<?php
$con = mysql_connect("localhost","root","");
mysql_select_db("image", $con);
if ($_SERVER['REQUEST_METHOD'] == 'POST')
{
$result = mysql_query("SELECT * FROM image ORDER BY file_name DESC LIMIT 1");
$row = mysql_fetch_array($result);
$src = '"'.$row['file_name'].'"';
$targ_w = $targ_h = 300;
$jpeg_quality = 90;
$img_r = imagecreatefromjpeg($src);
$dst_r = ImageCreateTrueColor( $targ_w, $targ_h );
imagecopyresampled($dst_r,$img_r,0,0,$_POST['x'],$_POST['y'],
$targ_w,$targ_h,$_POST['w'],$_POST['h']);
header('Content-type: image/jpg');
imagejpeg($dst_r,null,$jpeg_quality);
exit;
}
?>
我可以通过echoing从数据库中检索file_name,但是我无法在此部分附加文件$img_r = imagecreatefromjpeg($src);
这是导致错误的地方?
任何想法?
答案 0 :(得分:2)
嗯...因为您无缘无故地在文件名中添加引号?
将$src = '"'.$row['file_name'].'"';
更改为$src = $row['filename'];
。