Question

<?php include"dbconn.php";
    if(isset($_POST["submit_image"])){
        $imgdate = $_POST["image_date"];
        $imgalbum = $_POST["image_album"];  

$check = getimagesize($_FILES["image"]["tmp_name"]);
    if($check != false){
        $image = $_FILES['image']['tmp_name'];
        $imgContent = addslashes(file_get_contents($image));

    $insert = "INSERT INTO image_db (img_pic, img_date, img_album) VALUES ('$imgContent', '$imgdate', '$imgalbum')";
    $result = mysqli_query($conn,$insert);
    //check if the row were updated
    if (mysql_affected_rows($result)>0){
    echo "File uploaded successfully.";
    }else{
    echo "File upload failed, please try again.";
    } 
}else{
   echo "Please select an image file to upload.";}}?>

我称之为upload.php的顶级php文件

我称之为upload.html的底部

表单将在upload.html上，然后当用户点击它时将重定向到upload.php

我的dbconn.php已连接到表

<!DOCTYPE html>
<head>
	<title>Upload picture</title>
</head>
<body>
<form method="POST" action="upload.php" enctype="multipart/form-data">
	<input type="file" name="image">
	<label for="text">Image Date :</label>
	<input type="text" name="image_date">
		<label for="text">Image Album:</label>
		<input type="text" name="image_album">

	<input type="submit" name="submit_image" value="Upload">
</form>
</body>
</html>

我坚持到这里并继续说我的mysql_affected_rows（）期望参数1是资源，给定布尔值并继续无法上传

Answer 1

试试这个： mysqli_affected_rows ，没有 mysql_affected_rows

在将数据发送到mysql服务器之前，您需要使用 filter_input 。 http://php.net/manual/en/function.filter-input.php

$insert = "INSERT INTO image_db (img_pic, img_date, img_album) VALUES ('" . $imgContent . "', '" . $imgdate ."', '" . $imgalbum . "')";

为何无法上传图片

1 个答案: