simple ajax replace but with many links

alapimba · 11-05-2007, 03:57 PM

Hello, can anyone help me with this function?

i'm doing a replace, the content appears inside the div "foo".

now i want to have like 3 or 4 links with the funcione replace and each one replace his own content.

my code so far and working is:

Code:

<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>
<script type="text/javascript">
var http = false;
if(navigator.appName == "Microsoft Internet Explorer") {
  http = new ActiveXObject("Microsoft.XMLHTTP");
} else {
  http = new XMLHttpRequest();
} 
function replace() {
  http.open("GET", "conteudo.php?id=1", true);
  http.onreadystatechange=function() {
    if(http.readyState == 4) {
      document.getElementById('foo').innerHTML = http.responseText;
    }
  }
  http.send(null);
}
</script> 
 
</head>
<body>
<div class="accordion_content">
    <h2><p><a href="javascript:replace()">link 1</a></p>
<div id="foo">
  here appear the content
</div></h2>
</body>

shyam · 11-06-2007, 04:51 AM

use function parameters

Code:

function replace(where, what) {
  http.open("GET", "conteudo.php?id=" + what, true);
  http.onreadystatechange=function() {
    if(http.readyState == 4) {
      document.getElementById(where).innerHTML = http.responseText;
    }
  }
  http.send(null);
}

Code:

<a href="javascript:replace('foo', 1)">link 1</a>

alapimba · 11-06-2007, 11:53 AM

thank you