Problem with query

Anon

Hello,

I am having a problem with my query that takes the input from a user on a form(code not attached but know it works like a champ) uses the data to do a simple select query and then displays it to a table. The query return's nothing.....however when i hardcode the 'CaseNo' it brings back results......so I probably have some stupid syntax error but after sitting here all day writing this I just can't figure out what is wrong.

I even trid to do a generic select query with passing the CaseNo and nothing so I am stumped.....

Thanks for you help in advance...
{}{}Code Starts below{}{}{}{}

<!DOCTYPE HTML PUBLIC "-//SoftQuad Software//DTD HoTMetaL PRO 6.0::19990601::extensions to HTML 4.0//EN" "hmpro6.dtd">

  <HEAD> 
	 <TITLE>Fiche Database results</TITLE> 
  </HEAD> 
  <BODY> 
 <?PHP 

$connection=mysql_connect (localhost, username, pasword);
if($connection ==false) {
echo mysql_errno().":".mysql_error()."<br>";
exit;
}

$query="Select CaseNo,CaseName,DateFiled,DateTermed,DateReopen from case where CaseNo like 'CaseNo'";
$result=mysql_db_query("fichedb",$query);


if ($result) {


$numOFRows=mysql_num_rows($result);
for($i=0;$i<$numOFRows; $i++){
$CaseNo=mysql_result($result, $i, "CaseNo");
$CaseName=mysql_result($result, $i, "CaseName");
$DateFiled=mysql_result($result, $i, "DateFiled");
$DateTermed=mysql_result($result, $i, "DateTermed");
$DateReopen=mysql_result($result, $i, "DateReopen");

echo "<table border=2 width=98%>";
echo "<tr><td width=10%>$CaseNo</td><td width=11%>$CaseName</td><td width=13%>$DateFiled</td><td width=11%>$DateTermed</td><td width=12%>$DateReopen</td>
</tr>";
}
echo "</table>";
}
else{
echo mysql_errno().":".mysql_error()."<BR>";
}
mysql_close();

?>		 
</BODY>

Anon

Your query:

"Select CaseNo,CaseName,DateFiled,DateTermed,DateReop
en from case where CaseNo like 'CaseNo'"

appears to be lloking for all rows from case where the CaseNo = 'CaseNo'. I ssuspect you don't have any Cases thare numbered 'CaseNo'.

What you probably meant to do is something like this:

"Select CaseNo,CaseName,DateFiled,DateTermed,DateReopen from case where CaseNo like '$CaseNo'"

That will have php parsing the reference to $CaseNo in order to insert the value held in that variable.

Anon

Write : where CaseNo like '$CaseNo'
and not: where CaseNo like 'CaseNo'

If CaseNo is the name of the text input on the form

JBL

Anon

I'm guessing CaseNo is an integer, so you would want to search for an "exact" case number. That is, not search for a number that is "like" the one retrieved from the form. Maybe, I'm guessing wrong but in your query you should use '$CaseNo' so php puts in the variable's value. If I'm correct then the query should be like this: "select ... where CaseNo = '$CaseNo'".

Anon

Thanks everyone for there help I just forgot the $ !!!!!!