My JavaScript not working.iUnsolved!
Posted: Mon Feb 20, 2006 10:13 pm
When ever I move my mouse over the image it just says an error in the status bar.
I don't know javascript just got this off the net but any help is appreciated.
EDITED: This script has been changed and what it does is it doesn't display any pictures. Why is this please help me I'd really really appreciate it.
Code: Select all
<?
//set two variables with the value of the database name and table name
$db_name = "";
$table_name = "";
//make a connection to the database
$connection = @mysql_connect("", "", "") or die(msql_error());
//set a variable to select a database I guess
$db = @mysql_select_db($db_name, $connection) or die(mysql_error());
$sql_sel = "SELECT picture, webaddress FROM $table_name ORDER BY picture";
$result = @mysql_query($sql_sel,$connection) or die(mysql_error());
?>
<html>
<head>
<title>Home Page (example)</title>
<SCRIPT LANGUAGE="JavaScript">
// ############## SIMPLE BROWSER SNIFFER
if (document.layers) {navigator.family = "nn4"}
if (document.all) {navigator.family = "ie4"}
if (window.navigator.userAgent.toLowerCase().match("gecko")) {navigator.family = "gecko"}
// ######### popup text
descarray = new Array(
"This site has some of the greatest scripts around!",
);
overdiv="0";
// ######### CREATES POP UP BOXES
function popLayer(a){
if(!descarray[a]){descarray[a]="<font color=red>This popup (#"+a+") isn't setup correctly - needs description</font>";}
if (navigator.family == "gecko") {pad="0"; bord="1 bordercolor=black";}
else {pad="1"; bord="0";}
desc = "<table cellspacing=0 cellpadding="+pad+" border="+bord+" bgcolor=000000><tr><td>\n"
+"<table cellspacing=0 cellpadding=3 border=0 width=100%><tr><td bgcolor=ffffff><center><font size=-1>"
+descarray[a]
+"\n</td></tr></table>\n"
+"</td></tr></table>";
if(navigator.family =="nn4") {
document.object1.document.write(desc);
document.object1.document.close();
document.object1.left=x+15;
document.object1.top=y-5;
}
else if(navigator.family =="ie4"){
object1.innerHTML=desc;
object1.style.pixelLeft=x+15;
object1.style.pixelTop=y-5;
}
else if(navigator.family =="gecko"){
document.getElementById("object1").innerHTML=desc;
document.getElementById("object1").style.left=x+15;
document.getElementById("object1").style.top=y-5;
}
}
function hideLayer(){
if (overdiv == "0") {
if(navigator.family =="nn4") {eval(document.object1.top="-500");}
else if(navigator.family =="ie4"){object1.innerHTML="";}
else if(navigator.family =="gecko") {document.getElementById("object1").style.top="-500";}
}
}
// ######## TRACKS MOUSE POSITION FOR POPUP PLACEMENT
var isNav = (navigator.appName.indexOf("Netscape") !=-1);
function handlerMM(e){
x = (isNav) ? e.pageX : event.clientX + document.body.scrollLeft;
y = (isNav) ? e.pageY : event.clientY + document.body.scrollTop;
}
if (isNav){document.captureEvents(Event.MOUSEMOVE);}
document.onmousemove = handlerMM;
// End -->
</head>
<body>
<?
while ($row = mysql_fetch_array($result)) {
$icon = $row['picture'];
$webaddress = $row['webaddress'];
$display_block .= "<a href=\"$webaddress\" onmouseover='popLayer(0)' onmouseout='popLayer()'><img src=\"http:///$icon\" width=20 height=20 border=0></a>";
}
?>
<? echo "$display_block"; ?>
</body>
</html></head>
<body>
<? echo "$display_block"; ?>
</body>
</html>EDITED: This script has been changed and what it does is it doesn't display any pictures. Why is this please help me I'd really really appreciate it.