PHP内部的HTML不能使PHP功能工作 [英] Can't make PHP function work inside HTML inside PHP
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问题描述
我写了这段代码,它从一个链接中得到一个图像,它根据你的位置而变化:
I wrote this code, it gets an image from a link that varies according to where you are:
<img src='http://chusmix.com/Imagenes/grupos/<?php echo substr(get_search_query(), 1); ?>.jpg'>
如果PHP条件证明是正确的,我想让代码运行,但是我无法让它工作。看来该函数不会返回值,而是会以文本方式进行链接。我的意思是字面上 http://chusmix.com/Imagenes/grupos/ .jpg。但是代码本身是正确的。
I want to make that code run if a PHP condition proves true, but I cannot make it work. It seems that the function doesn't return a value instead it takes the link textually. I mean it goes to http://chusmix.com/Imagenes/grupos/.jpg literally. However the code works correctly by itself.
这是PHP代码:
This is the PHP code:
<?php
$search=get_search_query();
$first=$search[0];
if ($first=="#"){
echo "<html>";
echo "<img src='http://chusmix.com/Imagenes/grupos/<?php echo substr(get_search_query(), 1); ?>.jpg'>";
}
?>
推荐答案
您已经在php标签中。因此,不需要<?php
和?>
。
You are already inside the php tag. So there is no need for <?php
and ?>
.
Try:
Try:
echo "<img src='http://chusmix.com/Imagenes/grupos/".substr($search,1).".jpg'>";
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