phpフォームelse ifステートメント

Question

クリックしたときにデータを切り替えるボタンがあります。ページが読み込まれるとき、私はすべての "新しい"項目をロードしたい。ページが読み込まれたときに「新規」が読み込まれますが、フォームを送信して「使用済み」または「古い」を取得した場合でも、「新規」アイテムは引き続き表示されます。私はPHPに新しいです。ありがとう！

<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input type='submit' name='Outdated' value="Outdated" /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> 

ここはphpです。これは今だけ1つの条件が動作します試してみてください

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else if(isset($_POST['Outdated'])){ 
    $sql = "SELECT name FROM table_all_items WHERE name LIKE '%Outdated%' "; 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

：

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    if(isset($_POST['Outdated'])){ 
     $sql = "SELECT name FROM table_all_items WHERE name LIKE '%Outdated%' "; 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

あなたがように、すべてのこれらの条件を変更してください間違った方法であれば、他の条件を使用しています。

Answer 2

あなたは

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input type='submit' name='my_submit' value="Outdated" /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

とサーバ側ここ

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if($_POST['my_submit'] == 'Outdated'){ 
    $sql = "SELECT name FROM table_all_items WHERE name LIKE '%Outdated%' "; 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

Answer 3

はの問題を回避するために、一つの可能​​なアプローチで同じ名前を使用して、あなたのHTMLページに異なる値

をテストする必要があります複数の条件が成立していると真である

<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'; 
    } 
}