php inside javascript function

manamino

hi can u help me to pass php values to javascript array,
actually I have many drop down boxes
I need onchange of one of them to change the values of the others according to the value of the first one
and since I dont need to submit the form to take the values and use them in php , I tried at the first of my program to put all the data in javascript arrays and then onchange I use javascript function to search for my values. this is a logical solution but it is not working with me, can u find the problem
this is my program:

search.php
<?php
$db = mysql_connect('localhost','root', 'memo1982');
mysql_select_db ("collab1",$db);
$sql=mysql_query("select from student",$db);
?>
<script language='javascript'>
function lolo()
{
p=0;
var id=new array();
var fname=new array();
var lname=new array();
var gender=new array();
var level=new array();
var type=new array();
<?
$sql1=mysql_query("select from sanction s, student st, officer o where s.Id=st.Id and s.Oid=o.Oid",$db);
while($row = mysql_fetch_array($sql1, MYSQL_NUM))
{?>
id[p]=<?=$row["Id"]?>;
p=p+1;
<?
}?>
document.write(p); //i made this one only to see if the function is working but it is not working
}
</script>

<title>Disciplinary Sanctions Search</title>
<img src='ltu.bmp' width="974" height="72"/>
<Style TYPE="text/css">
.boldtable, .boldtable TD, .boldtable TH
{
font-size:12pt;
color:#000066;
}
</style>

</head>
<body background='bg.jpg'>
<h1><font color="#000080">Search<hr></font></h1>
<form name='search' method='post' action='search2.php'>
<table class='boldtable'>
<tr>
<td bgcolor='99cccc'>Student ID:</td>
<td>
<select name='sid' class='boldtable' onchange="lolo()">
<option value='All'>All
<?while ($row = mysql_fetch_array($sql, MYSQL_NUM)) {?>
<option value='<?= $row[0]?>'><?= $row[0]?>
<?

}?>
</select>
</td>
</tr>

<tr>
<td bgcolor='99cccc'>First Name:</td>
<td>
<select name='fn' class='boldtable'>
<option value='All'>All
<?
$sql=mysql_query("select * from student",$db);
while ($row = mysql_fetch_array($sql, MYSQL_NUM)) {?>
<option value='<?= $row[1]?>'><?= $row[1]?>
<?

}?>
</select>
</td>
</tr>

<tr>
<td bgcolor='99cccc'>Gender:</td>
<td>
<select name='sex' size='3' class='boldtable'>
<option value='All' selected>All
<option value='M'>Male
<option value='F'>Female
</select>
</td>
</tr>

<tr>
<td bgcolor='99cccc'>Level:</td>
<td>
<select name='level' size='6' class='boldtable'>
<option value='All' selected>All
<option value='G'>G-Grad
<option value='Sr'>Sr-Senior
<option value='J'>J-Junior
<option value='S'>S-Soph
<option value='F'>F-Fresh
</select>
</td>
</tr>

<tr>
<td bgcolor='99cccc'>Date From:</td>
<td>
Month:<select name='monthf' class='boldtable'>
<option value='0'>
<?
for ($i=1;$i<=12;$i++) {?>
<option value='<?= $i?>'><?= $i?>
<?

}?>
</select>

Day:<select name='dayf' class='boldtable'>
<option value='0'>
<?
for ($i=1;$i<=31;$i++) {?>
<option value='<?= $i?>'><?= $i?>
<?

}?>
</select>

Year:<select name='yearf' class='boldtable'>
<option value='0'>
<?

for ($i=1990;$i<=2010;$i++) {?>
<option value='<?= $i?>'><?= $i?>
<?

}?>
</select>

</td>
</tr>

<tr>
<td bgcolor='99cccc'>Date To:</td>
<td>
Month:<select name='monthto' class='boldtable'>
<option value='0'>
<?
for ($i=1;$i<=12;$i++) {?>
<option value='<?= $i?>'><?= $i?>
<?

}?>
</select>

Day:<select name='dayto' class='boldtable'>
<option value='0'>
<?
for ($i=1;$i<=31;$i++) {?>
<option value='<?= $i?>'><?= $i?>
<?

}?>
</select>

Year:<select name='yearto' class='boldtable'>
<option value='0'>
<?

for ($i=1990;$i<=2010;$i++) {?>
<option value='<?= $i?>'><?= $i?>
<?

}?>
</select>

</td>
</tr>

<tr>
<td bgcolor='99cccc'>Student Conduct Officer:</td>
<td>
<select name='officer' class='boldtable'>
<option value='All'>All
<?
$sql1=mysql_query("select * from officer",$db);
while ($row = mysql_fetch_array($sql1, MYSQL_NUM)) {?>
<option value='<?= $row[0]?>'><?= $row[1]." ".$row[2]?>
<?

}?>

<tr>
<td bgcolor='99cccc'>Violation Type:</td>
<td>
<select name='type' class='boldtable'>
<option value='All' selected>All
<option value='DW'>DW-Warning
<option value='DP'>DP-Probation
<option value='LP'>LP-Loss Privileges
<option value='LAC'>LAC-Loss Acad Credit
<option value='F'>F-Fines
<option value='RES'>RES-Restitution
<option value='DS'>DS-Sanction
<option value='US'>US-Univ. Suspend
<option value='UE'>UE-Univ. Expel
<option value='IS'>IS-Interim Suspend
<option value='ISH'>ISH-Interim Suspend Housing
<option value='TR'>TR-Temp Reassign
<option value='TRP'>TRP-Temp Restrict-Personic Contact
<option value='WDP'>WDP-Withdraw Prior to Disposition
</select>
</td>
</tr>

<tr>
<td bgcolor='99cccc'>Section:</td>
<td>
<input type='text' name='section'/>
</td>
</tr>
</table>
<br>
<input type='hidden' name='page' value='search1'/>
<input type="submit" value="Search"/>
<input type="reset" value="Reset"/>

</body>
</html>

thorpe

it might be best of you narrow the problem down a little, no one is going to rattle through all that code to find something that isn't working. by the way.... what does nto working mean? be more specific with your questions.

you might want to try using the boards [ php ] tages too, makes it much easier to read.

thorpe

um... just noticed something. ytou might want to move all that stuff at the top into the <head> of the document. that is where you should define javascript functions and arrays.

manamino

dont worry about all that stuff
see only the script where I have the problem
<script language='javascript'>
function lolo()
{
p=0;
var id=new array();
var fname=new array();
var lname=new array();
var gender=new array();
var level=new array();
var type=new array();
<?
$sql1=mysql_query("select * from sanction s, student st, officer o where s.Id=st.Id and s.Oid=o.Oid",$db);
while($row = mysql_fetch_array($sql1, MYSQL_NUM))
{?>
id[p]=<?=$row["Id"]?>;
p=p+1;
<?
}?>
document.write(p); //i made this one only to see if the function is working but it is not working
}
</script>

everything else is working carefully
i only need help for this script, also I put them inside head tag it is the same( Not working) that means when i change the value of the drop down nothing is appened. please if u know something help me

ahundiak

Just a guess but this:

while($row = mysql_fetch_array($sql1, MYSQL_NUM))
{?>
id[p]=<?=$row["Id"]?>;
p=p+1;
<?

Should be something like:

$p = 0;
while($row = mysql_fetch_array($sql1, MYSQL_ASSOC))
{?>
id[<?= $p ?>]=<?=$row["Id"]?>;
<? $p++; } ?>

manamino

I tried this but no way
please if u know any other way to make this let me know
and also I think it doesnot understand the array because if I deletd all the php stuff and let the array like that :
function lolo()
{

var id=new array();
var fname=new array();
var lname=new array();
var gender=new array();
var level=new array();
var type=new array();
id[0]='pkpp';
document.write(id[0]); }
also error on page and I dont know why, please help coz this is a part of a project I should post it to the doctor this week

Weedpacket

That's because "array" is supposed to be "Array". You should use a browser that will give you a decent error message - something along the lines of "array is not defined" - that will give you some clue what the problem is.