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HomeBackend DevelopmentPHP Tutorialphp新页面无法获取post值

在A.php中有个表单,submit会打开一个新的页面B.php,A网页不关闭。

然后在B.php里$_POST['name'];就是获取不到A.php中的值,但是在A.php中,能够在网页输出$_POST['name']。

求各位大神指教,怎么才能在新打开的网页B.php中取得A中表单的值?


回复讨论(解决方案)

看看你的表单代码

看看你的表单代码


这是A.php中的:
    	<form action='' method='POST' >    		<input type='hidden' value=<?php echo $node->getId() ?> name='nodeIdinfo'>    		<input type="submit" value="AP信息" name='showNdinfo' onclick="manageAp('<?php echo gethostbyname($_SERVER["SERVER_NAME"]); ?>')">    	</form>    </td>  </tr><?php endforeach ?></table><script type="text/javascript" >   function manageAp(url){      window.open("http://"+url+"/phpinfo.php","","fullscreen=1");   }

这是B.php中的:
<?php	if ($_POST['showNdinfo']){		echo $_POST['nodeIdinfo'];	}else {		echo "failed";	}?>

看看你的表单代码


--------------------------------------------------------
现在的情况是:
我点击A中的“AP信息”按钮,能成功在新窗口打开B(即phpinfo.php)网页,但是无法显示POST的值。

如果我把B中代码放到A中,则能在A网页中显示POST的值。

<form action='<?php echo gethostbyname($_SERVER["SERVER_NAME"]); ?>'   method='POST' target=window.open("","","fullscreen=1")>            <input type='hidden' value=<?php echo $node->getId() ?> name='nodeIdinfo'>            <input type="submit" value="AP信息" name='showNdinfo'>        </form>

<form action='<?php echo gethostbyname($_SERVER["SERVER_NAME"]); ?>'   method='POST' target=window.open("","","fullscreen=1")>            <input type='hidden' value=<?php echo $node->getId() ?> name='nodeIdinfo'>            <input type="submit" value="AP信息" name='showNdinfo'>        </form>


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这样的话,B的地址就不对了。


<form action='<?php echo gethostbyname($_SERVER["SERVER_NAME"]); ?>'   method='POST' target=window.open("","","fullscreen=1")>            <input type='hidden' value=<?php echo $node->getId() ?> name='nodeIdinfo'>            <input type="submit" value="AP信息" name='showNdinfo'>        </form>


---------------------------------------------------------------------------------------------------------------------
本来,B的地址是:http://gethostbyname($_SERVER["SERVER_NAME"]); /phpinfo.php

少了个 /phpinfo.php 你自己补上就是了


<form action='<?php echo gethostbyname($_SERVER["SERVER_NAME"]); ?>'   method='POST' target=window.open("","","fullscreen=1")>            <input type='hidden' value=<?php echo $node->getId() ?> name='nodeIdinfo'>            <input type="submit" value="AP信息" name='showNdinfo'>        </form>


---------------------------------------------------------------------------------------------------------------------
我说具体些吧,A地址是:192.168.28.154/node/index;B地址是:192.168.28.154/phpinfo.php
而如果这么改的话,点击之后地址栏跳转到:192.168.28.154/node/192.168.28.154

少了个 /phpinfo.php 你自己补上就是了


-------------------------------------------------------------
嗯,多谢,我懂的还是太少了。

少了个 /phpinfo.php 你自己补上就是了


-------------------------------------------------------
顺便还有个问题请教下,我希望弹出的窗口大小是:300*400的,该怎么写?
window.open第三个参数“height=300, width=400, toolbar=no, menubar=no, scrollbars=no, resizable=no, location=no, status=no”写了之后没反应。
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