站點搜索顯示整個表格而不是搜索結果？

我有一個數據庫，員工信息顯示到表中，我設法使搜索框，我已連接到我的數據庫，現在一切都是問題，當我搜索名稱，即「丹尼爾」，它只是給我退回了我已經擁有的桌子，所以那裏有兩張桌子，而不是「丹尼爾斯」信息。站點搜索顯示整個表格而不是搜索結果？

在此先感謝非常感謝任何幫助！

我的繼承人代碼 -

<!doctype html> 
    <html> 
     <head> 
      <title></title> 
      <link rel="stylesheet" href="../Css/index_stylesheet.css"> 
     </head> 
    <body> 
    <p> 

    <?php 
     $form = "<html> 
     <h1>Search!</h1> 
     <form method=\"get\"> 
      <label>Name: 
      <input type=\"text\" name=\"keyname\" /> 
      </label> 
      <input type=\"submit\" value=\"Search\" /> 
     </form> 
     </body> 
     </html>"; 
      //capture search term and remove spaces at its both ends if there is any 

      if(!empty($_GET['keyname'])){ 
      $keyname = $_GET['keyname']; 

      $searchTerm = trim($keyname); 

      //database connection info 
      $host = "localhost"; //server 
      $db = "development_suite"; //database name 
      $user = "root"; //dabases user name 
      $pwd = ""; //password 

      //connecting to server and creating link to database 
      $link = mysqli_connect($host, $user, $pwd, $db); 

      //MYSQL search statement 
      $query = "SELECT firstname ,surname,address,mobile,email 
         FROM development_suite.eeg_staff;"; 

      $results = mysqli_query($link,$query); 

      /* check whethere there were matching records in the table 
      by counting the number of results returned */ 
      if(mysqli_num_rows($results) >= 1){ 
       $output = "$row_1"; 
       while($row = mysqli_fetch_array($results)) 
       { 
        $output .= "First Name: " . $row['firstname'] . "<br />"; 
        $output .= "Surname: " . $row['surname'] . "<br />"; 
        $output .= "Address: " . $row['address'] . "<br />"; 
        $output .= "Mobile: " . $row['mobile'] . "<br />"; 
        $output .= "Email: " . $row['email'] . "<br /><br />"; 



       } 
      }else{ 
       $output = "There was no matching record for the name " . 
       strip_tags($searchTerm);    
      } 

     } else { 
      $output = "Enter name you are searching for."; 
     } 


      echo "$form\n$output"; 
     ?> 
    </body> 
</html>

來源

2014-06-27 Calvin Taylor

您必須在您的查詢中添加where子句。喜歡的東西...

$query = "SELECT firstname ,surname,address,mobile,email 
        FROM development_suite.eeg_staff 
        WHERE firstname LIKE '" . $_REQUEST['keyname'] . "';";

最好的問候，內博伊沙

來源

2014-06-27 13:13:43

感謝喜的幫助!!：d –

我應該把在這行有什麼關係嗎？出現錯誤，「警告：mysqli_num_rows（）期望參數1爲mysqli_result，布爾在C：\ wamp \ www \ manholeform \ Application \ includes 49包含\ search_function.php中給出」 –

第49行是if語句，以if （mysqli_num_rows（$ results） –

站點搜索顯示整個表格而不是搜索結果？

回答

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