Question

嗨，我不确定我在做什么：D 我正在使用php和js从我的数据库中获取数据并创建一个用户视图表。我想我需要将所有的回声放在变量或类似的东西中。但我不确定。感谢每一位人士的支持。
我的js fucntion

＆＃13;

function dataT(){
    $.ajax({
      type:'Post',
      url:'filltable.php',
      data:{context :' showusers'},
      success:function (data){
        $('#content').html(data);
      }
    });

    alert(data);
  }

＆＃13;

＆＃13;
我的HTML代码

＆＃13;

  <body onload="dataT()">
  
    <div class="row">
    <div class="container">
      <div class="col-sm-3"></div>
      <div class="col-sm-6">
        <h3>tabla</h3>
        <br>
        <table id="dt" class="table table-hover ">
          <thead>
            <th>Id</th>
            <th>UN</th>
            <th>Mail</th>
            <th>Actions</th>
          </thead>
          <tbody id="content">

          </tbody>
        </table>
      </div>
      <div class="col-sm-3"></div>

    </div>
  </div>

＆＃13;

＆＃13;
这是我用PHP获取的地方

＆＃13;

<?php
$servername = "localhost";
$username = "root";
$password = "";

  try {
       $conn = new PDO("mysql:host=$servername;dbname=kyo", $username, $password);
       // set the PDO error mode to exception
       $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);


       if($_POST['context']=='showusers'){
         fillT($conn);
       }

       //end try
       }
    catch(PDOException $e)
       {
       echo "Connection failed: " . $e->getMessage();
       }



       function fillT(){
         $sql = "SELECT `username`, `mail`, id FROM `users` ";
         $res = mysqli_query($conn,$sql);

         if(!$res){
           die("Error!!! ... D:");
         }else{
           while ($data = mysqli_fetch_assoc($res) ) {
             ?>
             <tr>
               <td><?php echo $row['id']; ?></td>
               <td><?php echo $row['username']; ?></td>
               <td><?php echo $row['mail']; ?></td>
             </tr>
             <?php
           }
           echo json_encode($data);
         }

         mysqli_free_result($res);
         mysqli_close($conn);
       }
?>

＆＃13;

Answer 1

试试这个......

  <?php
    $servername = "localhost";
    $username = "root";
    $password = "";

  try {
       $conn = new PDO("mysql:host=$servername;dbname=kyo", $username, $password);
       // set the PDO error mode to exception
       $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

       //end try
       }
    catch(PDOException $e)
       {
       echo "Connection failed: " . $e->getMessage();
       }
?>

<body>

    <div class="row">
    <div class="container">
      <div class="col-sm-3"></div>
      <div class="col-sm-6">
        <h3>tabla</h3>
        <br>
        <table id="dt" class="table table-hover ">
          <thead>
            <th>Id</th>
            <th>UN</th>
            <th>Mail</th>
            <th>Actions</th>
          </thead>
          <tbody id="content">


  <?php
$stmt = $conn->prepare("SELECT `username`, `mail`, id FROM `users` "); 
    $stmt->execute();

// set the resulting array to associative
$result = $stmt->setFetchMode(PDO::FETCH_ASSOC); 
          foreach($stmt->fetchAll() as $row){
             ?>
             <tr>
               <td><?php echo $row['id']; ?></td>
               <td><?php echo $row['username']; ?></td>
               <td><?php echo $row['mail']; ?></td>
               <td><?php echo "actions"; ?></td>
             </tr>
             <?php
         }
?>
          </tbody>
        </table>
      </div>
      <div class="col-sm-3"></div>

    </div>
  </div>

使用Php和Ajax从数据库中获取数据

1 个答案: