Parse error: syntax error, unexpected T_VARIABLE

[june_c21]

can someone tell me what wrong with my wrong code? i get this error "Parse error: syntax error, unexpected T_VARIABLE line 2"

 

<? php
$host = 'localhost';
$user = 'root';
$password = 'admin';
$dbase = 'claim';

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

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


 $name = $_GET['name'];

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

}

echo "</table>";

?>

[Rohan Shenoy]

In the first line should it be

<?php

instead of

<? php

Try removing the space there.

[BrianM]

Rohan, now you want to help with the script I posted!?

[june_c21]

thanks rohan

 

but now the parse error: syntax error, unexpected T_VARIABLE  is at line 26 which is this line

 

$query = "SELECT report1.id, month, start_date, end_date, user.name, report1.staff_no,details,amount,gl_code,user.bank,user.acc_no FROM user,report1 WHERE user.name = "$name"";

[conker87]

$query = "SELECT report1.id, month, start_date, end_date, user.name, report1.staff_no,details,amount,gl_code,user.bank,user.acc_no FROM user,report1 WHERE user.name = '$name'";

[BrianM]

$query = "SELECT report1.id, month, start_date, end_date, user.name, report1.staff_no, details, amount, gl_code, user.bank, user.acc_no FROM user, report1 WHERE user.name = '$name'";

 

That should work.

[june_c21]

now there is no error. thanks for all the help .

but why when i wrote

 

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

}

 

it came out all the data instead of where the staff_no = $name only?

[conker87]

Try mysql_fetch_array instead.

[june_c21]

it still the same. all the data will come out instead of match report1.staff_no = '$name' only. why?

[BrianM]

$name = $_POST['name'];

$query = "SELECT report1.id, month, start_date, end_date, user.name, report1.staff_no, details, amount, gl_code, user.bank, user.acc_no FROM user, report1 WHERE report1.staff_no = '$name'";

$result = mysql_query($query, $dblink);

while($myrow = mysql_fetch_array($result)) {
echo "<tr><td>" . $myrow[0] . "</td>";
  	echo "<td>" . $myrow[1] . "</td>";
  	echo "<td>" . $myrow[2] . "</td>";
  	echo "<td>" . $myrow[3] . "</td>";
  	echo "<td>" . $myrow[4] . "</td>";
   	echo "<td>" . $myrow[5] . "</td>";
   	echo "<td>" . $myrow[6] . "</td>";
   	echo "<td>" . $myrow[7] . "</td>";
echo "<td>" . $myrow[8] . "</td>";
echo "<td>" . $myrow[9] . "</td>";
echo "<td>" . $myrow[10] . "</td></tr>";
}

 

Try that.

[june_c21]

Brian,

 

the result still the same. all data came out when i click on submit button.

[BrianM]

So state again what exactly it is that your wanting the page to do when you click the submit button.

[june_c21]

ok, first this is my html coding where user can chose the staff_no and click on submit. once submit, it will find in the database the same staff_no that match the one user select. if it is yes, it will display the records base on the staff_no .

 

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

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

  $query = "SELECT staff_no, name from user order by name ";
      		$result = mysql_query($query);
		while($myrow = mysql_fetch_row($result))
		{
			echo "<OPTION VALUE=\"" . $myrow[0] . "\">" . $myrow[1]."</OPTION>";
		}
?>
            </select></td>
    </tr>
  </table>
  <p> </p>
  <p>
    <input type="submit" name="Submit" value="Submit">
  </p>
</form></body>
</html>

 

<?php
$host = "localhost";
$user = "root";
$password = "admin";
$dbase = "claim";

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

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

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

?>

[BrianM]

<html>
<head>
<script src="selectuser.js"></script>
</head>
<body>
<form id="form1" name="form1" method="post" action="checking.php">
  <p><strong>KEV Claim</strong></p>
  <table width="497" border="0">
    <tr>
      <td width="284">Claim</td>
      <td width="203"><select name="name" id="name" onChange="showUser(this.value)">

<?php

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

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

mysql_select_db($dbase, $dblink);

$query = "SELECT staff_no, name FROM user ORDER BY name";
      		
$result = mysql_query($query);

while($myrow = mysql_fetch_row($result)) {

echo "<option value=\"" . $myrow[0] . "\">" . $myrow[1]."</option>";
}

?>
        </select></td>
    </tr>
  </table>
  <p> </p>
  <p>
    <input type="submit" name="Submit" value="Submit">
  </p>
</form>
</body>
</html>

 

<?php

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

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

mysql_select_db($dbase, $dblink);

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

$name = $_POST['name'];

$query = "SELECT report1.id, month, start_date, end_date, user.name, report1.staff_no, details, amount, gl_code, user.bank, user.acc_no FROM user, report1 WHERE report1.staff_no = '$name'";

$result = mysql_query($query, $dblink);

while($myrow = mysql_fetch_array($result)) {
echo "<tr><td>" . $myrow . "</td>";
  	echo "<td>" . $myrow . "</td>";
  	echo "<td>" . $myrow . "</td>";
  	echo "<td>" . $myrow . "</td>";
  	echo "<td>" . $myrow . "</td>";
   	echo "<td>" . $myrow . "</td>";
   	echo "<td>" . $myrow . "</td>";
   	echo "<td>" . $myrow . "</td>";
echo "<td>" . $myrow . "</td>";
echo "<td>" . $myrow . "</td>";
echo "<td>" . $myrow . "</td></tr>";
}

echo "</table>";

?>

[june_c21]

brian, did u changed anything in the code?

[BrianM]

Yes, what are the results?

[june_c21]

the result still the same.... can someone please help me.....

 

brian, thanks for your code... but it doesn't work..

[BrianM]

Yes, what were the results?

[june_c21]

it came out all the username from database and changed their stuff_no into same staff_no. (250 records)

[beansandsausages]

If you only want one result, Limit it to just one result.

[june_c21]

how to do so?

[beansandsausages]

$query = "SELECT staff_no, name FROM user ORDER BY name LIMIT 0,1 ";

 

sumit like that,

[june_c21]

but that haven't solve my problem. i need to output the user that have the same staff no not all the user.

i wrote

 

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 = '$name'

 

 

but it output all the staff.

[beansandsausages]

 

try sumit like :

 

select * from `users` WHERE `report1.staff_no`='{$name}'

 

see if that works

[june_c21]

no .. it still doesnt work. please help and guide me more. i am new. thanks in advances