Question

我想要做的是，如果用户填写表单中的信息并单击提交按钮，它将把数据存储在数据库中。在我测试它的数据库连接没问题。

我现在的问题是它根本不会保存，任何人都可以帮我解决代码错误吗？

的index.php：

<?php include_once('header.php'); ?>
<html lang="en">
<head>
<script src="bootstrap/js/jquery-1.11.0.min.js"></script>
<style>
#Picture {
    width: 120px;
    height: 120px;
}
#imgInp { display:none;}
</style>
<script type="text/javascript">
$(document).ready(function() {
    var currentSrc = $('#Picture').attr('src');
    if(currentSrc==null || currentSrc==""){        
    $('#Picture').attr('src','http://i38.photobucket.com/albums/e149/eloginko/profile_male_large_zpseedb2954.jpg');

   $("#Picture").on('click', function() {
       $("#imgInp").trigger('click')}
   )}


    function readURL(input) {
        if (input.files && input.files[0]) {
            var reader = new FileReader();

            reader.onload = function (e) {
                $('#Picture').attr('src', e.target.result);
            }

            reader.readAsDataURL(input.files[0]);
        }
    }

    $("#imgInp").change(function(){
        readURL(this);
    });

});

</script>
</head>
<body>
<form method="post" action="index.php" enctype="multipart/form-data">   
<img id="Picture" name="Picture" data-src="#" /> <br />
<input type='file' id="imgInp" accept="image/*" /><br />
Name: <input type="text" id="name" name="name" /><br />
Age: <input type="text" id="age" name="age" /><br />
Address: <input type="text" id="address" name="address" /><br />
<input type="radio" name="gender" id="gender" value="male" />Male
<input type="radio" name="gender" id="gender" value="Female" />Female<br />
<input type="submit" name="add" value="Add" />
</form>
</body>
</html>

的header.php：

<?php
$host = "localhost";
$user = "root";
$pass = "";
$db = "test";

$dbc = new PDO("mysql:host=" . $host . ";dbname=" . $db, $user, $pass);

if(isset($_POST['add'])){

    $name = isset($_POST['name']) ? $_POST['name'] : null;
    $age = isset($_POST['age']) ? $_POST['age'] : null;
    $address = isset($_POST['address']) ? $_POST['address'] : null;
    $gender = isset($_POST['gender']) ? $_POST['gender'] : null;
    $picture = isset($_FILES['Picture']) ? $_FILES['Picture'] : null;

    $q = "INSERT INTO students(name, age, address, gender, profile_pic) VALUES(:name, :age, :address, :gender, :Picture)";

    $query = $dbc->prepare($q);
    $query->bindParam(':name', $name);
    $query->bindParam(':age', $age);
    $query->bindParam(':address', $address);
    $query->bindParam(':gender', $gender);
    $query->bindParam(':Picture', $picture);

    $results = $query->execute();

}
?>

Answer 1

改变这个 -

<form method="post" action="index.php" enctype="multipart/form-data">

到此 -

<form method="post" action="header.php" enctype="multipart/form-data">

我在PDO中的插入数据不起作用

1 个答案: