Question

我有一个按钮，可以在单击时切换我的数据。当页面加载时，我希望它加载所有＆＃34; New＆＃34;项目。＆＃34;新＆＃34;在页面加载时加载，但是当我提交表单以获得＆＃34;使用＆＃34;或者＆＃34;过时＆＃34;，它仍然适用于＆＃34;新的＆＃34;项目。我是php的新手。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input type='submit' name='Outdated' value="Outdated" /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    if(isset($_POST['Outdated'])){
        $sql = "SELECT name FROM table_all_items WHERE name LIKE '%Outdated%' ";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

您正在以错误的方式使用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else if(isset($_POST['Outdated'])){
    $sql = "SELECT name FROM table_all_items WHERE name LIKE '%Outdated%' ";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';
    }

现在试试这个条件只有一个。

Answer 2

您应该使用相同的名称并测试不同的值

在你的html页面

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

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

if( $_POST['my_submit'] == 'Outdated'){
    $sql = "SELECT name FROM table_all_items WHERE name LIKE '%Outdated%' ";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';
    }

和服务器端

 <header>
  <nav>
    <ul class="site-nav">
      <li><a href="#">home</a></li>
      <li><a href="#">about</a></li>
      <li><a href="#">gallery</a></li>
    </ul>
  </nav>
</header>

Answer 3

这是一种可能的方法，可以避免多重条件的问题在发生时成为现实

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input type='button' name='Outdated' value="Outdated" /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

<script type='text/javascript'>
    var col=document.querySelectorAll('form > input[type="button"]');
    if( col )for( var n in col )if( col[n].nodeType==1 )col[n].addEventListener('click',function(e){
        document.getElementById( 'action' ).value=this.value;
        this.parentNode.submit();
    }.bind( col[n] ),false);
</script>

if( isset( $_POST['action'] ) ){
    $term=filter_input( FILTER_POST, 'action', FILTER_SANITIZE_STRING );
    $sql="select `name` from `table_all` where `name` like '%{$term}%';";
    if( in_array($term,array('New','Used','Outdated') ) ){
        include '/../includes/product-layout.php';
    }
}

Php表单提交if else声明

3 个答案: