如何使用 PHP 打印功能打印图像
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How To Print Images with PHP print function
提问by dagogodboss
Please help me out. I have a form that collects picture from an upload field in a client form and the picture to be printed out on the user page. this work perfectly well on localhost and it looks like the below Ilamini Ayebatonye Thanks A lot
请帮帮我。我有一个表单，它从客户端表单的上传字段中收集图片，并将图片打印在用户页面上。这在本地主机上工作得很好，它看起来像下面的 Ilamini Ayebatonye 非常感谢
NAME: Name Surname
EMAIL ADDRESS: [email protected]
HOME ADDRESS: 123 Random Street, FakeTown
PHONE NUMBER: 012345678911
QUALIFICATION: I.C.T Certificate
SECTOR: CCNA
INTEREST: please put in your interest
Please Print this slip, attached your C.V and photocopies of your credentials and come with it to the day of the seminar.
请打印此单据，附上您的简历和证书复印件，并在研讨会当天随身携带。
but the problem with is that when I uploaded the site and try registering the picture won't show again,
但问题是当我上传网站并尝试注册时，图片不会再次显示，
what i see is the below
我看到的是下面
ECO9JA-CREATING JOBS IN ICT
Backgrounds_15670_zps11bc080d.png
NAME: Name Surname
EMAIL ADDRESS: [email protected]
HOME ADDRESS: 123 Random Street, FakeTown
PHONE NUMBER: 012345678911
QUALIFICATION: I.C.T Certificate
SECTOR: CCNA
INTEREST: please put in your interest
Please Print this slip, attached your C.V and photocopies of your credentials and come with it to the day of the seminar.
请打印此单据，附上您的简历和证书复印件，并在研讨会当天随身携带。
THANKS FOR REGISTERING
感谢注册
the script is below for the image variable the form upload and the print fxn are as follows:
表单上传和打印 fxn 的图像变量的脚本如下：
VARIABLE
多变的
<?php
$file= "file.csv";
$name = $_POST['name'];
$email = $_POST['email'];
$address = $_POST['address'];
$phone = $_POST['phone'];
$qualification = $_POST['qualification'];
$sector = $_POST['sector'];
$interest = $_POST['interest'];
$image= $_POST['image'];
<input type="hidden" name="MAX_FILE_SIZE" value="500">
<P>
<br>
<input type="file" name="image" value="upload" >
PRINT THIS CODE ON USERPAGE
在用户页面上打印此代码
<?php print"<img src=\"$image\" width=\"100px\" height=\"100px\"\/>";?>
回答by Aklesh Singh
Use this just copy and paste in your project.
使用它只需复制并粘贴到您的项目中。
$contents= file_get_contents('http://localhost/elgg_social/data/36/641483447202.jpg');
$expires = 14 * 60*60*24;
header("Content-Type: image/jpeg");
header("Content-Length: " . strlen($contents));
header("Cache-Control: public", true);
header("Pragma: public", true);
header('Expires: ' . gmdate('D, d M Y H:i:s', time() + $expires) . ' GMT', true);
echo $contents;
exit;
回答by mickzer
The problem must lie when you are posting the path for the image file. I have tested the following php with an image "test.png" and it works.
当您发布图像文件的路径时，问题一定出在。我已经使用图像“test.png”测试了以下 php 并且它有效。
<?php
$image="test.png";
print"<img src=\"$image\" width=\"100px\" height=\"100px\"\/>";
?>
回答by Mubashir
Looks like there is problem with your image attributes, remove "px". it should be like this
看起来您的图像属性有问题，请删除“px”。应该是这样的
<?php print"<img src=\"$image\" width=\"100\" height=\"100\"\/>";?>
回答by Vignesh Pichamani
Based on Uploading your photo
基于上传你的照片
<input type="file" name="image" value="upload" >
It fetches the file path itself. you could move the uploaded file to your folder and the corresponding file path is to save in db.
它获取文件路径本身。您可以将上传的文件移动到您的文件夹中，相应的文件路径保存在db中。
If you are uploading the path as looks like
如果你上传的路径看起来像
C:\Users\Public\Pictures\Sample Pictures\Desert.jpg
If you are posted as $image = $post['image']
如果您被发布为 $image = $post['image']
Echo the path it returns to error or null
回显它返回错误或空的路径
<img src="C:\Users\Public\Pictures\Sample Pictures\Desert.jpg"/>
If sometimes works but most time won't works as i feel this type of problem.
如果有时有效但大多数时间不起作用，因为我觉得这种问题。
Thanks.
谢谢。
回答by syd619
Based on the reference I commented on your php script you should have a code similar to this:
根据我对你的 php 脚本的评论，你应该有一个类似于这样的代码：
$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"))
&& ($_FILES["file"]["size"] < 20000)
&& in_array($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"];
}
}
}
else
{
echo "Invalid file";
}
Take a close look line by line.
You can set the maximum file size for your image upload at: $_FILES["file"]["size"] < 20000
一行一行地仔细观察。您可以在以下位置设置图像上传的最大文件大小：$_FILES["file"]["size"] < 20000
Furthermore this will code will make sure that the user is uploading an image and not something else.
此外，这将代码将确保用户正在上传图像而不是其他内容。
Finally based on your need and this script after successful upload
最后根据你的需要和成功上传后的这个脚本
move_uploaded_file($_FILES["file"]["tmp_name"],
"upload/" . $_FILES["file"]["name"]);
echo "Stored in: " . "upload/" . $_FILES["file"]["name"];
Then you could do this to show the image:
然后你可以这样做来显示图像：
echo '<img src="upload/'.$_FILES["file"]["name"].'">';
回答by Siddharth Shukla
header('Content-Type: image/jpeg');
print $response;
