<?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;
}
?>
我能够呼应到从数据库中检索的FILE_NAME,但我无法将文件附加在这个部分$img_r = imagecreatefromjpeg($src);
这是它会导致错误? 有什么想法?无法附加文件名
运行此脚本时是否生成错误消息? – 2012-02-10 18:33:44