php 在php中上传后如何显示图像？
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How to display image after upload in php?
提问by user3214024
after uploading the image to a folder. how to display the image..
将图像上传到文件夹后。如何显示图像..
its my upload.php
它是我的upload.php
<?php
$allowedExts = array("gif", "jpeg", "jpg", "png");
$temp = explode(".", $_FILES["file"]["name"]);
$extension = end($temp);
if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/jpg")
|| ($_FILES["file"]["type"] == "image/pjpeg")
|| ($_FILES["file"]["type"] == "image/x-png")
|| ($_FILES["file"]["type"] == "image/png"))
&& ($_FILES["file"]["size"] < 20000000000000000000000)
&& in_array($extension, $allowedExts))
{
if ($_FILES["file"]["error"] > 0)
{
echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
}
else
{
echo "Upload: " . $_FILES["file"]["name"] . "<br>";
echo "Type: " . $_FILES["file"]["type"] . "<br>";
echo "Size: " . ($_FILES["file"]["size"] / 1024) . " kB<br>";
echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br>";
if (file_exists("upload/" . $_FILES["file"]["name"]))
{
echo $_FILES["file"]["name"] . " already exists. ";
}
else
{
move_uploaded_file($_FILES["file"]["tmp_name"],
"upload/" . $_FILES["file"]["name"]);
echo "Stored in: " . "upload/" . $_FILES["file"]["name"]."<br>";
$image=$_FILES["file"]["name"]; /* Displaying Image*/
$img="upload/".$image;
echo '<img src= "upload/".$img>';
}
}
}
else
{
echo "Invalid file";
}
?>
the image is uploading fine. but the image is not displaying. a small box only showing.
图片上传正常。但图像没有显示。一个只展示的小盒子。
this part not working...
这部分不起作用...
/* Displaying Image*/
$image=$_FILES["file"]["name"];
$img="upload/".$image;
echo '<img src= "upload/".$img>';
how to display the image after upload success?
上传成功后如何显示图片？
回答by Manu
Try this:
尝试这个：
header('Content-Type: image/jpeg');
readfile($imgPath);
回答by Mario Radomanana
As you've already put "upload" in $img
因为您已经在 $img 中放置了“上传”
$img="upload/".$image;
You don't need to put it in src anymore
你不再需要把它放在 src 中
echo '<img src= "'.$img.'">';
回答by php_learner
I have tried your code and its working finely just need to change the line
我已经试过你的代码，它工作得很好，只需要改变行
echo'<img src= "upload/".$img>';
to
到
echo'<img src="'.$img.'">';
回答by Alex94
Don't put $img in quotations. When you do it the output is simply $img, not upload/...
不要将 $img 放在引号中。当你这样做时，输出只是 $img，而不是上传/...
echo < img src=" ' ,$img ,' ">;
回声 < img src=" ' ,$img ,' ">;
回答by putvande
You mixed up some quotes. It should be :
你混淆了一些报价。它应该是 ：
$image = $_FILES["file"]["name"];
$img = "upload/".$image;
echo "<img src=\"upload/$img\">";
回答by ElendilTheTall
You need:
你需要：
echo '<img src="upload/'.$img.'"/>';
echo '<img src="upload/'.$img.'"/>';
回答by Varun Pradhan
We have to Put enctype = "multipart/form-data"inside the form
我们必须把enctype = "multipart/form-data"表格放在里面
<html> <body> <form action = "" method = "POST" enctype = "multipart/form-data"> <input type = "file" name = "image" /> <input type = "submit"/> <ul> <li>Sent file: echo $_FILES['image']['name']; <li>File size: echo $_FILES['image']['size']; <li>File type: echo $_FILES['image']['type'] <li><img src=" echo 'images/'.$file_name; "> </ul> </form> </body> </html>
<html> <body> <form action = "" method = "POST" enctype = "multipart/form-data"> <input type = "file" name = "image" /> <input type = "submit"/> <ul> <li>Sent file: echo $_FILES['image']['name']; <li>File size: echo $_FILES['image']['size']; <li>File type: echo $_FILES['image']['type'] <li><img src=" echo 'images/'.$file_name; "> </ul> </form> </body> </html>
We Have
我们有
//file_upload.php if(isset($_FILES['image'])){$errors= array(); $file_name = $_FILES['image']['name']; $file_size = $_FILES['image']['size']; $file_tmp = $_FILES['image']['tmp_name']; $file_type = $_FILES['image']['type']; $file_ext=strtolower(end(explode('.',$_FILES['image']['name']))); $extensions= array("jpeg","jpg","png"); if(in_array($file_ext,$extensions)=== false){ $errors[]="extension not allowed, please choose a JPEG or PNG file."; } if($file_size > 2097152) { $errors[]='File size must be excately 2 MB'; } if(empty($errors)==true) { move_uploaded_file($file_tmp,"images/".$file_name); echo "Success"; }else{ print_r($errors); } } ?>
//file_upload.php if(isset($_FILES['image'])){$errors= array(); $file_name = $_FILES['image']['name']; $file_size = $_FILES['image']['size']; $file_tmp = $_FILES['image']['tmp_name']; $file_type = $_FILES['image']['type']; $file_ext=strtolower(end(explode('.',$_FILES['image']['name']))); $extensions= array("jpeg","jpg","png"); if(in_array($file_ext,$extensions)=== false){ $errors[]="extension not allowed, please choose a JPEG or PNG file."; } if($file_size > 2097152) { $errors[]='File size must be excately 2 MB'; } if(empty($errors)==true) { move_uploaded_file($file_tmp,"images/".$file_name); echo "Success"; }else{ print_r($errors); } } ?>
回答by Yagnik joshi
Just do this
就这样做
echo '<img src="upload/'.$file_name.' "/>
We have to take the $file_namevariable because..
""(double quotes) can only return text in the variable - it cannot return binary number to show image!
我们必须取$file_name变量，因为..
""（双引号）只能返回变量中的文本——它不能返回二进制数来显示图像！
