Question

我对PHP和mySql非常陌生，但我正在尝试学习项目。我已经按照本教程http://www.formget.com/ajax-image-upload-php/来上传图像作为MySQL数据库中blob表中的列，其中包含图像大小，id等行。

我有一个单独的数据表，我在其中为各个用户帐户创建列（每个帐户都有一行用于用户名，密码等）。我在这些列中创建了一行来存储blob。

我不需要教程为其图像创建的所有行（image_type，size等），但实际上只需要图像源（图像行）。我需要将此图像插入到我的帐户列中的图像的ROW中（取决于登录的帐户），不要为每个图像创建新列。我不知道如何使用我的代码来解决这个问题。这是我的HTML表单的JavaScript：

$(document).ready(function (e) {
    //To transfer clicks to divs
     $(".upload-button").on('click', function() {
       $("#file").click();
    });
    $(".save").on('click', function() {
       $(".submit").click();
    });


        $("#uploadimage").on('submit',(function(e) {
        e.preventDefault();


        $.ajax({
        url: "upload.php", // Url to which the request is send
        type: "POST",             // Type of request to be send, called as method
        data: new FormData(this), // Data sent to server, a set of key/value pairs (i.e. form fields and values)
        contentType: false,       // The content type used when sending data to the server.
        cache: false,             // To unable request pages to be cached
        processData:false,        // To send DOMDocument or non processed data file it is set to false
        success: function(data)   // A function to be called if request succeeds
        {

        }
        });
        }));

        // Function to preview image after validation
        $(function() {
        $("#file").change(function() {
         // To remove the previous error message
        var file = this.files[0];
        var imagefile = file.type;
        var match= ["image/jpeg","image/png","image/jpg"];
        if(!((imagefile==match[0]) || (imagefile==match[1]) || (imagefile==match[2])))
        {
        $('.userimg').attr('src','noimage.png');

        return false;
        }
        else
        {
        var reader = new FileReader();
        reader.onload = imageIsLoaded;
        reader.readAsDataURL(this.files[0]);
        }
        });
        });
        function imageIsLoaded(e) {
        $("#file").css("color","green");
        $('#image_preview').css("display", "block");
        $('.userimg').attr('src', e.target.result);
        $('.userimg').attr('width', '250px');
        $('.userimg').attr('height', '230px');
        };
});

然后引用upload.php，这是需要进行更改的地方：

<?php

if(isset($_FILES["file"]["type"]))
{
$validextensions = array("jpeg", "jpg", "png");
$maxsize = 99999999;
$temporary = explode(".", $_FILES["file"]["name"]);
$file_extension = end($temporary);
if ((($_FILES["file"]["type"] == "image/png") || ($_FILES["file"]["type"] == "image/jpg") || ($_FILES["file"]["type"] == "image/jpeg")
) && ($_FILES["file"]["size"] < $maxsize)//Approx. 100kb files can be uploaded.
&& in_array($file_extension, $validextensions)) {
if ($_FILES["file"]["error"] > 0)
{
echo "Return Code: " . $_FILES["file"]["error"] . "<br/><br/>";
}
else
{
if (file_exists("images/" . $_FILES["file"]["name"])) {
echo $_FILES["file"]["name"] . " <span id='invalid'><b>already exists.</b></span> ";
}
else
{
$sourcePath = $_FILES['file']['tmp_name']; // Storing source path of the file in a variable
$targetPath = "images/".$_FILES['file']['name']; // Target path where file is to be stored

 $size = getimagesize($_FILES['file']['tmp_name']);
 /*** assign our variables ***/
 $type = $size['mime'];
 $imgfp = fopen($_FILES['file']['tmp_name'], 'rb');
 $size = $size[3];
 $name = $_FILES['file']['name'];



 /*** check the file is less than the maximum file size ***/
 if($_FILES['file']['size'] < $maxsize )
 {
 /*** connect to db ***/
 $dbh = new PDO("mysql:host=localhost;dbname=sqlserver", 'username', 'password');

 /*** set the error mode ***/
 $dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

 /*** our sql query ***/
 $stmt = $dbh->prepare("INSERT INTO imageblob (image_type ,image, image_size, image_name) VALUES (? ,?, ?, ?)");

 /*** bind the params ***/
 $stmt->bindParam(1, $type);
 $stmt->bindParam(2, $imgfp, PDO::PARAM_LOB);
 $stmt->bindParam(3, $size);
 $stmt->bindParam(4, $name);

 /*** execute the query ***/
 $stmt->execute();
 $lastid = $dbh->lastInsertId(); 
 //Move uploaded File
 move_uploaded_file($sourcePath,$targetPath) ; // Moving Uploaded file
 if(isset($lastid))
 {
 /*** assign the image id ***/
 $image_id = $lastid;
     try {
     /*** connect to the database ***/
     /*** set the PDO error mode to exception ***/
     $dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

     /*** The sql statement ***/
     $sql = "SELECT image, image_type FROM imageblob WHERE image_id=$image_id";

     /*** prepare the sql ***/
     $stmt = $dbh->prepare($sql);

     /*** exceute the query ***/
     $stmt->execute(); 

     /*** set the fetch mode to associative array ***/
     $stmt->setFetchMode(PDO::FETCH_ASSOC);

     /*** set the header for the image ***/
     $array = $stmt->fetch();
     /*** check we have a single image and type ***/
     if(sizeof($array) == 2)
     {
         //To Display Image File from Database
         echo '<img src="data:image/jpeg;base64,'.base64_encode( $array['image'] ).'"/>';

     }
     else
     {
     throw new Exception("Out of bounds Error");
     }
     }
     catch(PDOException $e)
     {
     echo $e->getMessage();
     }
     catch(Exception $e)
     {
     echo $e->getMessage();
     }
     }
     else
     {
     echo 'Please input correct Image ID';
     }
 }
 else
 {
 /*** throw an exception is image is not of type ***/
 throw new Exception("File Size Error");
 }
}
}
}
else
{
echo "<span id='invalid'>***Invalid file Size or Type***<span>";
}
}

?>

我试图删除对图像大小，类型等的引用，因为我认为这些是不必要的，但这会产生错误。我已经倾注了其他SO帖子，但无法理解如何简单地将图像插入到mysql数据库中EXISTING列中的行中。我只能为图像创建新列。

我该如何做到这一点？

Answer 1

通常，由于各种原因，将文件存储在数据库中not通常为recommended。虽然该规则有一些例外，但您应该确定这是最适合您的解决方案。

相反，首先要问为什么要这样做以及为什么在磁盘上存储文件然后跟踪数据库中的 filename 并不是首选的解决方案。一旦你可以证明你的用例，（如果你可以证明你的用例是合理的）那么你应该问如何实现它。

将图像插入MySQL表中的列中的行而不是单个列？

1 个答案: