php 如何正确使用 content-type:image/jpeg？
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how to use the content-type:image/jpeg properly?
提问by user1645213
I've created a page "student_picture.php" using the content-type:image/jpeg. im having problem when I want to add other text in that same page..
我使用 content-type:image/jpeg 创建了一个页面“student_picture.php”。当我想在同一页面中添加其他文本时遇到问题..
here is the sample of my code:
这是我的代码示例：
<?php
session_start();
if(!isset($_SESSION["StudentNo"])){
header("location:login.php");
}
$StudentNo = $_SESSION['StudentNo'];
require("includes/connection.php");
$sql = "SELECT StudentPicture from dbo.Students where StudentNo = '$StudentNo'";
$stmt = sqlsrv_query( $conn, $sql );
if( $stmt === false) {
die( print_r( sqlsrv_errors(), true) );
}
while( $row = sqlsrv_fetch_array( $stmt, SQLSRV_FETCH_ASSOC) ) {
$img = $row['StudentPicture'];
if ($img == null ) {
echo "<img src='img/default_pic.gif'>";
} else {
$img = trim($img);
header('Content-Type: image/jpeg');
echo $img;
}
echo $StudentNo;
}
?>
The image is successfully displaying but the echo $StudentNo is not displaying.. can anyone help me with my prob? thanks in advance.
图像已成功显示，但 echo $StudentNo 未显示.. 任何人都可以帮助我解决问题吗？提前致谢。
回答by Kliptu
You might need to use following code :
您可能需要使用以下代码：
<?php
$image = 'http://www.example.com/image.jpg';
$info = getimagesize($image);
header('Content-Type: '.$info['mime']);
echo file_get_contents($image);
exit;
?>
回答by Lucas
I think that you cannot display text while using the Content-Typeof image/jpeg, or any other kind of image format that I know of. Text can only be displayed with text/?, or other exceptions such as application/pdf.
我认为您无法在使用Content-Typeofimage/jpeg或我所知道的任何其他类型的图像格式时显示文本。文本只能显示为text/?，或其他例外，例如application/pdf。
If you don't know how to display the image on a separate page while using a php file, just use:
如果您不知道如何在使用 php 文件时在单独的页面上显示图像，只需使用：
<img src="path/to/yourphpfile.php" />
Just like any other image.
就像任何其他图像一样。
Hope that helped.
希望有所帮助。
回答by Mihai Iorga
You have to exityour script, because the headeris sent to the client browser. If the $_SESSION["StudentNo"]is not present the script will process and will try to ouput your image.
你必须使用exit你的脚本，因为它header被发送到客户端浏览器。如果$_SESSION["StudentNo"]不存在，脚本将处理并尝试输出您的图像。
<?php
session_start();
if(!isset($_SESSION["StudentNo"])){
header("location:login.php");
die();
}
And that is not a proper way to output an image, echo "<img src='img/default_pic.gif'>";with header('Content-Type: image/jpeg');is not ok, either you return without that header or you read the image:
这是不以输出图像取之有道，echo "<img src='img/default_pic.gif'>";用header('Content-Type: image/jpeg');也不行，要么你不回这个头，或者你阅读的图像：
if ($img == null){
$img = 'img/default_pic.gif';
} else {
$img = trim($img);
}
header('Content-Type: image/jpeg');
readfile($img);
or you lose the header tag
或者你失去了标题标签
if ($img == null){
$img = "<img src='img/default_pic.gif'>";
} else {
$img = "<img src='".trim($img)."'>";
}
echo $img;
So your script could be:
所以你的脚本可能是：
<?php
session_start();
if(!isset($_SESSION["StudentNo"])){
header("location:login.php");
die();
}
$StudentNo = $_SESSION['StudentNo'];
require("includes/connection.php");
$sql = "SELECT StudentPicture from dbo.Students where StudentNo = '$StudentNo'";
$stmt = sqlsrv_query( $conn, $sql );
if( $stmt === false) {
die( print_r( sqlsrv_errors(), true) );
}
while( $row = sqlsrv_fetch_array( $stmt, SQLSRV_FETCH_ASSOC) ) {
$img = $row['StudentPicture'];
if ($img == null){
$img = "<img src='img/default_pic.gif'>";
} else {
$img = "<img src='".trim($img)."'>";
}
echo $img;
echo $StudentNo;
}
?>
I don't know the functions inside your DB connection, but I think you should use something like $row = sqlsrv_fetch_row( $stmt, SQLSRV_FETCH_ASSOC);and make without that while
我不知道您的数据库连接中的功能，但我认为您应该使用类似的东西$row = sqlsrv_fetch_row( $stmt, SQLSRV_FETCH_ASSOC);并在没有它的情况下制作while
回答by Mike
You are outputting an image when it is found in the database, you would either need to include a dynamic img via
当在数据库中找到图像时，您正在输出图像，您需要通过以下方式包含动态 img
<img src="anotherscript.php">
Or generate a jpg file using one of the built in image libraries and storing its file name in that database rather than storing the raw data
或者使用内置图像库之一生成 jpg 文件并将其文件名存储在该数据库中，而不是存储原始数据
