Question

您好我一直在编写用于将文件上传到服务器的代码并将值存储在Mysql数据库中我能够上传文件但面临在MYSQL服务器中插入值的问题

我正在从HTML表单中检索值，并且我已成功获取值并能够从文件中回显插入代码的表格部分需要帮助

    <?php
require "test.php";
$username=$_POST['username'];
$filename=$_FILES['uploadedfile']['name'];
$language=$_POST['language'];
$comment=$_POST['comment'];

$user_id=$_POST['user_id'];
$filenames=$_FILES['uploadedfile']['name'];

$category=$_POST['category'];
$subcategory=$_POST['subcategory'];
$comment=$_POST['comment'];
$language=$_POST['language'];
$duration=$_POST['duration'];


$domain ='example.com/store/upload/';
$path=$domain.$category.'/'.$filenames;
//$path=$domain.$category.'/'.$filenames;


// Where the file is going to be placed
$target_path = "upload/".$category.'/';

/* Add the original filename to our target path.
Result is "uploads/filename.extension" */
$target_path = $target_path . basename( $_FILES['uploadedfile']['name']);

if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) {
    echo "The file ".  basename( $_FILES['uploadedfile']['name']).
    " has been uploaded";
} else{
    echo "There was an error uploading the file, please try again!";
    echo "filename: " .  basename( $_FILES['uploadedfile']['name']);
    echo "target_path: " .$target_path;
}



echo $filenames."<br />";
echo $domain."<br />";
echo $category."<br />";
echo $path."<br />";
echo $filename."<br />";
echo $language."<br />";
echo $comment."<br />";
echo $subcategory."<br />";
echo $duration."<br />";
echo $user_id."<br />";


// query
try{


$sql="INSERT INTO vup_file(filename,path,category,sub-category,user_id,comment,language,duration)
VALUES (:filename,:path,:category,:subcategory,:user_id,:comment,:language,:duration)";
$query=$conn->prepare($sql);

$query->execute(array(':filename'=>$filename,':path'=>$path,':category'=>$category,':subcategory'=>$subcategory,':user_id'=>$user_id,':comment'=>$comment,':language'=>$language,':duration'=>$duration));
echo 'Inserted';


}catch(PDOException $e)
{
echo 'ERROR OCCURED : '.$e->getMessage();
}


?>

Answer 1

查询中的sub-category列包含连字符。它需要用反引号进行转义。

`sub-category`

在打开连接后立即添加$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);，这会发出错误信号。

SQL正在将其评估为数学问题（减号）。

转换为＆＃34; sub＆＃34; minus ＆＃34; category＆＃34;。

您可以选择使用下划线将列重命名为sub_category，而无需将其转义。

洞察

如果查询/问题中显示的是您的列被称为sub_category而不是sub-category，那么您需要在查询中将其更改为sub_category。

或者，如果它被称为subcategory。只有你知道你的专栏叫什么。

无法在PHP中使用pdo插入数据值

1 个答案: