Question

你好，我有一个问题，我看不到文章a1我认为我有一个错误，我认为问题是在if isset帖子...

`<input type="button" name="g1" value="click">
       <?php
       try{
       $db = new PDO('mysql:host=localhost;dbname=search','root','');
       }catch(PDOexeption $e){
       echo "erreur 404";
       }

       if(isset($_POST['g1'])){

           $insert = $db->exec("UPDATE g2 SET a1 = a1 + 1 WHERE id = '15'");

           $query = $db->query("SELECT * FROM g2");

              while($article = $query->fetch())
           {
            echo $article['a1'];
           }
    };

 ?>'

Answer 1

如果我理解正确您正在尝试从表单发送输入值，请使用此方法。 HTML PART

 <form action="#" method="post">
 <input type="submit" name="g1" value="Click" />
 </form>

PHP SIDE：

   try{
   $db = new PDO('mysql:host=localhost;dbname=search','root','');
   }catch(PDOexeption $e){
   echo "erreur 404";
   }

   if(isset($_POST['g1'])){

       $insert = $db->prepare("UPDATE g2 SET a1 = a1 + 1 WHERE id = '15'");
       $exe    = $insert->exe();

       $query = $db->prepare("SELECT * FROM g2");
       $sel   = $query->exe();
       foreach ($query as $select) {
             echo $select['a1'];
       }   


};

我希望它能帮到你

我对PDO有误

1 个答案: