PHP MySQL从数据库显示图像
[英]PHP MySQL display image from database
问题描述
Hi I have form to make new article with informations and image. 
嗨，我有用信息和图像制作新文章的表格。 I susccesfuly save all info and image to database (i guess). 我成功地将所有信息和图像保存到数据库（我猜）。 And when I want to display info and image so only info works. 当我想显示信息和图像时，仅信息有效。 See in picture. 如图所示。
Any help? 有什么帮助吗？ Thanks a lot 非常感谢
Inserting_post.php Inserting_post.php
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>APK Market</title>
<link rel="stylesheet" type="text/css" href="../style/style.css">
<link href="../bootstrap/css/bootstrap.css" rel="stylesheet" media="screen">
<script src="../bootstrap/js/bootstrap.min.js"></script>
<script src="../bootstrap/js/bootstrap.js"></script>
</head>
<body>
<form method="post" action="insert_post.php" enctype="multipart/form-data">
<div class="new_post">
<div class="headtitle">Insert new post</div>
<div class="new_title">
<p>New title</p>
<input type="text" name="title">
</div>
<div class="new_author">
<p>Author</p>
<input type="text" name="author">
</div>
<div class="new_keywords">
<p>Keywords</p>
<input type="text" name="keywords">
</div>
<div class="new_image">
<p>Image</p>
<input type="file" name="image">
</div>
<div class="new_content">
<textarea name="content" cols="20" rows="8"></textarea>
</div>
<div class="submit">
<input type="submit" name="submit" value="OK">
</div>
</div>
</form>
<script src="https://code.jquery.com/jquery.js"></script>
<script src="../bootstrap/js/bootstrap.js"></script>
</body>
</html>
<?php
include("../includes/connect.php");
if(isset($_POST['submit']))
{
$games_date = date('y-m-d-h');
$games_title = $_POST['title'];
$games_author = $_POST['author'];
$games_keywords = $_POST['keywords'];
$games_image = $_FILES['image']['name'];
$games_tmp = $_FILES['image']['tmp_name'];
$games_content = $_POST['content'];
if($games_title=="" or $games_author=="" or $games_keywords==""
or $games_content=="")
{
echo"<script>alert('any field is empty')</script>";
exit();
}
else
move_uploaded_file($games_tmp,"../uploaded_images/$games_image");
$insert_query= "insert into games(games_title,games_date,games_author,games_image,
games_keywords,games_content)
values ('$games_title','$games_date','$games_author','$games_image',
'$games_keywords','$games_cont ent')";
}
if(mysql_query($insert_query))
{
echo "<center><h1>Post published seccesfuly!</h1></center>";
}
?>
display page:
<div class="content">
<?php
include('connect.php');
$select_posts = "select * from games";
$run_posts = mysql_query($select_posts);
while($row=mysql_fetch_array($run_posts))
{
echo '<p class="games_title_result">' .$games_title = $row['games_title'];
echo '<p class="games_image_result"><img src="<?php echo $row["games_image"];?>';
echo '<p class="games_content_result">' .$games_content = $row['games_content'];
echo '<p class="games_date_result">' .$games_date = $row['games_date'];
echo '<p class="games_author_result">' .$games_author = $row['games_author'];
}
?>
</div>
3 个解决方案
解决方案1
2 已采纳 2014-02-15 14:49:43
Here's my answer: 这是我的答案：
echo '<p class="games_image_result"><img src="uploaded_images/'.$row["games_image"].'" />';
You got a syntax error where you include a <?php and ?> inside your php, and on your echo at that. 您遇到了语法错误，其中在<?php内包含<?php和?> ，并在此处回显。
So when you look at your img's src, it would have those. 因此，当您查看img的src时，它将具有这些内容。
Also, you forgot to close your img tag and your other p tags. 另外，您忘记了关闭img标签和其他p标签。
解决方案2
0 2014-02-15 14:44:03
Your image tag isn't closed properly and the path to the image is not complete. 您的图片标签未正确关闭，并且图片的路径不完整。 You are only specifying the filename. 您仅指定文件名。 I am not sure about your directory structure 我不确定您的目录结构
echo '<p class="games_image_result"><img src="uploaded_images/<?php echo $row["games_image"];?>" />';
解决方案3
0 2014-02-15 14:51:07
Display Image from DataBase using php 使用PHP从数据库显示图像
# SQL Statement
$sql = "SELECT `idimage` FROM `yourTableName` WHERE id=" . mysqli_real_escape_string($_GET['id']) . ";";
$result = mysqli_query("$sql") or die("Invalid query: " . mysqli_error());
# Set header
header("Content-type: image/your image name which type of your image");
echo mysqli_result($result, 0);
}
else
echo 'Please check the ID!';
?>
I wish it's may be help you 希望对您有帮助
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