[SOLVED] Help Me With These

june_c21 · May 4, 2008

why my code can't execute? when i execute, it output all the staff_no, name instead of specific staff_no only

<?php
//$id=$_GET["id"];
$staff_no = $_POST['staff_no'];

$host = 'localhost';
$user = 'root';
$password = '';
$dbase = 'claim';

$dblink = mysql_connect($host,$user,$password);
mysql_select_db($dbase,$dblink);

$sql="SELECT report1.staff_no, user.name FROM report1,user WHERE report1.staff_no = '".$staff_no."'";
$result = mysql_query($sql);

echo "<table border='1'>
<tr>

<th>Staff No</th>
<th>Name</th>

</tr>";
while($row = mysql_fetch_array($result))
{
echo "<tr>";

echo "<td>" . $row[0] . "</td>";
echo "<td>" . $row[1] . "</td>";

echo "</tr>";
}
echo "</table>";

?>

tronicsmasta · May 4, 2008

You are using an array here... you need to specify what part of the array with a variable.

while($row = mysql_fetch_array($result))
{
echo "<tr>";

echo "<td>" . $row[0] . "</td>";
echo "<td>" . $row[1] . "</td>";

echo "</tr>";
}
echo "</table>";

should be:

while($row = mysql_fetch_array($result))
{
     $report1_staffno = $row['report1.staffno'];
     $username = $row['user.name'];
     echo "<tr>";
     echo "<td>$report1_staffno</td>";
     echo "<td>$username</td>";
     echo "</tr>";
}
echo "</table>";

june_c21 · May 4, 2008

sorry .. is still doesn't work. it came out blank only

sasa · May 4, 2008

you don't set relatin on your tablese

tronicsmasta · May 4, 2008

There isn't any syntax errors...

the only think i can think of is the . in the table

eg

report1.staffno

user.name

june_c21 · May 4, 2008

ok, i test and this code got error

SELECT report1.staff_no, user.name FROM report1,user WHERE report1.staff_no = '$staff_no'

when i replace sk200023 into $staff_no

all the name in the database will came out and they auto change their staff_no into sk200023. why?

tronicsmasta · May 4, 2008

report1,user

has a coma.... is that suppose to be a . ? if its a coma, put report1,user into single quotes 'report1,user'

june_c21 · May 4, 2008

report1 is a table name

user is another table name

i need to match report1.staff_no (databse) with user input staff_no

and then based on both table got staff_no, i will take the name from user table that match with the user input staff_no

Barand · May 4, 2008

As sasa said, you haven't set the join relationship in the query

$sql="SELECT report1.staff_no, user.name 
       FROM report1 
           JOIN user ON report1.staff_no = user.staff_no
       WHERE report1.staff_no = '".$staff_no."'";

june_c21 · May 4, 2008

barand,

why it came out blank instead of showing the staff_no and name? anything i miss out?

june_c21 · May 4, 2008

<?php
//$id=$_GET["id"];
$staff_no = $_POST['staff_no'];

$host = 'localhost';
$user = 'root';
$password = '';
$dbase = 'claim';

$dblink = mysql_connect($host,$user,$password);
mysql_select_db($dbase,$dblink);

$sql="SELECT report1.staff_no, user.name 
       FROM report1 
           JOIN user ON report1.staff_no = user.staff_no
       WHERE report1.staff_no = '.$staff_no.'";

$result = mysql_query($sql);

echo "<table border='1'>
<tr>

<th>Staff No</th>
<th>Name</th>

</tr>";

while($row = mysql_fetch_array($result))
{
     $report1_staffno = $row['report1.staff_no'];
     $username = $row['user.name'];
     echo "<tr>";
     echo "<td>$report1_staffno</td>";
     echo "<td>$username</td>";
     echo "</tr>";
}
echo "</table>";
?>

this is the user input code

<html>
<head>
<script src="selectuser.js"></script>
</head>
<body><form id="form1" name="form1" method="post" action="checking.php">
  <p><strong>KEV Tool Store - Loan Transaction Page</strong></p>
  <table width="497" border="0">
    <tr>
      <td width="284">Choose</td>
      <td width="203"><select name="staff_no" onChange="showUser(this.value)">
        	  <?php
 $host     = 'localhost';
$user     = 'root';
$password = '';
$dbase    = 'claim';

$dblink = mysql_connect($host, $user, $password);
mysql_select_db($dbase, $dblink);

$query = "SELECT staff_no
FROM user
ORDER BY staff_no
";
      		
		$result = mysql_query($query);
while($row = mysql_fetch_array($result)) {
$staffno = $row['staff_no'];
echo "<option value=\"$staffno\">$staffno</option>";
     }
?>
</select></td>
    </tr>
    <tr>
      <td colspan="2"><p>
<div id="txtHint"><b>info will be listed here.</b></div>
</p> </td>
    </tr>
  </table>
  <p> </p>
  <p>
    <input type="submit" name="Submit" value="Submit">
  </p>
</form></body>
</html>

this whole code just can't get me the name and staff no when user input their staff_no . why? i been trying for 3 days....

Barand · May 4, 2008

Check for error

$result = mysql_query($sql) or die(mysql_error());

june_c21 · May 4, 2008

try already. no error

Barand · May 4, 2008

Then it could be your data.

It will only return rows where there are records on both tables which match on staff_no.

You can see which have matches and which dont by running this query. (if there is a match it will show a name)

$sql="SELECT report1.staff_no, user.name 
       FROM report1 
           LEFT JOIN user ON report1.staff_no = user.staff_no";

june_c21 · May 4, 2008

now i redo the html file again

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>
</head>

<body>
<form id="form1" name="form1" method="post" action="checking2.php">
  <label>Name 
  <input name="staff_no" type="text" id="staff_no" />
  </label>
  <p>
    <label>
    <input type="submit" name="Submit" value="Submit" />
    </label>
  </p>
</form>
</body>
</html>

php file

<?php
// echo " New Entry <a href=add.html>click here</a><br></br>";
$host = "localhost";
$user = "root";
$password = "";
$dbase = "claim";

$dblink = mysql_connect($host,$user,$password);
mysql_select_db($dbase,$dblink);
$staff_no = $_POST['staff_no'];

echo "<table border=3 align=center width=90%>";
echo "<tr><td width=12% align=center><span class=style5>ID</td></span>";
echo "<td width=12% align=center><span class=style5>Month</td></span>";
echo "<td width=12% align=center><span class=style5>Start Date </td></span>";
echo "<td width=12% align=center><span class=style5>End Date</td></span>";
echo "<td align=center><span class=style5>Name</td></span>";
echo "<td align=center><span class=style5>Staff No</td></span>";
echo "<td align=center><span class=style5>Details</td></span>";
echo "<td align=center><span class=style5>Amount</td></span>";
echo "<td align=center><span class=style5>GL code</td></span>";
  echo "<td width=12% align=center><span class=style5>Bank </td></span>";
echo "<td align=center ><span class=style5>Account No. </td></span>";

echo "<td align=center ><span class=style5>Total </td></span>";
echo "<td align=center colspan=3><span class=style5>Action </td></span></tr>";


$query = "SELECT report1.id, month, start_date, end_date, user.name, report1.staff_no,details,amount,gl_code,user.bank,user.acc_no FROM user,report1 WHERE report1.staff_no = '$staff_no'";
$result = mysql_query($query, $dblink);
while($myrow = mysql_fetch_row($result))
{
echo "<tr><td><span class=style2>" . $myrow[0] . "</span</td>";
    echo "<td><span class=style2>" . $myrow[1] . "</span</td>";
    echo "<td><span class=style2>" . $myrow[2] . "</span</td>";
    echo "<td><span class=style2>" . $myrow[3] . "</td></span>";
    echo "<td><span class=style2>" . $myrow[4] . "</span></td>";
    echo "<td><span class=style2>" . $myrow[5] . "</span></td>";
    echo "<td><span class=style2>" . $myrow[6] . "</span></td>";
    echo "<td><span class=style2>" . $myrow[7] . "</span></td>";
echo "<td><span class=style2>" . $myrow[8] . "</span></td>";
echo "<td><span class=style2>" . $myrow[9] . "</span></td>";
echo "<td><span class=style2>" . $myrow[10] . "</span></td>";


$query1 = "SELECT SUM(amount) FROM report1 where staff_no = '$myrow[5]'";
$result1 = mysql_query($query1, $dblink);
while ($myrow1 = mysql_fetch_row($result1))
{
echo "<td><span class=style2>" .$myrow1[0]. "</span></td>";

  }
}
echo "</table>";
?>

why when i input one particular staff_no, it display blank when i click on submit button

june_c21 · May 4, 2008

thanks barand. topic solved!!!!!

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