从php程序的弹出窗口中调用另一个php程序

Question

I am trying to call another php program in a pop up window within a php program.

I want to click on the word allowance and bring up a window with the allowance details. The lookallow program is working great but need to call it from a click on the word allowance.

This code works and opens another window and closes the first. I want a pop up window to overlay the current window.

echo "<td> <a href='lookallow.php?ordr=$ORDER &bord=$BORD'> Allowance </a> </td>";

I tried this to get the popup windso but keep getting syntax errors when trying to do the popup window. Could someone please show me the correct syntax to do the pop up window?

echo "<td> <a href='#' onclick='window.open("lookallow.php?ordr=$ORDER &bord=$BORD","","width=200,height=100");'>Allowance</td>";

Much Thanks for your help!

Answer 1

You have errors on your syntax (check the quotes):

echo "<td> <a href='#' onclick='window.open("lookallow.php?ordr=$ORDER &bord=$BORD","","width=200,height=100");'>Allowance</td>";

Correct way:

echo "<td> <a href='#' onclick='window.open('lookallow.php?ordr=$ORDER &bord=$BORD','','width=200,height=100');'>Allowance</td>";

;)

Answer 2

您在代码中关闭“”。而是尝试如下操作：

<td> <a href='#' onclick='window.open(<?php echo "lookallow.php?ordr=$ORDER&bord=$BORD"; ?>,"","width=200,height=100");'>Allowance</td>