help me..data tak boleh display

beih107 · #1 (**permalink**) 24-04-2007, 11:43 AM

hi sume..saya ada mslh mengenai data x boleh display kat interface sistem..tp dalam database data tersebut dah masuk..n then data tersebut hanya boleh display kat page lain bukan page yg spatutnye..query untuk page ni adalah gabungan 3 table..so,apa yg saya perlu buat atau ubah?

coding interface

<? include("odbc.php");
include("banner.php");
include("homestaff.php");
?>

<html>

<head>

<script language="JavaScript" src="calendar.js"></script>
<title>borrow file record</title>

</head>

<body>

<br>
<table width="100%" cellpadding="2" cellspacing="0" border="0" bordercolor="#000000" align="center">

<tr>
<td align="right">
<a href="staff.php"><strong>Sign Up</strong></a>
</td>
</tr>

<tr align="center">
<td height="400" valign="top">

<form method="POST" name="form" action="borrowprocess.php">
<br><table align="center" bgcolor="#FF9966">
<tr>
<td width="104" height="36"><strong>Borrow Date :</strong></td>
<td width="183"><input type="text" name="borrowdate">
<a href="javascript:show_calendar('form.borrowdate'); " onMouseOver="window.status='borrowdate';return true;" onMouseOut="window.status='';return true;"><img src="images/b_calendar.png" width=16 height=16 border=0 align="middle"></a></td>
</tr>

<tr>
<td height="29"><strong>Staff No :</strong></td>
<td><select name="staffno">
<option value="">Please Select Staff No</option>
<?
$sqlst = mysql_query("SELECT * FROM staff");
while($staf = mysql_fetch_array($sqlst))
{ ?>
<option value="<? echo $staf['staff_id'] ?>"<?php if($code['staff_id'] == $code1) echo ' selected="selected"'; ?>><? echo $staf['staffno'];?> </option>
<? } ?>
</select>
<a href="staffrecord.php?staff_id=<? echo $result['staff_id'];?><? echo $result['staff_name'];?>');">[details]
</a>

</tr>

<tr>
<td><strong>File Name:</strong></td>
<td><select name="file">
<option value="">Select File</option>
<?
$sqlst = mysql_query("SELECT * FROM item");
while($code = mysql_fetch_array($sqlst))
{ ?>
<option value="<? echo $code['item_id'] ?>"<?php if($code['item_id'] == $code1) echo ' selected="selected"'; ?>><? echo $code['item_name'];?> </option>
<? } ?>
</select></td>
</tr>

<tr>
<td><strong>Return Date :</strong></td>
<td><input type="text" name="returndate">
<a href="javascript:show_calendar('form.returndate'); " onMouseOver="window.status='returndate';return true;" onMouseOut="window.status='';return true;"><img src="images/b_calendar.png" width=16 height=16 border=0 align="middle"></a></td>

</tr>

<tr>
<td colspan="2" align="center"><input type="submit" name="hantar" value="ADD">
<input type="reset" name="padam" value="CANCEL"></td></tr>
</table>
</form>

<table align="center" border="1" bordercolor="#000000" width="100%">

<tr bgcolor="#FF5C0F">
<td colspan="7" align="center"><strong>BORROW FILE RECORD</strong></td>
</tr>

<tr>
<td width="7%" align="center"><strong>No</strong></td>
<td width="13%" align="center"><strong>Borrow Date </strong></td>
<td width="13%" align="center"><strong>Staff No</strong></td>
<td width="18%" align="center"><strong>File Name </strong></td>
<td width="12%" align="center"><strong>Return Date </strong></td>
<td width="20%" align="center"><strong>Action</strong></td>
</tr>

<?

$i=1;
$sqlstmt =mysql_query("SELECT item.*, borrow.*, staff.*
FROM staff RIGHT JOIN (item RIGHT JOIN borrow ON item.item_id=borrow.item_id)
ON staff.staff_id=borrow.item_id");

while($result = mysql_fetch_array($sqlstmt))

?>

<tr>
<td height="31"><? echo $i;?> </td>
<td><? echo date('d-m-Y',strtotime($result['borrowdate']));?> </td>
<td><? echo $result['staffno'];?> </td>
<td><? echo $result['item_name'];?> </td>
<td><? echo date('d-m-Y',strtotime($result['returndate']));?> </td>

<td align="center"><p><a href="deleteborrow.php?borrow_id=<? echo $result['borrow_id'];?>" onClick="return confirm('Are You Sure To Delete? \n <? echo $result['staffno'];?>');"><img src="images/butdel.jpg" border="0"></a> 
<a href="editborrow.php?borrow_id=<? echo $result['borrow_id'];?>" onClick="return confirm('Are You Sure To Edit? \n <? echo $result['staffno'];?>');"><img src="images/butedit.jpg" border="0"></a>
</p>
</td>

</tr>

</table>

</td>
</tr>

<tr>
<td> </td>
</tr>

</table>

</body>
</html>

process

<?php
include("odbc.php");
echo $borrow=$_POST['borrowdate'];
echo $borrow = substr($borrow,6,4)."_".substr($borrow,3,2)."_".su bstr($borrow,0,2);
echo $staff = $_POST['staffno'];
echo $file= $_POST['file'];
echo $return = $_POST['returndate'];
echo $return = substr($return,6,4)."_".substr($return,3,2)."_".su bstr($return,0,2);

$query_add_data = "INSERT INTO borrow(borrowdate,staff_id,item_id,returndate) VALUES ('$borrow','$staff','$file','$return')";
$result = mysql_query($query_add_data);

echo "<META HTTP-EQUIV=Refresh CONTENT=\"3; URL=borrowrecord.php\">";
?>

genzy · #2 (**permalink**) 06-05-2007, 05:19 AM

Which table data do you mean??
'Dah masuk' - When? Is the data just inserted on that page? or already exist?

Check whether the query results in the page per what you want and query criteria, in the MySQL console or GUI front-end tool to analyse the query first.

Tip: It is a good practice to put comments for a certain block of coding statements/SQL queries. Coding documentation is very important. If not, future maintenance will be a very painful nightmare.

polturgiest · #3 (**permalink**) 18-05-2007, 10:08 AM

hie

instead of using

PHP Code:


			
$sqlst = mysql_query("SELECT * FROM staff");

try debugging it using

PHP Code:


			
$sqlst = mysql_query("SELECT * FROM staff") or die(mysql_error());

from there can get the mysql error msg..hope this help

note: dont forget to remove it later

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