我试图使多个图像上传与数据连接,所以图像名称应保存在数据库中,现在当我上传图像,它是上传,也显示在文件夹中,但没有在我的数据库中上传,并说对不起,数据无法更新!
请帮帮我,我已经试了一天了。这是我的上传代码
<?php
require_once 'dbconfig.php';
require_once 'common_functions.php';
if (isset($_POST["sub"])) {
$msg = "";
if (count($_FILES["user_files"]) > 0) {
$folderName = "uploads/";
$counter = 0;
for ($i = 0; $i < count($_FILES["user_files"]["name"]); $i++) {
$file_name=$_FILES["user_files"]["name"][$i];
if ($_FILES["user_files"]["name"][$i] <> "") {
$ext = strtolower(pathinfo($file_name,PATHINFO_EXTENSION)); // get image extension
$file_name = $folderName . rand(10000, 990000) . '_' . time() . '.' . $ext;
$valid_extensions = array('jpeg', 'jpg', 'png'); // valid extensions
if(in_array($ext, $valid_extensions)){
$filename=basename($file_name,$ext);
$newFileName=$filename.$ext;
if (!move_uploaded_file($_FILES["user_files"]["tmp_name"][$i],"Uploads/".$newFileName)) {
$msg .= "Failed to upload" . $_FILES["user_files"]["name"][$i] . ". <br>";
$counter++;
}
if(!isset($msg)) {
$stmt = $DB_con->prepare('INSERT INTO products SET productname=:productname');
$stmt->bindParam(':productname',$newFileName);
if($stmt->execute()){
$msg = "Your Post ' ".$newFileName." ' has been successfully uploaded , <a target='_blank' href='post.php?name=".$newFileName." '>View</a>.";
// redirects video view page after 5 seconds.
} else {
$msg = "error while inserting....";
}
} else {
$msg = "Sorry Data Could Not Updated !";
}
} else {
$msg = "Sorry, only JPG, JPEG, PNG files are allowed.";
}
}
}
}
} else {
$msg = errorMessage("You must upload atleast one file");
}
?>
1条答案
按热度按时间wqsoz72f1#
“insert”语句中没有“set”。尝试。。。
将来,您可以通过触发mysqli错误并显示以下内容来节省一天左右的时间: