点击链接时需要显示弹出页面

Question

//function
function maps(url) {    
   window.open(url, "_blank", "toolbar=yes, scrollbars=yes, resizable=yes, top=500, left=500,width=400, height=400");
}

php page

 $display.= "<td><a  href='#' onclick='maps('google.com')'>"visit "</a></td>";

I have checked the function without passing url variable and it works but when i used url varibale its not working. I want this function to use everywhere.

help me

Answer 1

try this once..

change

onclick='maps('google.com')'

to

onclick='maps(\'google.com\')'

Answer 2

There is problem in concatenate the url string inside function. Try to give it inside \\" .

And give absolute url as parameter.

$display = "<td><a href='#' onclick='maps(\"http://www.google.com\");'>Visit</a></td>";

Answer 3

您以错误的方式连接字符串。使用下面的代码将起作用

$display= "<td><a  href='#' onclick=\"maps('google.com')\">visit</a></td>";