如何在PHP中显示数据库中的图像？

Question

我写下以下代码来显示数据库中的图像

<?php
mysql_connect("localhost","root","");
mysql_select_db("test");
$qry = "select id from image";
$res = mysql_query($qry) or die(mysql_error());
while($row = mysql_fetch_array($res))
{
  echo "<img src=image.php?id='$row[0]'>";
}
?>

Image.php

<?php
$query = mysql_query("SELECT img FROM images WHERE id = ".$_GET['id']);
$row = mysql_fetch_array($query);
if($type=="pjpeg")
$type="jpeg";
header("Content-type:$type");
echo $row['img'];
?>

但这不起作用。它显示空白图像图标。

Answer 1

您也可以use base64 encoding构建图片。像

<img src="data:image/jpeg;base64,/9j/4AAQSkZJRgABAQEAWgBaAAD/4gxYSUNDX1BST0ZJTEUAAQEAAAxITGlubwIQAAB..." />

UPDATE，base64编码示例

你可以轻松地做到这一点：

<?php
$imageId = intval($_GET["id"]);

$query = mysql_query("SELECT img FROM images WHERE id = ". $imageId);
$row = mysql_fetch_array($query);

$mime = null;
// place $type init. here
if ($type=="pjpeg") // <<< where do you get $type btw?
    $mime = "image/jpeg";

$b64Src = "data:".$mime.";base64," . base64_encode($row["img"]);
echo '<img src="'.$b64Src.'" alt="" />';
?>

Answer 2

jpeg不是有效的内容类型，应该是image/jpeg

Answer 3

如上所述 +确保将replace = true参数传递给标题函数。

header( 'Content-Type: image/jpeg', true );