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View Full Version : how to call 2 funcions on click?



alapimba
12-20-2007, 01:15 PM
Hello
I'm with a few problems with some functions.
It works if i have only one, but when i write the second an call it it don't work.

My function:

<?php do { $numero = $row_rs_anos['id']; ?>
function titulo<?php echo $numero ?>() {
http.open("GET", "titulo<?php echo $numero ?>.html", true);
http.onreadystatechange=function() {
if(http.readyState == 4) {
document.getElementById('titulo').innerHTML = http.responseText;
initLightbox();
}
}
http.send(null);
}
function replace<?php echo $numero ?>() {
http.open("GET", "teste<?php echo $numero ?>.html", true);
http.onreadystatechange=function() {
if(http.readyState == 4) {
document.getElementById('imgs').innerHTML = http.responseText;
initLightbox();
}
}
http.send(null);
}

<?php } while ($row_rs_anos = mysql_fetch_assoc($rs_anos)); ?>

as you see the funcione titulo and the function replace are basicly the same, just change the page they open from and the div where it will appear.. i don't know if i should write it twice as i did or theres any better way..

I'm calling it with:

<a href="javascript:replace<?php echo $row_rs_anos2['id']; ?>(); titulo<?php echo $row_rs_anos2['id']; ?>()"><?php echo $row_rs_anos2['ano']; ?></a>

What i'm doing wrong and how to fix it?
Thanks

Kor
12-20-2007, 10:58 PM
It's a tricky way to send 2 requests nearly at the same time. Have you tried to call the second function after the first gets it's response?


function replace<?php echo $numero ?>() {
http.open("GET", "teste<?php echo $numero ?>.html", true);
http.onreadystatechange=function() {
if(http.readyState == 4) {
document.getElementById('imgs').innerHTML = http.responseText;
initLightbox();
titulo<?php echo $numero ?>();
}
}
http.send(null);
}

and:


<a href="#" onclick="replace<?php echo $row_rs_anos2['id']; ?>();return false"><?php echo $row_rs_anos2['ano']; ?></a>

alapimba
12-21-2007, 12:10 AM
yeah that way it works.
Thanks :)



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