Why getting parse error in PHP at line 20?

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  • Rana Chakra
    New Member
    • Sep 2010
    • 11
    #1

    Why getting parse error in PHP at line 20?

    Code:
    <?php
$username="root";
$password="";
$database="lko_phone";
$host="localhost";
$msg="";
mysql_connect($host,$username,$password);
@mysql_select_db($database);

if($_SERVER["REQUEST_METHOD"] == "POST")
{
  $myusername=$_POST['unm'];
  $mypassword=$_POST['pwd'];

  $sql="SELECT * FROM login WHERE id='$myusername' and pwd='$mypassword'";
  $result=mysql_query($sql);
  $row = mysql_fetch_assoc($result)

  // Mysql_num_row is counting table row
  $num = mysql_num_rows($result);

  // If result matched $myusername and $mypassword, table row must be 1 row

  if($count==1)
  {

    session_start();

    $_SESSION['UID'] = $row['id'];
    $_SESSION['PWD'] = $row['pwd'];
    $_SESSION['RGT'] = $row['access'];

    header("location:Welcome.php");
  }
  else
  {
    msg="Wrong user name and/or Password"
  }

}

mysql_free_result($result);

?>

<html>
<body background="Images/Back15.jpg" text='black'>
<center>
<table width=100% border=0>
<tr>
   <td align=center><img src="Images/Logo.JPG"></td>
   <td><font face="courier new" size="5" color="orange"><strong>Welcome To: LII - ESS maintenance</strong></font>
   <br><font face="courier new" size="2" color="yellow">Employee-System-Software Maintenance</font></td>
</tr>
</table>
</center>

<form action="index.php" method=POST>
<center>
<table width="200" cellpadding="0" cellspacing="0" border="0">
<tr>
  <td width="14"><img src="TableImage/TL.png" width=14 height=23 border="0" /></td>

  <td background="Images/Back1.jpg" colspan="2" align="center">
	<font face='arial' size='4' color='maroon'><strong>
	Enter Login Information
	</font>
  </td>

<td width="14"><img src="TableImage/TR.png" width=14 height=23 border="0" /></td>

</tr>

<tr>

<td background="Images/Back1.jpg">&nbsp</td>
<td background="Images/Back1.jpg">&nbsp</td>
<td background="Images/Back1.jpg">&nbsp</td>
<td background="Images/Back1.jpg">&nbsp</td>

</tr>

<tr>

<td background="Images/Back1.jpg"></td>

<td background="Images/Back1.jpg" align="right"><font face='verdana' size='4'>Username:</font></td>
<td background="Images/Back1.jpg"><input type='text' name='unm'/></td>

<td background="Images/Back1.jpg"></td>

</tr>

<tr>

<td background="Images/Back1.jpg"></td>

<td background="Images/Back1.jpg" align="right"><font face='verdana' size='4'>Password:</font></td>
<td background="Images/Back1.jpg"><input type='password' name='pwd'></td>

<td background="Images/Back1.jpg"></td>

</tr>

<tr>

<td background="Images/Back1.jpg">&nbsp</td>
<td background="Images/Back1.jpg">&nbsp</td>
<td background="Images/Back1.jpg">&nbsp</td>
<td background="Images/Back1.jpg">&nbsp</td>

</tr>

<tr>

<td background="Images/Back1.jpg">&nbsp</td>

<td background="Images/Back1.jpg" align="center" colspan="2"><input type='submit' name='log' value='  login  '></td>

<td background="Images/Back1.jpg">&nbsp</td>

</tr>

<tr>

<td><img src="TableImage/BL.png" width=14 height=23 border="0" /></td>
<td background="Images/Back1.jpg">&nbsp</td>
<td background="Images/Back1.jpg">&nbsp</td>
<td><img src="TableImage/BR.png" width=14 height=23 border="0" /></td>

</tr>
</table>
</center>
<p align=Center><?php echo $msg ?></p>
</form>
</body>
</html>
    Tags: None
    • vjayis
      New Member
      • Mar 2008
      • 134
      #2
      semicolon missing in line 17.,

      try applying it.

      Code:
       $row = mysql_fetch_assoc($result);

        Comment

        • Rana Chakra
          New Member
          • Sep 2010
          • 11
          #3
          Modified - Now getting in Line 38

          Code:
          <?php
$username="root";
$password="";
$database="lko_phone";
$host="localhost";
$msg="";
mysql_connect($host,$username,$password);
@mysql_select_db($database);

if($_POST)
{
  $myusername=$_POST['unm'];
  $mypassword=$_POST['pwd'];

  $sql="SELECT * FROM login WHERE id='$myusername' and pwd='$mypassword'";
  $result=mysql_query($sql);
  $row = mysql_fetch_assoc($result);

  // Mysql_num_row is counting table row
  $num = mysql_num_rows($result);

  echo $num;
  // If result matched $myusername and $mypassword, table row must be 1 row

  if($num==1)
  {

    session_start();

    $_SESSION['UID'] = $row['id'];
    $_SESSION['PWD'] = $row['pwd'];
    $_SESSION['RGT'] = $row['access'];

    header("location:Welcome.php");
  }
  else
  {
    msg="Wrong user name and/or Password";
  }

}

mysql_free_result($result);

?>

<html>
<body background="Images/Back15.jpg" text='black'>
<center>
<table width=100% border=0>
<tr>
   <td align=center><img src="Images/Logo.JPG"></td>
   <td><font face="courier new" size="5" color="orange"><strong>Welcome To: LII - ESS maintenance</strong></font>
   <br><font face="courier new" size="2" color="yellow">Employee-System-Software Maintenance</font></td>
</tr>
</table>
</center>

<form action="index.php" method=POST>
<center>
<table width="200" cellpadding="0" cellspacing="0" border="0">
<tr>
  <td width="14"><img src="TableImage/TL.png" width=14 height=23 border="0" /></td>

  <td background="Images/Back1.jpg" colspan="2" align="center">
	<font face='arial' size='4' color='maroon'><strong>
	Enter Login Information
	</font>
  </td>

<td width="14"><img src="TableImage/TR.png" width=14 height=23 border="0" /></td>

</tr>

<tr>

<td background="Images/Back1.jpg">&nbsp</td>
<td background="Images/Back1.jpg">&nbsp</td>
<td background="Images/Back1.jpg">&nbsp</td>
<td background="Images/Back1.jpg">&nbsp</td>

</tr>

<tr>

<td background="Images/Back1.jpg"></td>

<td background="Images/Back1.jpg" align="right"><font face='verdana' size='4'>Username:</font></td>
<td background="Images/Back1.jpg"><input type='text' name='unm'/></td>

<td background="Images/Back1.jpg"></td>

</tr>

<tr>

<td background="Images/Back1.jpg"></td>

<td background="Images/Back1.jpg" align="right"><font face='verdana' size='4'>Password:</font></td>
<td background="Images/Back1.jpg"><input type='password' name='pwd'></td>

<td background="Images/Back1.jpg"></td>

</tr>

<tr>

<td background="Images/Back1.jpg">&nbsp</td>
<td background="Images/Back1.jpg">&nbsp</td>
<td background="Images/Back1.jpg">&nbsp</td>
<td background="Images/Back1.jpg">&nbsp</td>

</tr>

<tr>

<td background="Images/Back1.jpg">&nbsp</td>

<td background="Images/Back1.jpg" align="center" colspan="2"><input type='submit' name='log' value='  login  '></td>

<td background="Images/Back1.jpg">&nbsp</td>

</tr>

<tr>

<td><img src="TableImage/BL.png" width=14 height=23 border="0" /></td>
<td background="Images/Back1.jpg">&nbsp</td>
<td background="Images/Back1.jpg">&nbsp</td>
<td><img src="TableImage/BR.png" width=14 height=23 border="0" /></td>

</tr>
</table>
</center>
<p align=Center><?php echo $msg ?></p>
</form>
</body>
</html>

            Comment

            • Rana Chakra
              New Member
              • Sep 2010
              • 11
              #4
              SOlved it. wrote msg="...." instead of $msg="..."
              Thanks....

                Comment

                • vjayis
                  New Member
                  • Mar 2008
                  • 134
                  #5
                  yeah., u missed '$' symbol there .,

                  thanks

                    Comment

                    • Rana Chakra
                      New Member
                      • Sep 2010
                      • 11
                      #6
                      Thanks for replying

                      Thanks to you too for paying kind attention. will come back if got stuck anywhere.

                        Comment

                        • Rana Chakra
                          New Member
                          • Sep 2010
                          • 11
                          #7
                          mysql_fetch_ass oc() expects parameter 1 to be resource.. line 58

                          mysql_fetch_ass oc() expects parameter 1 to be resource at line 58... Please help
                          Code:
                          <?php
session_start();
if($_SESSION['UID'])
  {
   if($_SESSION['RGT'] == "R")
      header("location:NoAccess.php");
  }
else
      header("location:NoAccess.php");
$message="";
$username="root";
$password="";
$database="lko_phone";
$host="localhost";
mysql_connect($host,$username,$password);
@mysql_select_db($database);
?>
<html>
<head>
	<title>Delete Employee Record</title>
</head>
<body background="Images/Back5.jpg" text="white" link="white" alink="white" vlink="white">
<center>
<table background="Images/Back10.jpg" width="100%" border="0">
<tr>
   <td><img src="Images/Logo.JPG"></td>
   <td  align=center><font face="courier new" size="5" color="orange"><strong>Welcome To: LKO - TelWin</strong></font>
   <br><font face="courier new" size="2" color="yellow">Online Telephone Address Book/Call-Register Maintenance</font></td>
   <td align=right>
	<?php
	  $welcome=$_SESSION['UID'];
	  if(strcmp($welcome,"mbchaudhuri")==0)
	  {
	   $query = sprintf("SELECT * FROM notes");
	   $result=mysql_query($query);
	   $num=mysql_num_rows($result);
	   $message=sprintf("<font face=verdana color=aqua size=2>You have :<strong>".$num." Notes to see</strong></font><br>");
	  }
	  else
	   $message="";
	  echo "<font face=verdana size=2 color=yellow>Welcome: ".$welcome."<br>";
	  echo "<font face=verdana color=orange size=2>Today is :<strong>".date("d-M-Y, l")."</strong></font><br>";
	  echo $message;
	  echo "<a href=Logout.php>Logout</a>";
?>
</table>
</center>
<form action=SearchEmployee.php>
<h2>Deleted the following employee record..<?php echo $_POST['esrl']?></h2>
<?php
  if($_POST && !empty($_POST['Delete']))
  {
    $r=$_POST['esrl'];
    $query=sprintf("select * from employee where srl = $r");
    echo $query;
    $result=mysql_query($query);

    while($row = mysql_fetch_assoc($result))
    {
	echo "<br><strong>Employee Name:".$row['ename'];
	echo "<br>Designation  :".$row['desig'];
	echo "<br>Department   :".$row['dept'];
	echo "<br>Direct No    :".$row['d_no'];
	echo "<br>Extension    :".$row['extn'];
	echo "<br>Mobile No    :".$row['mobile'];
	echo "<br>Residential  :".$row['res_no'];
	echo "<br>Other Contact:".$row['other_no'];
	echo "</strong>";
    }
    echo mysql_error();
    echo "<br><br><input type=submit value='Back to search page' >";
    mysql_query("delete from emloyee where srl=$r");
  }

?>
</form>
</body>
</html>
                          Last edited by Rana Chakra; Sep 20 '10, 12:41 PM. Reason: Wrong subject

                            Comment

                            • Dormilich
                              Recognized Expert Expert
                              • Aug 2008
                              • 8694
                              #8
                              see http://bytes.com/topic/php/insights/...d-mysql-result

                                Comment

                                • Rana Chakra
                                  New Member
                                  • Sep 2010
                                  • 11
                                  #9
                                  Yep.. solved the problem. I changed the code as follows:
                                  Code:
                                  $query="select * from employee where srl ='". $r."'";
$result=mysql_query($query);
while($row = mysql_fetch_assoc($result))
    {
	echo "<br><strong>Employee Name:".$row['ename'];
	echo "<br>Designation  :".$row['desig'];
	echo "<br>Department   :".$row['dept'];
	echo "<br>Direct No    :".$row['d_no'];
	echo "<br>Extension    :".$row['extn'];
	echo "<br>Mobile No    :".$row['mobile'];
	echo "<br>Residential  :".$row['res_no'];
	echo "<br>Other Contact:".$row['other_no'];
	echo "</strong>";
    }
$query="delete from employee where srl ='". $r."'";
    mysql_query($query);
                                  Now working perfectly. I do not why it is taking string value, but, in table it srl is defined as integer.

                                    Comment

                                    • Dormilich
                                      Recognized Expert Expert
                                      • Aug 2008
                                      • 8694
                                      #10
                                      you pass srl as a string. (omit the ')

                                        Comment

                                        • Rana Chakra
                                          New Member
                                          • Sep 2010
                                          • 11
                                          #11
                                          code under if is working for i%3==1 .... help

                                          I just want to display three columns in one row. But, when value of $i is 1 or 4, the statements under if is firing. Dont know why it is happening, and how to solve it.
                                          Code:
                                          $i=0;
   while ($row = mysql_fetch_assoc($result))
   {
    if(($i%3)==0)
    {
      echo "<tr>";
      echo "<td><input type=radio name=JobNo value=".$row['job_no']." />"</td>";
      echo "<td><font face=verdana size=2>".$row['job_no']."</font></td>";
      echo "</tr>";
    }
    else
    {
      echo "<td><input type=radio name=JobNo value=".$row['job_no']." />"</td>";
      echo "<td><font face=verdana size=2>".$row['job_no']."</font></td>";
    }

    $i=$i+1;
   }

                                            Comment

                                            • Rana Chakra
                                              New Member
                                              • Sep 2010
                                              • 11
                                              #12
                                              OK got it.... <tr> and </tr> is firing for only one column...... Sorry for the sily question.

                                                Comment

                                                • Rana Chakra
                                                  New Member
                                                  • Sep 2010
                                                  • 11
                                                  #13
                                                  This code removed the problem....
                                                  Code:
                                                  <center><table border="1" bordercolor="orange">
  <tr>
	<td bgcolor="black">&nbsp</td>
	<td bgcolor="black"><font color="orange" face="verdana" size="2"><b>Job Number</b></font></td>
	<td bgcolor="black">&nbsp</td>
	<td bgcolor="black"><font color="orange" face="verdana" size="2"><b>Job Number</b></font></td>
	<td bgcolor="black">&nbsp</td>
	<td bgcolor="black"><font color="orange" face="verdana" size="2"><b>Job Number</b></font></td>
<?php
  $query = sprintf("SELECT * FROM job order by job_no");
  $result=mysql_query($query);
  $i=3;
  
   while ($row = mysql_fetch_assoc($result))
   {
    if(($i%3)==0) echo "</tr><tr>";

      echo "<td><input type=radio name=JobNo value=".$row['job_no']." />->".($i%3)."</td>";
      echo "<td><font face=verdana size=2>".$row['job_no']."</font></td>";

    $i=$i+1;
   }
  mysql_free_result($result);
  echo "</tr>";
?>
</table>

                                                    Comment

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