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php - 我确保数据库有图像并且数据库连接正确，但我无法显示图像

转载 作者：行者123 更新时间：2023-11-29 12:39:44 25 4
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我的显示代码是:

 <?php
 include "file_constants.php";
 // just so we know it is broken
 error_reporting(E_ALL);
 // some basic sanity checks
 if(isset($_GET['id']) && is_numeric($_GET['id'])) {
     //connect to the db
     $link = mysql_connect("$host", "$user", "$pass")
     or die("Could not connect: " . mysql_error());

     // select our database
     mysql_select_db("$db") or die(mysql_error());

     // get the image from the db
     $sql = "SELECT image FROM test_image WHERE id=" .$_GET['id'] . ";";

     // the result of the query
     $result = mysql_query("$sql") or die("Invalid query: " . mysql_error());

     // set the header for the image
     header("Content-type: image/jpeg");

     echo mysql_result($result, 0);
     // close the db link
     mysql_close($link);
 }
 else {
     echo 'Please use a real id number';
 }
?>

我确保数据库有图像并且数据库连接正确。我可以将图像从 php 上传到 phpmyadmin (MYSQL)。但是，我无法显示图像。(http:///file_display.php?id=1)有人可以帮我在php中显示图像吗？非常感谢!

文件插入代码:

<html>
<head><title>File Insert</title></head>
<body>
<h3>Please Choose a File and click Submit</h3>

<form enctype="multipart/form-data" action=
"<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input type="hidden" name="MAX_FILE_SIZE" value="10000000" />
<input name="userfile" type="file" />
<input type="submit" value="Submit" />
</form>

<?php

// check if a file was submitted
if(!isset($_FILES['userfile']))
{
    echo '<p>Please select a file</p>';
}
else
{
    try {
    $msg= upload();  //this will upload your image
    echo $msg;  //Message showing success or failure.
    }
    catch(Exception $e) {
    echo $e->getMessage();
    echo 'Sorry, could not upload file';
    }
}

// the upload function

function upload() {
    include "file_constants.php";
    $maxsize = 10000000; //set to approx 10 MB

    //check associated error code
    if($_FILES['userfile']['error']==UPLOAD_ERR_OK) {

        //check whether file is uploaded with HTTP POST
        if(is_uploaded_file($_FILES['userfile']['tmp_name'])) {    

            //checks size of uploaded image on server side
            if( $_FILES['userfile']['size'] < $maxsize) {  

               //checks whether uploaded file is of image type
              //if(strpos(mime_content_type($_FILES['userfile']['tmp_name']),"image")===0) {
                 $finfo = finfo_open(FILEINFO_MIME_TYPE);
                if(strpos(finfo_file($finfo, $_FILES['userfile']['tmp_name']),"image")===0) {    

                    // prepare the image for insertion
                    $imgData =addslashes (file_get_contents($_FILES['userfile']['tmp_name']));

                    // put the image in the db...
                    // database connection
                    mysql_connect($host, $user, $pass) OR DIE (mysql_error());

                    // select the db
                    mysql_select_db ($db) OR DIE ("Unable to select db".mysql_error());

                    // our sql query
                    $sql = "INSERT INTO test_image
                    (image, name)
                    VALUES
                    ('{$imgData}', '{$_FILES['userfile']['name']}');";

                    // insert the image
                    mysql_query($sql) or die("Error in Query: " . mysql_error());
                    $msg='<p>Image successfully saved in database with id ='. mysql_insert_id().' </p>';
                }
                else
                    $msg="<p>Uploaded file is not an image.</p>";
            }
             else {
                // if the file is not less than the maximum allowed, print an error
                $msg='<div>File exceeds the Maximum File limit</div>
                <div>Maximum File limit is '.$maxsize.' bytes</div>
                <div>File '.$_FILES['userfile']['name'].' is '.$_FILES['userfile']['size'].
                ' bytes</div><hr />';
                }
        }
        else
            $msg="File not uploaded successfully.";

    }
    else {
        $msg= file_upload_error_message($_FILES['userfile']['error']);
    }
    return $msg;
}

// Function to return error message based on error code

function file_upload_error_message($error_code) {
    switch ($error_code) {
        case UPLOAD_ERR_INI_SIZE:
            return 'The uploaded file exceeds the upload_max_filesize directive in php.ini';
        case UPLOAD_ERR_FORM_SIZE:
            return 'The uploaded file exceeds the MAX_FILE_SIZE directive that was specified in the HTML form';
        case UPLOAD_ERR_PARTIAL:
            return 'The uploaded file was only partially uploaded';
        case UPLOAD_ERR_NO_FILE:
            return 'No file was uploaded';
        case UPLOAD_ERR_NO_TMP_DIR:
            return 'Missing a temporary folder';
        case UPLOAD_ERR_CANT_WRITE:
            return 'Failed to write file to disk';
        case UPLOAD_ERR_EXTENSION:
            return 'File upload stopped by extension';
        default:
            return 'Unknown upload error';
    }
}
?>
</body>
</html>

SQL 是:

create table test_image (
id              int(10)  not null AUTO_INCREMENT PRIMARY KEY,
name            varchar(25) not null default '',
image           blob        not null
 );

教程是http://vikasmahajan.wordpress.com/2010/07/07/inserting-and-displaying-images-in-mysql-using-php/

最佳答案

如果您仅将图像名称保存在数据库中，这将在 HTML 中显示该图像

<?php
    $result = mysql_query("$sql") or die("Invalid query: " . mysql_error());
    while($row = mysqli_fetch_array($result)) {
        echo '<img src="www.yourdomain.com/your/directory/"'. $row["image "].'/>';
    }
?>

注意:如果只有一行，则不需要使用 while 循环

关于php - 我确保数据库有图像并且数据库连接正确，但我无法显示图像，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/26292647/

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