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php中的Window.location-javascript_Javascript_Php_Popup - Fatal编程技术网

php中的Window.location-javascript

php中的Window.location-javascript,javascript,php,popup,Javascript,Php,Popup,我在php程序中创建了一段代码: $dest = 'https://www.google.com'; $stay = 'currenturl'; echo '<pre>'; echo "<script language=\"javascript\">"; echo " window.location = '". $stay."';"; echo " var win = null

我在php程序中创建了一段代码:

$dest = 'https://www.google.com';

$stay = 'currenturl';
          echo '<pre>'; 
          echo "<script language=\"javascript\">";
          echo "    window.location = '". $stay."';";
          echo "    var win = null;";
          echo "    LeftPosition = (screen.width) ? (screen.width-995)/2 : 0;";
          echo "    TopPosition = (screen.height) ? (screen.height-640)/2 : 0;";
          echo "    settings = 'height=500,width=600,top='+TopPosition+',left='+LeftPosition+',scrollbars=no,resizable=yes,status=yes';";
          echo "    win = window.open('" . $dest . "','network',settings);";
          echo "</script>"; 
          echo '</pre>';    
          exit; 
$dest='1!'https://www.google.com';
$stay='currenturl';
回声';
回声“;
回显“window.location=”$留下来;
echo“var win=null;”;
回声“LeftPosition=(屏幕宽度)?(屏幕宽度-995)/2:0;”;
回声“TopPosition=(屏幕高度)?(屏幕高度-640)/2:0;”;
echo“设置='高度=500,宽度=600,顶部='+TopPosition+',左侧='+LeftPosition+',滚动条=否,可调整大小=是,状态=是';”;
echo“win=window.open”(“$dest.”、“network”、settings);”;
回声“;
回声';
出口
如果我运行这段代码,它将打开我的弹出窗口,即google,并刷新当前页面(因为这是我唯一的想法,允许我停留在同一页面上,否则它将变为空白…)。我想避免刷新,保持在同一页上


你知道如何更改这个“window.location”来实现它吗?

伙计,你不应该使用php代码通过注入js脚本来打开一个新窗口,这就是为什么当php代码将新内容输出到当前页面时,你要转到另一个页面的原因,你应该在JavaScript函数中拥有关于新窗口的所有逻辑。那么你应该保持原样,你能帮我吗?链接还是smth?我在javascript中是绿色的