张贴后的jQuery重定向

Question

I have 3 php files.

• file1.php contains PHP + HTML design and jQuery scripts.
• file2.php contains just PHP and mySQL processing

• file3.php generates a PDF using TCPDF, based on 2 variables that it receives in $_REQUEST, and displays the resulting PDF

  1. I call from a button in file1.php a $.post towards file2.php .
  2. Some data processing happens in file2.php (mysql inserts and updates) and I obtain var1 and var2 .
  3. When processing is done in file2.php I want to go and display the PDF that is created by file3.php based on var1 and var2

So the jQuery code in file1.php where I call post is:

        $("#saveinv").click(function() { // button is clicked
            var listview_array = new Array();
            // here I populate the array with values then...
            $.post("file2.php", 
               {json: JSON.stringify(listview_array)}, 
               function(data){alert("success"); 
            });
        });

In file2.php there is nothing to see. After much processing, I end up with 2 variables that help me call the file3.php

file3.php must be called with 2 params: var1 and var2 so something like:

file3.php?var1=5&var2=77

But I have no idea where to place the call or what exact syntax to use. I tried different solutions without success. Maybe I was doing it wrong and you can help me.

• I tried to place at the end of file2.php a header('Location: file3.php?var1=5&var2=77');

• I tried to :

  • add at the end of file2.php the line : return 'file3.php?var1=5&var2=77';
  • then replace the alert('success') in file1.php with window.location = data.redirect but I keep getting redirected to "undefined".

• If I do window.location='file3.php?var1=5&var2=77' in file1.php then all works perfect, but I need those variables to be sent from file2.php. They are not constants.

What should I do, and where?

Thank you

Answer 1

Just put your result as JSON in file2.php:

echo '({"var1":"1","va2":"2"})';

where you get it as data inside your ajax callback function in file1.php, then use eval

var args = eval(data);
window.location = 'file3.php?var1=' + args.var1 + '&var2=' + args.var2;

Answer 2

add at the end of file2.php the line : return 'file3.php?var1=5&var2=77';

You cannot use return , just use print or echo and it should work.