爲每一行分配一個按鈕的值

Question

我正在嘗試使用php，而且我對它真的很陌生。

我創建了一個頁面（使用php代碼）將信息發送到數據庫和一個顯示數據庫信息的頁面。

我的問題是：可以說我要輸入10個信息，但我只顯示5個信息，如何顯示其餘信息點擊每個按鈕的行？

我必須爲按鈕賦值嗎？我不明白我如何可以將按鈕關聯到該行。

基本上對於每一行我想顯示該特定行的信息的其餘部分點擊該特定按鈕。

任何幫助將appricciate。感謝

-----腳本，在數據庫中插入信息-----

<?php 

$name = mysqli_real_escape_string($link, $_POST['name']); 
$mail = mysqli_real_escape_string($link, $_POST['mail']); 
$number = mysqli_real_escape_string($link, $_POST['number']); 
$device = mysqli_real_escape_string($link, $_POST['device']); 
$model = mysqli_real_escape_string($link, $_POST['model']); 
$problem = mysqli_real_escape_string($link, $_POST['problem']); 

     $sql = "INSERT INTO cj (name, mail, number, device, model, problem) VALUES ('$name', '$mail', '$number', '$device', '$model', '$problem')"; 
     $result = mysqli_query($link, $sql); 
     // if query fails stop script and echo error 
    if($result === false) 
    {echo "ERROR: Could not able to execute $sql. " . mysqli_error($link); 
     exit; 
    }   
    header("location:add-customer.php?message=A job and customer has been added to the database"); 

    exit; 
     echo "You'll never see this"; 
    ?> 

-----------頁面顯示信息---- -

 <?php 
$sql = "SELECT * from "database name""; 
    echo " 
<table class='table'> 
    <thead> 
     <tr>"; 
/* Get field information for all columns */ 

    echo "<form action='' method=post>"; 
    echo "<tr class='info'> 

       <input type=hidden name=hidden value=" . $row['id'] . "> 
       <td>" . $row['id'] . "</td> 
       <td>" . $row['device'] . "</td> 
       <td>" . $row['model'] . "</td> 
       <td>" . $row['problem'] . "</td> 
       <td> 

      </td>   

       <td> 
       <a class='btn btn-primary btn-sm' data-toggle='modal' data-target='#myModal' value= " . $row['id'] . " > Info</a></form></td> 

      </tr>"; 
    echo "</form>"; 


} 
echo " 
    </tbody> 

</table>"; 

?> 

<div class="container"> 
    <!-- Trigger the modal with a button --> 
    <!-- Modal --> 
    <div class="modal fade" id="myModal" role="dialog"> 
    <div class="modal-dialog modal-lg"> 
    <!-- Modal content--> 
     <div class="modal-content"> 
     <div class="modal-header"> 
      <button type="button" class="close" data-dismiss="modal">&times;</button> 
      <h4 class="modal-title">Customer Information</h4> 
     </div> 

     <div class="modal-body"> 

     <?php 
include("../includes/connection.php"); 
if ($link->connect_errno > 0) { 
    die('Unable to connect to database [' . $link->connect_error . ']'); 
} 
$sql = "SELECT name,mail,number from databasename WHERE **???**; 


if (!$result = $link->query($sql)) { 

    die('There was an error running the query [' . $link->error . ']'); 
} 
echo " 
<table class='table'> 
    <thead> 
     <tr>"; 
/* Get field information for all columns */ 
while ($finfo = $result->fetch_field()) { 
    echo " 
     <th>" . $finfo->name . "</th>"; 
} 
echo " 
     </tr> 
    </thead> 
    <tbody>"; 
while ($row = $result->fetch_assoc()) { 
    echo "<tr class='info'> 

       <td>" . $row['name'] . "</td> 
       <td>" . $row['mail'] . "</td> 
       <td>" . $row['number'] . "</td> 

    </tr>"; 

... 

Answer 1

我找到了解決方法用ajax：

這裏是代碼：（Ajax調用第三PHP網頁，其中固定我的問題）

<script> 
    $(document).ready(function(){ 
     $('.get_info').click(function(){ 
     var job_id = $(this).parent().siblings('input[name=hidden]').val(); 
     $.ajax({ 
      url: 'THIRDPAGE.php', 
      data: 'job_id=' + job_id, 
      type: 'POST', 
      success: function(data){ 
      // console.log(data); 
      $('.info_data').html(data); 
      } 
     }); 
     }); 
    }); 
    </script> 

對於是oone到這裏只是評論，我可以更好地解釋我做了什麼。

Answer 2

連接到數據庫並使用PHP獲取所有10個信息。但只顯示5個信息。在那裏放一個按鈕，然後使用Javascript或Jquery顯示其他行。 （onclick事件）