将值传递到php中的下一页 [英] Pass a value to the next page in php

问题描述

任何人都可以帮助我...

我已经创建了此代码以传递值，但它仍返回0到下一页...我错过了一些代码还是我写错了代码....

在第一页中，我将值一个传递给了下一页.

anybody can help me please...

i had create this code to pass the value but it still return 0 to the next page...did i miss some code or i wrote the code wrong....

in the first page i had pass value one to next page..

<a href="kenal2.php?count=1"></a>

在第二页中，我检索了值一，并在值中加上了1 ...

在第二页的首页中.

in the second page i had retrieve the value one and i had add the value with 1...

in the top page of the second page..

        if (isset($_GET[''count'']))<br />
            $count = $_GET[''count''];<br />
        else<br />
            $count = 0;<br />
?>

在第二页底部...我已经使用该值进行了一些计算

at the bottom of the second page...i had done some calculation with the value

<?php echo $count;<br mode="hold" /?>      $sum = $count + 1; 
     ?>

所以我想将$ sum值传递给第三页...

so i want to pass the $sum value to the third page...

<a href="kenal3.php?count = $sum"></a>

在第三页的顶部...

in the top of third page...

        if (isset($_GET[''sum'';]))
            $sum = $_GET[''sum''];
        else
            $sum = 0;
?>

当我尝试显示值时...应该显示2但显示的值为0 ..似乎无法从第二页检索值...
为什么会这样...

when i try to show the value...it should come out with 2 but value 0 that had been show..seem that it could not retrieve the value from second page...
why this happen...

推荐答案

_GET [''count''])))< br/>

_GET[''count'']))<br />

count =

_GET [''count''];< br/> else< br/>

_GET[''count''];<br /> else<br />

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