用 php 回显图像
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Echo image with php
提问by user1919
I am trying to do something really simple with php but I can not find where I have the mistake! So, I want to echo an image like this (part of the code):
我正在尝试用 php 做一些非常简单的事情，但我找不到我出错的地方！所以，我想回显这样的图像（代码的一部分）：
$file_path[0] = "/Applications/MAMP/htdocs/php_test/image_archive/".$last_file[0];
echo $file_path[0];
echo "<br>";
echo "<img src=\".$file_path[0].\" alt=\"error\">";
I must have some kind of error when I echo the img tag but I can not find it. Any help would be appreciated.
当我回显 img 标签但我找不到它时，我一定有某种错误。任何帮助，将不胜感激。
<div class="content">
<h1> Map</h1>
<?php
include '/Applications/MAMP/htdocs/php_test/web_application_functions/last_file.php';
$last_file = last_file();
$file_path[0] = "/Applications/MAMP/htdocs/php_test/image_archive/".$last_file[0];
echo $file_path[0];
echo "<br>";
echo '<img src="' . $file_path[0] . '" alt="error">';
?>
<!-- end .content --></div>
回答by PassKit
You should use a path either relative to the server root, or the current file.
您应该使用相对于服务器根目录或当前文件的路径。
echo "<img src='/php_test/image_archive/" . $last_file[0] . "' alt='error'>";
回答by Popnoodles
echo "<img src=\".$file_path[0].\" alt=\"error\">";
is wrong
是错的
echo "<img src=\"".$file_path[0]."\" alt=\"error\">";
is what you think you're doing
是你认为你在做什么
It's useful to use single quotes so you don't make this mistake
使用单引号很有用，这样你就不会犯这个错误
echo '<img src="' . $file_path[0] . '" alt="error">';
Secondly
其次
/Applications/MAMP/htdocs/php_test/image_archive/looks like a path not a directory
/Applications/MAMP/htdocs/php_test/image_archive/看起来像路径而不是目录
Perhaps you mean
也许你的意思是
$file_path[0] = "/php_test/image_archive/".$last_file[0];
or
或者
$file_path[0] = "image_archive/".$last_file[0];
