我是PHP的新手。
我编写了以下PHP代码，以两种尺寸重新调整图像大小，效果很好。
我的问题
我需要在MySQL数据库中保存上传的图像路径。
路径
1）小图片的路径
2）大图像的路径
这是代码
<?php
error_reporting(0);
$change="";
$abc="";
define ("MAX_SIZE","400");
function getExtension($str) {
$i = strrpos($str,".");
if (!$i) { return ""; }
$l = strlen($str) - $i;
$ext = substr($str,$i+1,$l);
return $ext;
}
$errors=0;
if($_SERVER["REQUEST_METHOD"] == "POST")
{
$image =$_FILES["file"]["name"];
$uploadedfile = $_FILES['file']['tmp_name'];
if ($image)
{
$filename = stripslashes($_FILES['file']['name']);
$extension = getExtension($filename);
$extension = strtolower($extension);
if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif"))
{
$change='<div class="msgdiv">Unknown Image extension </div> ';
$errors=1;
}
else
{
$size=filesize($_FILES['file']['tmp_name']);
if ($size > MAX_SIZE*1024)
{
$change='<div class="msgdiv">You have exceeded the size limit!</div> ';
$errors=1;
}
if($extension=="jpg" || $extension=="jpeg" )
{
$uploadedfile = $_FILES['file']['tmp_name'];
$src = imagecreatefromjpeg($uploadedfile);
}
else if($extension=="png")
{
$uploadedfile = $_FILES['file']['tmp_name'];
$src = imagecreatefrompng($uploadedfile);
}
else
{
$src = imagecreatefromgif($uploadedfile);
}
echo $scr;
list($width,$height)=getimagesize($uploadedfile);
$newwidth=150;
$newheight=($height/$width)*$newwidth;
$tmp=imagecreatetruecolor($newwidth,$newheight);
$newwidth1=50;
$newheight1=($height/$width)*$newwidth1;
$tmp1=imagecreatetruecolor($newwidth1,$newheight1);
imagecopyresampled($tmp,$src,0,0,0,0,$newwidth,$newheight,$width,$height);
imagecopyresampled($tmp1,$src,0,0,0,0,$newwidth1,$newheight1,$width,$height);
$filename = "profile.pic/profile/big" .date('Y-m-d_His - '). $_FILES['file']['name'];
$filename1 = "profile.pic/header/small".date('Y-m-d_His - '). $_FILES['file']['name'];
imagejpeg($tmp,$filename,100);
imagejpeg($tmp1,$filename1,100);
imagedestroy($src);
imagedestroy($tmp);
imagedestroy($tmp1);
}}
}
//If no errors registred, print the success message
if(isset($_POST['Submit']) && !$errors)
{
// mysql_query("update {$prefix}users set img='$big',img_small='$small' where user_id='$user'");
$change=' <div class="msgdiv">Image Uploaded Successfully!</div>';
}
?>
这是表格
<!DOCTYPE HTML PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xml:lang="en" xmlns="http://www.w3.org/1999/xhtml" lang="en"><head>
<meta content="text/html; charset=UTF-8" http-equiv="Content-Type">
<meta content="en-us" http-equiv="Content-Language">
<title>picture demo</title>
<link href="file:///C|/Users/Rameen/Downloads/.css" media="screen, projection" rel="stylesheet" type="text/css">
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.0/jquery.min.js"></script>
<style type="text/css">
.help
{
font-size:11px; color:#006600;
}
body {
color: #000000;
background-color:#999999 ;
background:#999999 url(<?php echo $user_row['img_src']; ?>) fixed repeat top left;
font-family:"Lucida Grande", "Lucida Sans Unicode", Verdana, Arial, Helvetica, sans-serif;
}
.msgdiv{
width:759px;
padding-top:8px;
padding-bottom:8px;
background-color: #fff;
font-weight:bold;
font-size:18px;-moz-border-radius: 6px;-webkit-border-radius: 6px;
}
#container{width:763px;margin:0 auto;padding:3px 0;text-align:left;position:relative; -moz-border-radius: 6px;-webkit-border-radius: 6px; background-color:#FFFFFF;}</style>
</head><body>
<div align="center" id="err">
<?php echo $change; ?> </div>
<div id="space"></div>
<div id="container" >
<div id="con">
<table width="502" cellpadding="0" cellspacing="0" id="main">
<tbody>
<tr>
<td width="500" height="238" valign="top" id="main_right">
<div id="posts">
<img src="<?php echo $filename; ?>" /> <img src="<?php echo $filename1; ?>" />
<form method="post" action="" enctype="multipart/form-data" name="form1">
<table width="500" border="0" align="center" cellpadding="0" cellspacing="0">
<tr><Td style="height:25px"> </Td></tr>
<tr>
<td width="150"><div align="right" class="titles">Picture
: </div></td>
<td width="350" align="left">
<div align="left">
<input size="25" name="file" type="file" style="font-family:Verdana, Arial, Helvetica, sans-serif; font-size:10pt" class="box"/>
</div></td>
</tr>
<tr><Td></Td>
<Td valign="top" height="35px" class="help">Image maximum size <b>400 </b>kb</span></Td>
</tr>
<tr><Td></Td><Td valign="top" height="35px"><input type="submit" id="mybut" value=" Upload " name="Submit"/></Td></tr>
<tr>
<td width="200"> </td>
<td width="200"><table width="200" border="0" cellspacing="0" cellpadding="0">
<tr>
<td width="200" align="center"><div align="left"></div></td>
<td width="100"> </td>
</tr>
</table></td>
</tr>
</table>
</form>
</div>
</td>
</tr>
</tbody>
</table>
</div>
</div>
</body></html>
任何帮助将不胜感激。
答案 0 :(得分：0)
您应该为这两个图片尝试move_uploaded_file功能。在服务器中创建两个sepearte目录并遍历这两个文件并使用move_uploaded_file。
还在数据库中创建两个字段以存储两个路径。并将两部分都放入其中。
答案 1 :(得分：0)
$路径=＆＃39;上传/＆＃39; $ _ FILES [＆＃39;文件＆＃39;] [＆＃39;名称＆＃39;];
move_uploaded_file（$ _ FILES [＆＃39; file＆＃39;] [＆＃39; tmp_name＆＃39;]，$ path）;
mysql_query（＆＃34;插入tablename（文件路径）值（＆＃39; $ path＆＃39;）＆＃34;）;
答案 2 :(得分：0)
问题：我需要在MySQL数据库中保存上传的图像路径
你已经有了目录，所以我认为需要为此编写一个数据库查询。
$yourDir = "path to your directory";
$completePathtoYourFile = $yourDir . "/" . $imageName . ".jpg"; //do not append .jpg if already.
了解如何连接到mysql数据库。 DB CONNECTION
$insert = "INSERT INTO table_name (field1, field2) VALUES ('$completePathtoYourFile', .......) ";
$result = mysqli_query($connectionVariable, $insert);
if(!$result){
echo "Not Inserted..";
}
else{
echo "Saved..";
}
正如其他人所提到的，您需要将上传的文件从临时存储移动到目标路径为 $ yourDir 的目录。