2016-11-04 41 views
0

我試圖顯示上傳的圖像,它的網址後,它已被處理使用此代碼,但我有點卡住我將如何實現這一點,而不是頁面返回與「完成...」如何在php上傳後顯示生成的圖像?

http://llngg6czd-site.1tempurl.com/tester/index.php

的index.php

<!DOCTYPE html> 
<html> 
<head> 
<title>Upload Files using normal form and PHP</title> 
</head> 
<body> 
<form enctype="multipart/form-data" method="post" action="upload_image.php"> 
<div class="row"> 
<label for="image">Select a File to Upload</label><br /> 
<input type="file" name="image" /> 
</div> 
<div class="row"> 
<input type="submit" value="Upload" /> 
</div> 
</form> 
</body> 
</html> 

image_upload.php

<?php 
require_once('ImageManipulator.php'); 
if ($_FILES['image']['error'] > 0) { 
echo "Error: " . $_FILES['image']['error'] . "<br />"; 
} else { 
// array of valid extensions 
$validExtensions = array('.jpg', '.jpeg', '.gif', '.png'); 
// get extension of the uploaded file 
$fileExtension = strrchr($_FILES['image']['name'], "."); 
// check if file Extension is on the list of allowed ones 
if (in_array($fileExtension, $validExtensions)) { 
$newNamePrefix = time() . '_'; 
$manipulator = new ImageManipulator($_FILES['image']['tmp_name']); 
// resizing to 200x200 
$newImage = $manipulator->resample(200, 200); 
// saving file to uploads folder 
$manipulator->save('uploads/' . $newNamePrefix . $_FILES['image']['name']); 
echo 'Done ...'; 
} else { 
echo 'You must upload an image...'; 
} 
} 

Imagemanipulator.php

https://gist.github.com/philBrown/880506 

任何幫助將不勝感激。

回答

0

您將文件保存爲

'uploads/' . $newNamePrefix . $_FILES['image']['name'] 

因此,你可以簡單地通過回饋的IMG-標籤與 'SRC' 屬性表現出來:

echo '<img src="./uploads/' . $newNamePrefix . $_FILES['image']['name'] . '">';