对PHP也很新,我试图让php根据添加到页面的ID动态加载文件。我的想法是为多个页面设置一个文件...我有多个页面加载不同的侧边栏:
<?php
if ($PageType == 'about'){
echo "
<img src='http://test.com/images/img.png' alt='' title='' class='img' />
include('../include-images.html');
<a href='#'><img src='test.com/images/one.jpg' alt='' title='' class='img' /></a>
<br class='brclear'/>
";
}
如何在INCLUDE('../include-images.html');
内运行<?php if echo ""
的任何帮助?
答案 0 :(得分:1)
您无法在include
内执行echo
。只需分解echo
:
<?php
if ($PageType == 'about'){
echo "<img src='http://test.com/images/img.png' alt='' title='' class='img' />";
include('../include-images.html');
echo "<a href='#'><img src='test.com/images/one.jpg' alt='' title='' class='img' /></a>
<br class='brclear'/>";
}
答案 1 :(得分:0)
简单地打破回声并包含你需要的HTML,就是这样
xhrFields: { withCredentials: true }
或者更好,我更喜欢
if ($PageType == 'about'){
echo "<img src='http://test.com/images/img.png' alt='' title='' class='img' />";
include('../include-images.html');
echo "<a href='#'><img src='test.com/images/one.jpg' alt='' title='' class='img' /></a>
<br class='brclear'/>
";
}
答案 2 :(得分:0)
当它在字符串文字中时,你不能让PHP运行include
,所以这样做。
echo "<img src='http://test.com/images/img.png' alt='' title='' class='img' />";
include('../include-images.html');
echo "<a href='#'><img src='test.com/images/one.jpg' alt='' title='' class='img' /></a>
<br class='brclear'/>
";
此外,如果您从自己的网站获取图片,请不要使用src=
中的完整域名,只需使用相对路径
echo "<img src='images/img.png' alt='' title='' class='img' />";
include('../include-images.html');
echo "<a href='#'><img src='images/one.jpg' alt='' title='' class='img' /></a>
<br class='brclear'/>
";
然后,如果将代码移动到新域名,则无需更改脚本。
答案 3 :(得分:0)
<? if($PageType == 'about') { ?>
<img src='http://test.com/images/img.png' alt='' title='' class='img' />
<? include('../include-images.html'); ?>
<a href='#'><img src='test.com/images/one.jpg' alt='' title='' class='img' /></a>
<br class='brclear'/>
<? } ?>