PHP表单提交if else语句

Question

我有一个按钮，单击它即可切换我的数据。当页面加载时，我希望它加载所有"new"项目。页面加载时会加载"new"项目，但是当我提交表单以“使用”或“过时”时，它仍然会附加"new"项目。我是 php 新手。谢谢!

 <form action='' method='POST'>
            <input type='submit' name='New' value="New"/>
            <input type='submit' name='Used' value="Used"/>
            <input type='submit' name='Outdated' value="Outdated" />
          </form>

这是 php。

    if(isset($_POST['New'])){
        $sql = "SELECT name FROM table_all_items WHERE name LIKE '%New%' ";
        include '/../includes/product-layout.php';
        }

    if(isset($_POST['Used'])){
        $sql = "SELECT name FROM table_all_items WHERE name LIKE '%Used%' ";
        include '/../includes/product-layout.php';
        }

    if(isset($_POST['Outdated'])){
        $sql = "SELECT name FROM table_all_items WHERE name LIKE '%Outdated%' ";
        include '/../includes/product-layout.php';
        }

    else {
        $sql = "SELECT name FROM table_all_items WHERE name LIKE '%New%' ";
        include '/../includes/product-layout.php';
        }

Answer 1

您以错误的方式使用 if-else 条件，您应该将所有这些条件更改为:

if(isset($_POST['New'])){
    $sql = "SELECT name FROM table_all_items WHERE name LIKE '%New%' ";
    include '/../includes/product-layout.php';
    }

else if(isset($_POST['Used'])){
    $sql = "SELECT name FROM table_all_items WHERE name LIKE '%Used%' ";
    include '/../includes/product-layout.php';
    }

else if(isset($_POST['Outdated'])){
    $sql = "SELECT name FROM table_all_items WHERE name LIKE '%Outdated%' ";
    include '/../includes/product-layout.php';
    }

else {
    $sql = "SELECT name FROM table_all_items WHERE name LIKE '%New%' ";
    include '/../includes/product-layout.php';
    }

现在试试这个，只有一个条件有效。