首页 > 解决方案 > 如何将所有表记录保存到数据库php jquery json

问题描述

我设计了表格。只创建了两个字段拳头名,姓氏。输入拳头名和姓氏后单击添加按钮记录成功附加到表上。但是当我尝试将所有表记录保存到数据库中时。我没有收到任何错误,但记录没有保存到数据库中。编写我尝试过的代码,所以我写在下面。请任何人帮助我给我正确的解决方案,将记录添加到数据库中。谢谢

桌子

<div class="row">
        <div class="col-sm-12">
    <table class="table table-bordered" id="product">
        <caption> Products</caption>
        <thead>
        <tr>
             <th>Firstname</th>
            <th>Lastname</th>
        </tr>
        </thead>
        <form class="form-horizontal" id="product_list">
            <tbody> </tbody>
        </form>
    </table>
        </div>
</div>

形式

<div class="row">
    <div class="col-sm-12">
        <div class="col-md-1">
            <label>firstname</label>
            <input type="text" class="form-control" id="firstname" name="firstname">
        </div>

        <div class="col-md-1">
            <label>lastname</label>
            <input type="text" class="form-control" id="lastname" name="lastname">
        </div>
        <div class="col-md-1">
            <button type="button" id="save" class="btn btn-info" onclick="addProject()">Upadate
            </button>
            <button type="button" id="clear" class="btn btn-warning" onclick="save()">Save</button>
        </div>
    </div>
</div>

jQuery

function  addProject()
    {
        var firstname = $("#firstname").val();
        var lastname = $("#lastname").val();
        var markup = "<tr><td>" + firstname + "</td><td>" + lastname + "</td></tr>";
        $("table tbody").append(markup);
    }

    function save() {
        var table_data = [];
        $('table tbody tr').each(function(row,tr)
        {
            var sub = {
                'firstname' : $(tr).find('td:eq(0)').text(),
                'lastname' : $(tr).find('td:eq(1)').text(),
            };
            table_data.push(sub);
        });

        var _method;
            $.ajax({
                type : "POST",
                url : "save.php",
                dataType: "JSON",
                data :  {data:table_data},
                success: function (data) {
                    console.log(table_data);
                }

            });
    }

保存.php

<?php

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "sal";

$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

for($x = 0; $x < count($relative_list); $x++)
{
    $stm = $conn->prepare("INSERT INTO record(firstname,lastname)
         VALUES (?,?)");
    $stm->bind_param("ss",$firstname,$lastname);
    $firstname= $relative_list[$x]['firstname'];
    $lastname= $relative_list[$x]['lastname'];
    if ($stm->execute()) {
        echo 1;
    } else {
        echo $conn->error;
    }
    $stm->close();
}
?>

标签: phpjquery

解决方案


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