Php 无法将图片添加到数据库
为什么我的图片无法添加到数据库中,即使我创建了blob数据类型来存储变量Php 无法将图片添加到数据库,php,Php,为什么我的图片无法添加到数据库中,即使我创建了blob数据类型来存储变量$picture的文件 //prepare and bind >> $stmt = $conn->prepare("INSERT INTO user (name, email, gender, phonenum, address, ic_number, occupation, picture) VALUES (?, ?, ?, ?, ?,
$picture
的文件
//prepare and bind >>
$stmt = $conn->prepare("INSERT INTO user (name, email, gender, phonenum, address, ic_number, occupation, picture) VALUES
(?, ?, ?, ?, ?, ?, ?, ?)");
$stmt->bind_param("sssssssb", $name, $email, $gender, $phonenum, $address, $ic_number, $occupation, $picture);
// add into database >>
$name = $_POST['name'];
$email = $_POST['email'];
$gender = $_POST['gender'];
$phonenum = $_POST['phonenum'];
$address = $_POST['address'];
$ic_number = $_POST['ic_number'];
$occupation = $_POST['occupation'];
$picture = $_POST['picture'];
echo "New records created successfully";
$stmt->execute();
$stmt->close();
$conn->close();
}
?>
</div>
<div class="right">
<h3>Profile picture: </h3>
<img id="output" style="width:150px;height:200px;"/>
<br><input type="file" name="picture" placeholder="" onchange="loadFile(event)">
</div>
<!-- pictures-->
<script>
var loadFile = function (event) {
var output = document.getElementById('output');
output.src = URL.createObjectURL(event.target.files[0]);
};
</script>
//准备并绑定>>
$stmt=$conn->prepare(“插入用户(姓名、电子邮件、性别、电话号码、地址、ICU号码、职业、图片)值
(?, ?, ?, ?, ?, ?, ?, ?)");
$stmt->bind_参数(“SSB”、$name、$email、$gender、$phonenum、$address、$ic_number、$accountry、$picture);
//添加到数据库>>
$name=$_POST['name'];
$email=$_POST['email'];
$gender=$_POST['gender'];
$phonenum=$_POST['phonenum'];
$address=$_POST['address'];
$ic_编号=$_POST['ic_编号'];
$occulation=$_POST['occulation'];
$picture=$_POST['picture'];
回显“成功创建新记录”;
$stmt->execute();
$stmt->close();
$conn->close();
}
?>
个人资料图片:
您是否已检查值$\u POST['picture']是否包含数据
$\u文件保存有关附加文件的数据这应该是一条注释。为什么在执行查询之前要输出成功消息“New records created successfully”?为什么你还没有做任何事情来检查查询是否成功,如果没有询问数据库到底出了什么问题——尽管如何做很容易研究/阅读?@Carey这是一个不真实的说法。请阅读手册中的第一个示例,然后删除您的评论。