetical Order</option>
        </select>
</form>

如果您将代码更改为上述代码，我认为现在可以使用。

希望它有所帮助。

我的代码有什么问题？

4 个答案: