在PHP中顯示正確的輸出問題

Question

我試圖根據SELECT框和用戶輸入（兩者都是必需的）進行搜索選項。但是在下面的代碼中，正確和錯誤的輸入都顯示爲錯誤的輸入。有人可以解釋代碼中的錯誤嗎？

這裏是HTML

<form action="" method="POST"> 
<select name="selectOpt"> 
        <option>Select a list</option> 
        <option value="one">ID</option> 
        <option value="delName">Dealer Name</option> 
        <option value="medName">Medical Name</option> 

      </select> 
<input type="text" name="uinput" placeholder="Enter Search Key"/> 

      <button type="submit" name="submit">search</button> 
</form> 

PHP

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

    if(!empty($_POST['selectOpt']) && !empty($_POST['uinput'])){ 


     if($_POST['selectOpt']=='one'){ 
      $id = $_POST['selectOpt']; 

      if (!preg_match("/^[0-9]*$/",$id)){ 
       echo "not valid"; 

      }else{ 
       echo "valid"; 
      } 
     } 
    }else{ 
     echo "Enter Value"; 
    } 



} 

Answer 1

你的模式​​，被寫入到需要的零個或多個數字序列，不匹配任何可能的值，爲您的選擇框，它們都是字母串。

Answer 2

<form action="save.php" method="POST"> 
<select name="selectOpt"> 
        <option>Select a list</option> 
        <option value="one">ID</option> 
        <option value="delName">Dealer Name</option> 
        <option value="medName">Medical Name</option> 

      </select> 
<input type="text" name="uinput" placeholder="Enter Search Key"/> 

      <button type="submit" name="submit">search</button> 
</form> 

<?php 

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

    if(!empty($_POST['selectOpt']) && !empty($_POST['uinput'])){ 


     if($_POST['selectOpt'] == "one"){ 

      $id = $_POST['selectOpt']; 


      if (preg_match("/^[0-9]*$/",$id)){ 

       echo "not valid"; 

      }else{ 
       echo "valid"; 

        echo $input = $_POST['uinput']; 
      } 
     } 
     else{ 

      echo "PLEASE Enter Proper and Valid Search "; 

     } 

    } 
    else{ 
     echo "PLEASE Enter Value "; 
     } 



} 

?>