Open a php page from javascript function

Question

I have a php page, first.php and I want to open the page with passing some arguments from a javascript function. Could you please help me, thanks.

function() {
    var tableName = "<?= $p ?>"; //obtaining the value from another php file
    var checkB = checkbox_form.checkbux[counter].value;
    window.open('"http://localhost/first.php?q="+checkB+"&p="+tableName', '_self');
}

But I am not able to open the page, please help. Thanks in advance.

Timo Tijhof · Accepted Answer · 2012-01-25 03:40:43Z

4

As the previous answers state, you could easily use window.location to open a PHP page; however, you should always remember to escape your variables when using them in a URL, using the encodeURIComponent() JavaScript function:

window.location = "http://localhost/first.php?q=" + encodeURIComponent(checkB) + "&p=" + encodeURIComponent(tableName);

edited Jan 25, 2012 at 3:40

Timo Tijhof

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answered May 23, 2011 at 22:46

Freyja

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1 Comment

Timo Tijhof Over a year ago

Do not use escape()/unescape(). There's proper functions for different purposes. In this case, encodeURIComponent is appropriate.

Kyle Sletten · Accepted Answer · 2011-05-23 22:34:55Z

1

It's simpler than you'd think.

window.location = 'http://localhost/first.php?q=' + checkB + '&p=' + tableName;

answered May 23, 2011 at 22:34

Kyle Sletten

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Comments

Damien Goor · Accepted Answer · 2013-01-29 13:06:38Z

My understanding is that you need to open another tab or popup with some dynamic parameters. I have 2 solutions for this:

1- Attach some extra JS to the anchor user will click using the onMouseOver() event and feed the href with your computed URL. The target must be set to "_blank".

Example:

<a href="whateverPage.php" target="_blank" onMouseOver="this.href='myPage.php?myParam=' + myParamValue;">Goto new page</a>

Note that in this example 'myParamValue' needs to be global.

2- You want to open a new tab or pop up after an ajax request? In my case I want generate a new report PHP page on the server and want to open it immediately. Previous solution does not help.

Here is my solution to fool the pop-up blockers:

//this generates the new report page 
report = new ajaxReq("gentabrep.php", ajaxCallBackFunction);
//open the pop-up on user action/event which is normally allowed
w = window.open("", "");
//run ajax request, note I also pass the "w" pop-up reference to the request
report.request("connId=" + connId + "&file=" + file, "POST", [w, file]);

function ajaxCallBackFunction(returnedStr, status, params){
  //I feed the pop-up with the necessary javascript to redirect the page immediately
  params[0].document.writeln("<scr"+"ipt type='text/javaScript'>window.location='reports/" + params[1] + ".php';</scr"+"ipt>");
}

Senad Meškin · Accepted Answer · 2011-05-23 22:34:25Z

0

use

document.location = 'http://localhost/first.php?q='+checkB+'&p='+tableName;

answered May 23, 2011 at 22:34

Senad Meškin

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1 Comment

chella Over a year ago

I tried all the above three but the resulting page is the same page from which the function was called with the variable attached to it. from localhost/thirdPage.php?p=tableName to localhost/…

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