dougongnan2167 2010-11-24 18:04
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通过代理传递url属性

I am running a proxy so I can perform an ajax request on data through the url parameters. The proxy php looks like :

<?php
header('Content-type: application/xml');
$daurl = 'http://thesite.com/form.asp';
$handle = fopen($daurl, "r");
if ($handle) {
    while (!feof($handle)) {
        $buffer = fgets($handle, 4096);
        echo $buffer;
    }
    fclose($handle);
}
?>

I am hitting the proxy with ajax that ends up appending a parameter like :

$j.ajax({
            type: 'GET',
            url: 'sandbox/proxy.php',
            data: 'order=' + ordervalue,
            dataType: 'html',
            success: function(response) {
            $j("#result").html(response);
            }
        });

So the request is like sandbox/proxy.php?order=123

How can I grab that data (order=123) and append it to the $daurl variable (http://thesite.com/form.asp?order=123) so that I can have the proxy actually return something?

This is virgin territory for me so you can not over-explain =)

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2条回答 默认 最新

  • doubi4814 2010-11-24 18:09
    关注

    Simple.

    $daurl = 'http://thesite.com/form.asp';
    
    //if you only want 'order':
    if(isset($_GET['order'])) 
       $daurl .= '?order=' . $_GET['order'];
    
    //if you want the entire query string:
    
    if(strlen($_SERVER['QUERY_STRING']) > 0) 
       $daurl .= '?' . $_SERVER['QUERY_STRING'];
    ...
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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