I'm trying to display images by taking their file path from an sql table, but i'm having a lot of troubles.
我试图通过从sql表中获取他们的文件路径来显示图像，但是我遇到了很多麻烦。
Here is whats going on:
这里有个问题:
$image is a variable containing the text "itemimg/hyuna.png" which is path to an image.
$image是一个包含“itemimg/hyuna”文本的变量。png是图像的路径。
$image = 'itemimg/hyuna.png';
I assumed I would be able to display the image outside of the php block like so:
我假设我可以在php块外面显示图像，如下所示:
<img src= "<? $image ?>" alt="test"/>
This doesn't work though for some reason.
但出于某种原因，这行不通。
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:
所以我认为它可能无法在php块之外读取变量(我是初学者)，所以为了测试我做了:
<h1> "<? $image ?>" </h1>
It displays itemimg/hyuna.png as a h1 banner.
它显示itemimg / hyuna。作为h1的横幅。
Meaning it's accessing the varible fine.
这意味着它可以很好地访问变量。
So I thought maybe the path is wrong. So I tried:
所以我认为这条路可能是错的。所以我试着:
<img src= "itemimg/hyuna.png" alt="test"/>
This displays the image perfectly.
这将完美地显示图像。
So now I'm stuck scratching my head why the first bit of code displays nothing but the text "test" from "alt="
所以现在我一直在挠头为什么第一段代码只显示了alt=的文本"test"
Extra question: How do I go about assigning a value from an sql cell to a variable? I attempted the following with no luck:
额外问题:如何将sql单元格中的值分配给变量?我不走运地尝试了以下方法:
$q = "select * from item where id=$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
项目是一个包含collumn: image的表，其中包含到image的文件路径
4 个解决方案
#1
6
First of all, you should not use PHP Shorttags.
首先，您不应该使用PHP Shorttags。
When you use the PHP Shorttags you have to say:
当你使用PHP时，你必须说:
<img src="<?=$image ?>" alt="test" />
But i would encourage to escape the Content off the variable like this:
但是我鼓励大家把内容从变量中去掉
<img src="<?php echo htmlspecialchars($image); ?>" alt="test" />
Your extra question:
This should lead to an syntax error because the string could not be parsed, just use $image = $row['image'];
这将导致语法错误，因为字符串无法解析，只需使用$image = $row['image'];
#2
3
Try this
试试这个
<img src= "<?php echo $image ?>" alt="test"/>
< img src = " < ?php echo $image ?>" alt="test"/>
#3
2
try
试一试
<img src= "<?= $image ?>" alt="test"/>
or
或
<img src= "<? echo $image; ?>" alt="test"/>
#4
-1
try this
试试这个
<img src= "<?php echo $image ?>" alt="test"/>
#1
6
First of all, you should not use PHP Shorttags.
首先，您不应该使用PHP Shorttags。
When you use the PHP Shorttags you have to say:
当你使用PHP时，你必须说:
<img src="<?=$image ?>" alt="test" />
But i would encourage to escape the Content off the variable like this:
但是我鼓励大家把内容从变量中去掉
<img src="<?php echo htmlspecialchars($image); ?>" alt="test" />
Your extra question:
This should lead to an syntax error because the string could not be parsed, just use $image = $row['image'];
这将导致语法错误，因为字符串无法解析，只需使用$image = $row['image'];
#2
3
Try this
试试这个
<img src= "<?php echo $image ?>" alt="test"/>
< img src = " < ?php echo $image ?>" alt="test"/>
#3
2
try
试一试
<img src= "<?= $image ?>" alt="test"/>
or
或
<img src= "<? echo $image; ?>" alt="test"/>
#4
-1
try this
试试这个
<img src= "<?php echo $image ?>" alt="test"/>