Pass Variable From JavaScript Function on Open.WIndow

Question

I have to pass a string variable created in php to Javascript function to open form child using open.window. I have code like this

<html>
<head>
<script LANGUAGE="JavaScript" TYPE="text/javascript">
    function openWindow(v){
        WREF = window.open("test2.html?cad=e&cid="+v,"koreksi","width=550,height=350,top=50");
        if(!WREF.opener){ WREF.opener = this.window; }
    }
</script>
</head>

<body>
<?php $var='Red'; ?>
<a HREF="javascript:void(0);" onClick="openWindow(<?php $var='Red'; ?>);">Open the window</a>
</body>
</html>

By those code, i hope i can get an address like this http://localhost/adduser.php?cad=e&cid=red

Can you help me ? ( sorry i can't describe clearly for my q )

user4795581 · Accepted Answer · 2015-04-16 09:18:46Z

1

you can use <?php echo $var;?> to pass variable

this line should be:--

<a href="javascript:void(0);" onClick="openWindow('<?php echo $var;?>');">Open the window</a>

answered Apr 16, 2015 at 9:18

user4795581

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1 Comment

Cristik Over a year ago

This will result in a javascript error, as $var should be enclosed in quotes.

Cristik · Accepted Answer · 2015-04-16 09:18:28Z

0

You need to enclose the php variable in single quotes and echo it:
<a HREF="javascript:void(0);" onClick="openWindow('<?php echo $var ?>');">Open the window</a>

Edit: the single quotes are needed as the PHP echo will render the actual string, which will result in a javascript function call like: openWindow(http://localhost/adduser.php?cad=e&cid=red), which is not valid syntax; on the other hand openWindow('http://localhost/adduser.php?cad=e&cid=red') is.

answered Apr 16, 2015 at 9:18

Cristik

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4 Comments

agus priyo Over a year ago

Thank you for your advice, and i tried to changed my code like this : <body> <?php $var='Red'; ?> <a HREF="javascript:void(0);" onClick="openWindow('<?php echo $var; ?>');">Open the window</a> </body> unfortunately, the results of the link is not as I expected. This is the result link : localhost/adduser.php?cad=e&cid=<?php echo $var; ?> What is wrong on my code above ?

agus priyo Over a year ago

Thank you for response. I think there is not any interpreter. The first php file is the above script ( my question above ). And the second file is like this : <?php $cadd = mysql_real_escape_string($_GET['cad']); $cuid = mysql_real_escape_string($_GET['cid']); echo $cadd; echo $cuid; ?>

agus priyo Over a year ago

i used xampp. i acces the page from localhost/test1.html. Test1.html is the name of my php file

agus priyo Over a year ago

Yes .... thank you, you are correct. I changed test1.html to test1.php. Thank your advice

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