Question

我想从用户那里获取一张图片并插入数据库。用户应该在下面的输入字段中上传文件。另外，我想在需要时显示数据库中的图像。

<label>Photo</label>
<input type="file" name="photo" required>
<?php
mysql_connect("localhost", "root", ""); mysql_select_db("prs");
if(isset($_POST['submit'])){
    $photo = $_FILES['photo']['name'];
    {
        $query = mysql_query("insert into date (photo) values ('$photo')");
    }
}
?>

Answer 1

最好的方法是在数据库表中保存文件路径或文件名。并将您的文件保存在服务器中。

以下是根据您的需要上传和检索脚本的MULTIPLE示例示例。我将在这里使用3个文件。 front.html，upload.php，最后是preview.php

Front.html

此文件包含基本的html表单内容。

<form enctype="multipart/form-data"  action="upload.php" method="post">
<input id="uploadFile" type="file" name="files[]"  class="img" />
<button class="btn btn-primary" name="up_img">POST</button>
</form>

2.upload.php

if(isset($_POST['up_img'])){ $errors= array();

    foreach($_FILES['files']['tmp_name'] as $key => $tmp_name ){
        $file_name =$key.$_FILES['files']['name'][$key];
        $file_size =$_FILES['files']['size'][$key];
        $file_tmp =$_FILES['files']['tmp_name'][$key];
        $file_type=$_FILES['files']['type'][$key];

        if($file_size > 2097152){
            $errors[]='File size must be less than 2 MB';
        }
        $pType='I';
        $query="insert into img(user,FILE_NAME)values('$sUser','$file_name') ";
        $desired_dir="images_folder";
        if(empty($errors)==true){
            if(is_dir($desired_dir)==false){
                mkdir("$desired_dir", 0700);        // Create directory if it does not exist
            }
            if(is_dir("$desired_dir/".$file_name)==false){
                move_uploaded_file($file_tmp,"$desired_dir/".$file_name);
            }else{                                  // rename the file if another one exist
                $new_dir="$desired_dir/".$file_name.time();
                 rename($file_tmp,$new_dir) ;               
            }
         mysql_query($query);           
        }else{
                print_r($errors);
        }
    }
    if(empty($error)){
    header('Location:preview.php');
    }

    }

3.preview.php

现在我已成功将我的图像/图像上传到服务器并准备检索它们。记住我的图像文件现在存储在文件夹调用images_folder中，它们的路径保存在img表中。

<?php
$sql=mysql_query('SELECT * FROM img ');
while($row = mysql_fetch_array($sql))
{
$sPic=$row['FILE_NAME'];  

echo '<img src="images_folder/'.$sPic.'">';

}
?>

就是这样。这就是您将图像上传到服务器并检索它们的方式。

Answer 2

所以这里有一个小选项来上传图片并将图片保存到mysql中这不是刚刚打字的测试。

if(isset($_FILES['picture']) //checks for form post if not show form
    {
        $title = $_POST['title']; //get title
        $desc = $_POST['description']; //get description
        $target = "downloads/pics/".basename( $_FILES['uploaded']['name']);  //this gets the target folder to save pictures make sure folder exists
        $ok=1; //basic error set to 1
        //This is our size condition 
        if ($uploaded_size > 20000000) 
        { 
            $out = "Your file is too large.<br>"; //checks for file size this can be changed to w/e
            $ok=0; //if over error is set to 0
        } 
        if ($ok==0) 
        { 
            $out.= "Sorry your file was not uploaded"; 
            return $out;
        } 
        //If everything is ok we try to upload it 
        else 
        { 
            if(move_uploaded_file($_FILES['picture']['tmp_name'], $target)) //tires to move the uploaded picture to target
            { 
                $out = "The Picture: <b><u>".basename( $_FILES['picture']['name']). "</b></u> has been uploaded";  //if successfull then post was uploaded
                $sql = "INSERT INTO database SET title=\"".$title."\",picture=\"".basename( $_FILES['uploaded']['name'])."\",description=\"".$desc."\""; //sql string to insert the folder and picture name
                    if (!mysql_query($sql))
                    {
                        die('Error In Picture Upload: - SQL INSERT INTO - '.date('m-d-y').' - SQL ERROR: '.mysql_error()); //if mysql error post error
                    }
                    else
                    {
                        $out.="<br /><br />Return To Your <a href=\"return location\">Downloads</a>"; //if mysql is ok show the return link
                        return $out;
                    }
            } 
        }
    else // if form is not submited then show form
    {
        $out ="\t\t<h3>".ucwords('new picture')."</h3>\r\n";
        $out.="\t\t<form name=\"post\" action=\"".$_SERVER['PHP_SELF']."\" method=\"post\" enctype=\"multipart/form-data\">\r\n";
        $out.="\t\t\t<p>Title:<br /><input type=\"text\" class=\"inputtext\" name=\"title\" value=\"\" /></p>\r\n";
        $out.="\t\t\t<p><b>".ucwords('picture upload')."</b>: <input type=\"file\" name=\"picture\" /></p>\r\n";
        $out.="\t\t\t<p>Description:<br /><textarea id=\"text9\" name=\"description\"></textarea></p>\r\n";
        $out.="\t\t\t<p><input type=\"submit\" name=\"submit\" value=\"".ucwords('upload')."\" /></p>\r\n";
        $out.="\t\t</form>\r\n";
        echo $out;
    }

在MySql中保存和检索照片

2 个答案: