来个高手啊,一个困扰我很久的php问题,求帮助
两个php文件,一个为a.php,代码是
1 |
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另一个b.php,代码是
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网页输出图片路径是
求助把b.php文件的代码合并到a.php,网页输出用一个变量来代替,如

回复讨论(解决方案)
不明白你的意思!
a.php 只是定义了一个函数,b.php 中并没有调用它
不明白你的意思!
a.php 只是定义了一个函数,b.php 中并没有调用它



贴全相关代码
贴全相关代码
你好,谢谢你的回复,htm网页的代码是这样的
用于discuz列表页获取帖子图片的
贴全相关代码
版主啊,求帮忙 不知道你想表达什么
如果你是想让b.php 这段代码 能在a.php里运行 ,那就加在
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这段代码里,定义一个$array[‘tupian’] = (b.php 返回的值)
这样子,return $array 的时候,在html里,解析一下,就可以了
不知道你想表达什么
如果你是想让b.php 这段代码 能在a.php里运行 ,那就加在
1 |
|
这段代码里,定义一个$array[‘tupian’] = (b.php 返回的值)
这样子,return $array 的时候,在html里,解析一下,就可以了 我想表达的是能不能用变量$tupian代替b.php,然后html里解析直接是


不知道你想表达什么
如果你是想让b.php 这段代码 能在a.php里运行 ,那就加在
1 |
|
这段代码里,定义一个$array[‘tupian’] = (b.php 返回的值)
这样子,return $array 的时候,在html里,解析一下,就可以了 还有就是您说的这个代码具体怎么加,请贴一下代码,我php很初级,谢谢
dz好像有个
提问时最好表达简明准确,你的目的是什么,你碰到的问题是什么
dz好像有个
提问时最好表达简明准确,你的目的是什么,你碰到的问题是什么
谢了,大概就是你写的这个代码的意思,你的这个代码怎么加啊,
while ($rw=DB::fetch($rs)){
$array[]=$rw;
$tupian = b.php?aid=$o[aid];
}
我这样加直接就报错了
dz好像有个
提问时最好表达简明准确,你的目的是什么,你碰到的问题是什么
谢了,大概就是你写的这个代码的意思,你的这个代码怎么加啊,
while ($rw=DB::fetch($rs)){
$array[]=$rw;
$tupian = b.php?aid=$o[aid];
}
我这样加直接就报错了
我写的是语法不标准的示例,准确的是
$tupian = 'b.php?aid=' . $rw['aid'];
字符串要加引号

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