如何在數據庫中存儲文件的名稱和路徑？

Question

可以上傳文件，但是不能將路徑和文件名存儲到我的數據庫中。

這是我的表單html。

<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1, shrink-to-fit=no">
<meta name="description" content="">
<meta name="author" content="">

<form action="../model/uploadcredentials.php" method="POST" enctype="multipart/form-data">
    <input type="file" name="fileToUpload" placeholder="Upload your credentials">
    <button type="submit" name="submit">Upload</button>
</form>

這是我的動作文件

<?php
if(isset($_FILES["fileToUpload"]["name"]))
{
    $target_dir = "../assets/credentials/";
    $target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
    $uploadOk = 1;
    $imageFileType = strtolower(pathinfo($target_file,PATHINFO_EXTENSION));
    // Check if image file is a actual image or fake image
    if(isset($_POST["submit"])) {
        $check = getimagesize($_FILES["fileToUpload"]["tmp_name"]);
        if($check !== false) {
            echo "File is an image - " . $check["mime"] . ".";
            $uploadOk = 1;

        } else {
            echo "File is not an image.";
            $uploadOk = 0;
        }
    }
    // Allow certain file formats
    if($imageFileType != "jpg" && $imageFileType != "png" && $imageFileType != "jpeg"
    && $imageFileType != "gif" ) {
        echo "Sorry, only JPG, JPEG, PNG & GIF files are allowed.";
        $uploadOk = 0;
    }
    // Check if $uploadOk is set to 0 by an error
    if ($uploadOk == 0) {
        echo "Sorry, your file was not uploaded.";
    // if everything is ok, try to upload file
    } else {
        if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)) 
        {
            echo "The file ". basename( $_FILES["fileToUpload"]["name"]). " has been uploaded.";

            $file = stripcslashes($_FILES["fileToUpload"]["name"]);
            $file = mysqli_real_escape_string($con,$file);
            $query = $con->query("INSERT INTO `portfolio`(p.desc) VALUES('$file')");
        } else {
            echo "Sorry, there was an error uploading your file.";
        }
    }
}
?>

我正在嘗試獲取文件名以及將其存儲在數據庫中的路徑

Answer 1

始終值得檢查並報告錯誤。

在您的INSERT中， (p.desc)試圖查找(p.desc)名為p （通常是別名）的表。 嘗試...

$query = $con->query("INSERT INTO portfolio(desc) VALUES('$file')");

（無p. ）。

但是我仍然建議使用預處理語句，請閱讀如何在php中創建安全的mysql預處理語句？

如何在數據庫中存儲文件的名稱和路徑？

問題描述

1 個解決方案

解決方案1
0 2018-07-22 06:34:07

如何在數據庫中存儲文件的名稱和路徑？

問題描述

1 個解決方案

解決方案1 0 2018-07-22 06:34:07

解決方案1
0 2018-07-22 06:34:07