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 $node->getId() ?> name='nodeIdinfo'> <input type="submit" value="AP信息" name='showNdinfo'> </form>
---------------------------------------------------------------------------------------------------------------------
这样的话,B的地址就不对了。
<form action='<?php echo gethostbyname($_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

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