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Дмитрий Сухоцкий
Asked:2020-07-09 18:05:17 +0000 UTC2020-07-09 18:05:17 +0000 UTC

数据到ajax表

  • 772

我正在实现一个用于添加/删除记录的表。数据是通过表单写入到表中，但是在窗口重载后才显示（我想实现不重载）。我阅读了有关 ajax 的信息，我尝试使用它，但添加到表中不起作用。

在表中，我从数据库中推断出价值。通过单击添加字段，将打开一个模式窗口并捕获数据

<table class="table" id="tableID">

  <thead class="thead-light">
  <tr>
      <th scope="col">id <span class="active sortid">↓</span><span>↑</span></th>
      <th scope="col">Name <span class="active sortname">↓</span><span>↑</span></th>
      <th scope="col">Price <span class="active sortprice">↓</span><span>↑</span></th>
      <th scope="col">Rating <span class="active sortrating">↓</span><span>↑</span></th>
    </tr>
    </thead>
  <tbody>
      <?php
      $result = mysqli_query($link,"SELECT id,name,price,rating FROM product");
      while($row = mysqli_fetch_array($result)) {
        echo '<tr>
        <th scope="row">'.$row['id'].'</th>
        <td>'.$row['name'].'</td>
        <td>'.$row['price'].'</td>
        <td>'.$row['rating'].'</td>
        <td><button type="button" id="'.$row['id'].'" class="btn btn-primary edit">edit field</button></td>
        <td><button type="button" data_id="'.$row['id'].'" class="btn btn-danger delete">delete field</button></td>
    </tr>';
      }
        ?>

  </tbody>

</table>

<button type="button" class="btn btn-primary bye">add field</button>

<div class="modal" id="window">
  <div class="modal-dialog" role="document">
    <div class="modal-content">
      <div class="modal-header">
        <h5 class="modal-title">Modal title</h5>
        <button type="button" class="close" data-dismiss="modal" aria-label="Close">
          <span aria-hidden="true">&times;</span>
        </button>
      </div>
      <div class="modal-body">
      <form class="b-popup" id="popup1"  method="post" id="ajax_form" action="">
        <div class="form-group">
          <label for="name">name</label>
          <input type="text" class="form-control" name="name" id="name"  placeholder="Enter name">
          <label for="price">price</label>
          <input type="text" class="form-control" name="price" id="price"  placeholder="Enter price">
          <label for="rating">rating</label>
          <input type="text" class="form-control" name="rating" id="rating"  placeholder="Enter rating">
        </div>
        <button type="submit" class="btn btn-primary btn2" data-dismiss="modal" >Submit</button>
      </form>
      </div>
    </div>
  </div>
</div>

在这里它实现了 ajax 并将其传递给addfield.php以写入数据库

 $(document).ready(function() {
    $('#ajax_form').submit(function(event){
        event.preventDefault(); //останавливаем стандартную отправку
        var name = $("#name").val();
        var price = $("#price").val();
        var rating = $("#rating").val();
        $.ajax({
            type: "POST",
            url: "add_field.php",
            data: {"name": name, "price": price,"rating": rating},
            cache: false,
            success: 
               $('#tableID').append('<tr><td>price</td></tr><tr><td>".$name."</td></tr>');

            }

        });
    });

 });

addfield.php

<?php 
include ("db.php");
    $name = $_POST['name'];
    $price = $_POST['price'];
    $rating = $_POST['rating'];
    $link = db_connect();
    $result = mysqli_query($link,"INSERT INTO product (id,
                                               name,
                                               price,
                                               rating) 
                                    VALUE      (null,
                                               '$name',
                                               '$price',
                                               '$rating')");


?>
php

1 个回答

  1. Best Answer
    $(document).ready(function() {
    $('#ajax_form').submit(function(event){ // или .on('submit', если jquery > 1.7 
        event.preventDefault(); //останавливаем стандартную отправку
        var name = $("#name").val();
        var price = $("#price").val();
        var rating = $("#rating").val();
        $.ajax({
            type: "POST",
            url: "add_field.php",
            data: {"name": name, "price": price,"rating": rating},
            cache: false,
            success: function(data){
                $('#tableID').append('<tr><td>'+data.price+'</td></tr><tr><td>'+data.name+'</td></tr>');
            }
        });
    });
 });

    从服务器返回 json

    <?php
// .... конектимся к базе

$result = mysqli_query($link, "INSERT INTO product (name, price, rating) 
                                VALUE ('$name', '$price', '$rating')");
$id = mysqli_insert_id($link);

$result = mysqli_query($link, "SELECT id, name, price, rating FROM product WHERE id = '{$id}'");
$row = mysqli_fetch_array($result);

$arr = array('price' => $row['price'], 'name' => $row['name']);
echo json_encode($arr);
?>

    像这样的东西

    • 1

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