Dropdown select onchange to update multiple divs
I have a selectbox with an onchange event to update content with content from another file. It works fine. but what I want to do is update the content in two different divs at the same time when I select something from my selectbox.
So I added another function similar to the first one and added it to onchange in the tag.
The problem is that it still only updates one of the two. What am I doing wrong? Or is there an easier way to update the two at the same time with information from another page?
script on main page:
<script>
function showFirstDiv(str) {
if (str=="") {
document.getElementById("div1").innerHTML="";
return;
}
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
} else { // code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function() {
if (xmlhttp.readyState==4 && xmlhttp.status==200) {
document.getElementById("div1").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","infopage.php?info_div1="+str,true);
xmlhttp.send();
}
function showSecondDiv(str) {
if (str=="") {
document.getElementById("div2").innerHTML="";
return;
}
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
} else { // code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function() {
if (xmlhttp.readyState==4 && xmlhttp.status==200) {
document.getElementById("div2").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","infopage.php?info_div2="+str,true);
xmlhttp.send();
}
html on the main page
<div id="users">
<form>
<select name="users" onchange="showFirstDiv(this.value); showSecondDiv(this.value);">
<option value="">Select a person:</option>
<option value="1">Peter Griffin</option>
<option value="2">Lois Griffin</option>
<option value="3">Joseph Swanson</option>
<option value="4">Glenn Quagmire</option>
</select>
</form>
</div>
<!-- div1 -->
<div id="div1">
<p>info to be listed in div 1 here</p>
</div>
<!-- div2 -->
<div id="div2">
<p>info to be listed in div 2 here</p>
</div>
infopage.php
<?php
if ($_GET[info_div1] == '1')
{
echo 'info 1 to be showed in div 1';
}
if ($_GET[info_div1] == '2')
{
echo 'info 2 to be showed in div 1';
}
if ($_GET[info_div1] == '3')
{
echo 'info 3 to be showed in div 1';
}
if ($_GET[info_div1] == '4')
{
echo 'info 4 to be showed in div 1';
}
if ($_GET[info_div2] == '1')
{
echo 'info 1 to be showed in div 2';
}
if ($_GET[info_div2] == '2')
{
echo 'info 2 to be showed in div 2';
}
if ($_GET[info_div2] == '3')
{
echo 'info 3 to be showed in div 2';
}
if ($_GET[info_div2] == '4')
{
echo 'info 4 to be showed in div 2';
}
?>
Both functions use the same global variable xmlhttp
. Therefore, when you call the second function, it will overwrite this variable with its XMLHttpRequest object. You must use a local variable by declaring it with var
.
In general, you should always declare variables as local unless you really need global variables.
function showFirstDiv(str) {
var xmlhttp;
if (str=="") {
document.getElementById("div1").innerHTML="";
return;
}
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
} else { // code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function() {
if (xmlhttp.readyState==4 && xmlhttp.status==200) {
document.getElementById("div1").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","infopage.php?info_div1="+str,true);
xmlhttp.send();
}
function showSecondDiv(str) {
var xmlhttp;
if (str=="") {
document.getElementById("div2").innerHTML="";
return;
}
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
} else { // code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function() {
if (xmlhttp.readyState==4 && xmlhttp.status==200) {
document.getElementById("div2").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","infopage.php?info_div2="+str,true);
xmlhttp.send();
}