保存上传到数据库中服务器文件夹上的图像的路径

Question

这是一个脚本，用于在服务器文件夹上上传多个图像并在数据库中存储它们的路径，但是图像的路径只存储在一列中。

例如：我上传了3张图片，image1.jpg，image2.jpg，image3.jpg。 这些图像应该存储在3列中，即offimage1，offimage2，offimage3。

现在的问题是所有3个图像的路径仅存储在offimage1下。 存储在offimage1中的路径看起来像这样，

uploads/image1.jpg*uploads/image1.jpgimage2.jpg*uploads/image1.jpgimage2.jpgimage3.jpg

我希望图像应该以这种方式存储：

uploads/image1.jpg in colum offimage1 
uploads/image1.jpgimage2.jpg in colum offimage2 
uploads/image1.jpgimage2.jpgimage3.jpg in colum offimage3

HTML表单

 <form enctype="multipart/form-data" action="insert_image.php?id=<?php echo $_GET['id']; ?>" method="post">
        <div id="filediv"><input name="file[]" type="file" id="file"/></div><br/>
        <input type="button" id="add_more" class="upload" value="Add More Files"/>
        <input type="submit" value="Upload File" name="submit" id="upload" class="upload"/>
    </form>

insert_image.php

<?php
ob_start();
require 'connection.php';
if (isset($_POST['submit'])) {
    $j = 0; //Variable for indexing uploaded image 

    $target_path = "uploads/"; //Declaring Path for uploaded images
    for ($i = 0; $i < count($_FILES['file']['name']); $i++) {//loop to get individual element from the array

        $validextensions = array("jpeg", "jpg", "png");  //Extensions which are allowed
        $ext = explode('.', basename($_FILES['file']['name'][$i]));//explode file name from dot(.) 
        $file_extension = end($ext); //store extensions in the variable

        $target_path = $target_path . md5(uniqid()) . "." . $ext[count($ext) - 1];//set the target path with a new name of image
        $j = $j + 1;//increment the number of uploaded images according to the files in array       

      if (($_FILES["file"]["size"][$i] < 100000) //Approx. 100kb files can be uploaded.
                && in_array($file_extension, $validextensions)) {
            if (move_uploaded_file($_FILES['file']['tmp_name'][$i], $target_path)) {//if file moved to uploads folder
                //echo $j. ').<span id="noerror">Image uploaded successfully!.</span><br/><br/>';

                $file_name_all.=$target_path."*";
                $filepath = rtrim($file_name_all, '*'); 
                //echo $filepath;
                $officeid = $_GET['id'];
                $sql = "UPDATE register_office SET offimage='$filepath' WHERE id='$officeid' ";
                            if (!mysqli_query($con,$sql)) 
                                {
                                    die('Error: ' . mysqli_error($con));
                                }

            } else {//if file was not moved.
                echo $j. ').<span id="error">please try again!.</span><br/><br/>';
            }
        } else {//if file size and file type was incorrect.
            echo $j. ').<span id="error">***Invalid file Size or Type***</span><br/><br/>';
        }
    }
    header("Location: co_request_sent.php ");
}
mysqli_close($con); 
?>

如果有人能帮助我，我将不胜感激

Answer 1

您的所有三个图像的路径信息看起来像一个大字符串的原因：

uploads/image1.jpg*uploads/image1.jpgimage2.jpg*uploads/image1.jpgimage2.jpgimage3.jpg

是因为你循环遍历FILES数组并在每次迭代时连接$ target_path：

$target_path = $target_path . md5(uniqid()) . "." . $ext[count($ext) - 1];

如果您希望$ target_path仅保存当前图像的信息，请首先通过移动来重置变量：

$target_path = "uploads/";

在你的for循环中（所以向下移动两行）。

这是你问题的第一部分。 但是，在名为offimage1，offimage2等的列中存储有关图像的信息是一个非常糟糕的主意 。

这种事情应该存储在一行，而不是列。

所以也许你最好用一个名为“id”，“officeid”和“path”的列创建一个名为offimages的新表。

现在只需为每个图像插入一行（现在无论你有多少都没关系），并确保使用officeid将其链接到register_office表。