我想在mysql数据库中显示php上的图像。我使用以下代码:
$last_id = mysql_insert_id();
echo "Image Uploaded. <p /> Your image <p /> <img src=get.php?id=$last_id>";
<?php
mysql_connect("localhost","root","") or die(mysql_error());
mysql_select_db("image") or die(mysql_error());
$id = $_REQUEST['id'];
$result = mysql_query("SELECT * FROM image WHERE id = $id");
$image = $result['image'];
header("Content-type: image/jpeg");
echo $image;
?>
答案 0 :(得分:3)
首先,Don't use mysql_*
functions in new code。它们不再被维护and are officially deprecated。请参阅red box?转而了解prepared statements,并使用PDO或MySQLi - this article将帮助您确定哪个。如果您选择PDO here is a good tutorial。
试试这个:
<?php
$db=mysql_connect("localhost","root","") or die(mysql_error());
mysql_select_db("image",$db) or die(mysql_error());
$id = (int)$_REQUEST['id'];
$result = mysql_query("SELECT * FROM image WHERE id=$id",$db);
if(mysql_numrows($result)>=1){
$row = mysql_fetch_assoc($result);
header("Content-type: image/jpeg");
echo $row['image'];
}
?>
同时强>
echo "Image Uploaded. <p /> Your image <p /> <img src=get.php?id=$last_id>";
无效的HTML。
echo 'Image Uploaded. <p> Your image </p> <img src="get.php?id='.$last_id.'" />';