將圖像上傳到服務器
[英]upload image into server
問題描述
我想知道如何將圖像上傳到服務器目錄？ 有人可以教我嗎，因為我也想了解更多有關這件事的信息。 這是因為我對這種編碼不太熟悉。 我試圖將圖像上傳到數據庫，並且可以正常工作，但是圖像沒有發送到服務器。 僅將文件名插入數據庫。 請幫幫我 ！ 謝謝
<?php
include("db.php");
$id = $_POST['id'];
$bg = trim($_POST['com_bg']);
$ben = trim($_POST['com_benefit']);
$rem = trim($_POST['com_remark']);
//print_r($_FILES);
//echo "<br><br>";
$name = $_FILES['file']['name'];
$format = substr($name, strpos($name, '.'), strlen($name) - 1);
//if they DID upload a file...
if($_FILES['file']['name'])
{
//Check if the file is JPEG,PNG or GIF image
if(($_FILES["file"]["type"] == "image/jpeg")||($_FILES["file"]["type"] == "image/png")||($_FILES["file"]["type"] == "image/gif"))
{
$newname = dirname(__FILE__).'/upload/'.$name;
//Check if the file with the same name is already exists on the server
if (!file_exists($newname))
{
//move it to where we want it to be
move_uploaded_file($_FILES["file"]["tmp_name"],"upload/" . $_FILES["file"]["name"]);
$sql= "UPDATE company SET com_logo='$name' WHERE com_idx = '$id'";
$result = mysql_query($sql);
}
else
{
header("Location: profile_2.php?note=1");
//echo "Error: File ".$_FILES["file"]["name"]." already exists";
}
}
else
{
header("Location: profile_2.php?note=2");
//echo 'wrong extension ';
}
}
$sql= "UPDATE company SET com_intro='$bg',com_description='',com_benefit='$ben',com_remark='$rem' WHERE com_idx = '$id'";
$result = mysql_query($sql);
header("Location: profile_2.php");
?>
2 個解決方案
解決方案1
0 2013-04-15 06:29:21
首先，您應該在php.ini文件中檢查這些配置
file_uploads
upload_max_filesize
max_input_time
memory_limit
max_execution_time
post_max_size
其次，確保yopu要上傳文件的文件夾必須存在並且應該具有適當的權限
建議避免使用mysql_ *函數學習PDO或mysqli
解決方案2
0 2013-04-21 15:01:20
試試這個例子
<header>
<?php
function UploadOne($fname)
{
$uploaddir = 'uploadedfiles/';
if (is_uploaded_file($fname['tmp_name']))
{
$filname = basename($fname['name']);
$uploadfile = $uploaddir . basename($fname['name']);
if (move_uploaded_file ($fname['tmp_name'], $uploadfile))
$res = "File " . $filname . " was successfully uploaded and stored.<br>";
else
$res = "Could not move ".$fname['tmp_name']." to ".$uploadfile."<br>";
}
else
$res = "File ".$fname['name']." failed to upload.";
return ($res);
}
?>
</header>
<body>
<?php
if ($_FILES['picture']['name'] != "")
{
$res = UploadOne($_FILES['picture']);
$filname = $_FILES['picture']['name'];
echo ($res);
}
?>
<h1>UPLOADING FILES</h1>
<form name="fupload" enctype="multipart/form-data" action="upfiles.php" method="post">
<input type="file" name="picture" />
<input type="submit" value="Submit" />
</form>
</body>
將圖像上傳到mysql服務器
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