使用php变量显示图像 [英] Displaying an image using a php variable
问题描述
我试图通过从sql表中获取文件路径来显示图像,但是我遇到了很多麻烦。
这是什么:
$ image 是一个包含文本
itemimg / hyuna.png 这是图像的路径。
$ image ='itemimg / hyuna.png';
我假设我能够在php块之外显示图像,如下所示:
< img src =< ;? $ image?> ALT = 测试/>
尽管出于某种原因,这并不起作用。
所以我想也许它不能读取php块外的变量(我是一个初学者),所以为了测试我做了:
< H1> <?$ image?> < / H1>
它将 itemimg / hyuna.png
显示为一个h1横幅。
这意味着它正在访问变数罚款。
所以我想也许路径是错误的。所以我试过了:
< img src =itemimg / hyuna.pngalt =test/>
完美显示图片。
所以现在我不知道为什么第一个代码只显示 额外问题: 首先,不应该使用PHP Shorttags。 当您使用PHP Shorttags时,您必须说: 但是我会鼓励将内容从这个变量中跳出: 导致语法错误,因为字符串无法解析,只需使用 I'm trying to display images by taking their file path from an sql table, but i'm having a lot of troubles. Here is whats going on: I assumed I would be able to display the image outside of the php block like so: This doesn't work though for some reason. So I thought maybe it's not able to read the variable outside the php block(i'm a beginner), so for testing i did: It displays Meaning it's accessing the varible fine. So I thought maybe the path is wrong. So I tried: This displays the image perfectly. So now I'm stuck scratching my head why the first bit of code displays nothing but the text Extra question:
How do I go about assigning a value from an sql cell to a variable?
I attempted the following with no luck: First of all, you should not use PHP Shorttags. When you use the PHP Shorttags you have to say: But i would encourage to escape the Content off the variable like this:
This should lead to an syntax error because the string could not be parsed, just use 这篇关于使用php变量显示图像的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!test
中的文本alt =$
如何从sql单元格向变量赋值?
我尝试了以下方法,但没有运气:
$ q =select * from item where id = $ id ;
$ results = mysql_query($ q);
$ row = mysql_fetch_array($ results,MYSQL_ASSOC);
$ image =。$ row ['image']。;
项目
是一个带有列表的表格: image
其中包含图像的文件路径
< img src =<?= $ image?> alt =test/>
< img src =<?php echo htmlspecialchars($ image);?> alt =test/>
您的额外问题:
$ image = $ row ['image'];
$image
is a variable containing the text "itemimg/hyuna.png"
which is path to an image.$image = 'itemimg/hyuna.png';
<img src= "<? $image ?>" alt="test"/>
<h1> "<? $image ?>" </h1>
itemimg/hyuna.png
as a h1 banner.<img src= "itemimg/hyuna.png" alt="test"/>
"test"
from "alt="
$q = "select * from item where id=$id";
$results = mysql_query($q);
$row = mysql_fetch_array($results, MYSQL_ASSOC);
$image = ".$row['image'].";
item
is a table with a collumn: image
which contains file paths to images<img src="<?=$image ?>" alt="test" />
<img src="<?php echo htmlspecialchars($image); ?>" alt="test" />
Your extra question:
$image = $row['image'];