Question

我是PHP，MYSQL的新手，试图将某种数据添加到我用html和php代码制作的表中。我的数据库是在phpmyadmin，它只是办公用品和东西等随机列表。我将在下面提供该表的信息。现在我也做了一个插入，如果你单击它，你就会进入输入字段，你可以在其中添加或插入数据到数据库。我回应了如果实际插入插入数据的结果是什么，如果它没有工作弹出的东西。顺便说一句，我没有任何错误，所以我真的不确定问题是什么。还请你忍受我的英语不好，所以我无法正确解释。这是代码，有4个单独的文件：这是数据库的链接代码：

<?php
$servername = 'localhost';
$user ='mysql1_user';
$password ='ABCD1234';
$databasename ='mysqldb1';

$con= mysqli_connect($servername, $user, $password, $databasename);
if(!$con){
    die("Wrong Entry for database". mysqli_connect_error());

}
else{
    //echo "Correct Entry";

}
?>

这是主要代码：

<html>
    <head>
        <title>MYSQL</title>
        <script src="https://code.jquery.com/jquery-3.3.1.js"
          integrity="sha256-2Kok7MbOyxpgUVvAk/HJ2jigOSYS2auK4Pfzbm7uH60="
          crossorigin="anonymous"></script>
    </head> 
    <body>
    <a href="f1.php?page=insert">INSERT</a><br><br><br>


<?php
@$v1=$_GET['page'];
if($v1=='insert'){ 
    $insertp = <<<t1
    Name: <input id="name">
    Quantity: <input id="qty">
    Type: <input id="type">
    Price: <input id="price">
    <button id="add">Add</button>
    <script>
    $("add").click(function(){
        $.post("insert.php", {
            name: $("#name").val(),
            qty: $("#qty").val(),
            type: $("#type").val(),
            price: $("#price").val(),

        },
        function(result){
            $("body").append(result);

        })


    })
    </script>

t1;
    echo $insertp;
}
else{
require "config/con1.php";
echo "<input type='search'><button id='search'>Search</button>";
$sql1="SELECT *FROM products 
-- WHERE name='Pen' ORDER BY price DESC  WHERE type='Office' AND price BETWEEN 10 and 700 ";

$result= mysqli_query($con, $sql1);
$value = mysqli_num_rows($result); 
echo "<table border=2>";
while($line = mysqli_fetch_assoc($result)){
    echo "<tr><td>".$line ['name']. "</td><td>".$line["qty"]."</td><td>".$line['type']."</td><td>".$line['price']."</td><td><a href=edit.php?id=".$line['id']."><img src='pencil.png' width='50'></a></td><td><img src='delete.png' width='50' class='delete' id=".$line['id']."></td></tr>"; 
    }
    echo "</table>"; 
}
?>

<script>
$("#search").click(function(){
    var varval=$("input[type='search']").val();
    $.post("search.php", {vtext: varval}, function (value){
        $("table").html(value)

    })

})
    $(".delete").click(function(){
        var delid=$(this);
        $.post("delete.php", {deleteid:$(this).attr("id")}, function(result){
            $(delid).parent().parent().hide(2000)

        });

    })

</script>
    </body>
</html>

以下是插入代码的位置：

    <?php
require "config/con1.php";
$name=$_POST['name'];
$qty=$_POST['qty'];
$type=$_POST['type'];
$price=$_POST['price'];

$secure_name=mysqli_real_escape_string($con, $name);
$secure_qty=mysqli_real_escape_string($con, $qty);
$secure_type=mysqli_real_escape_string($con, $type);
$secure_price=mysqli_real_escape_string($con, $price);
$sql="INSERT INTO products(name, qty, type, price) VALUES ('$secure_name', '$secure_qty', $secure_type, $secure_price)";

if(mysqli_query($con, $sql)){
    echo "One detail has been inserted into the database";

}
else{
    echo "Wrong ". mysqli_error($con);
}
?>

Answer 1

$（'add'）应为$（'#add'）。如果要绑定ID，则需要添加＃。

Answer 2

将INSERT查询更改为：

$sql="INSERT INTO products(name, qty, type, price) VALUES ('$secure_name', '$secure_qty', '$secure_type', '$secure_price')";

试图将数据插入数据库表但没有任何反应？

2 个答案: