Annoying error

Sooz719

This page is simply meant to look up some information from a database and print it out in a table.

<?php session_start();
 include("header.inc");

//open db connection
$link=mysqli_connect('localhost','root','voBtGM','JobBase');


$stored_username = $_SESSION['userName'];

$select="SELECT JobTitle, Salary, Location, Description FROM Jobs WHERE Username = $stored_username;
$result=mysqli_query($link,$select);


?>
	<TABLE>
	  <TR><TH>Job Title</TH>
	      <TH>Salary</TH>
	      <TH>Location</TH>
	      <TH>Description</TH>
	  </TR>


<?while ($row = mysqli_fetch_array($result)) {
	  $Title= $row['JobTitle'];
	  $Sal= $row['Salary'];
	  $Location=$row['Location'];
	  $Description=$row['Description'];

?>
	  <TR><TD>$Title</TD>
	      <TD>$Sal</TD>
	      <TD>$Location</TD>
	      <TD>$Description</TD>
	  </TR>
	}


</TABLE>

<?mysqli_close($link);?>


</body>
</HTML>

On running it i get the following error :

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in C:\Program Files\Apache Group\Apache2\htdocs\AdvertisersHome.php on line 25

Can someone advise me why this is?

Thanks for any help

tripodkid

Looks like your missing a quote at the end of line 11.

Try this:

$select="SELECT JobTitle, Salary, Location, Description FROM Jobs WHERE
 Username = $stored_username; $result=mysqli_query($link,$select)";

thorpe

Should be...

$select="SELECT JobTitle, Salary, Location, Description FROM Jobs WHERE Username = $stored_username";
$result=mysqli_query($link,$select);

Sooz719

I have repaired this mistake and now get this error :
Parse error: syntax error, unexpected $end in C:\Program Files\Apache Group\Apache2\htdocs\AdvertisersHome.php on line 48

Any ideas?

tripodkid

You need to put PHP tags around your closing bracket.

<?php 
              } 
?>

Sooz719

Ok, thats fixed that problem......now i have another one.

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in C:\Program Files\Apache Group\Apache2\htdocs\AdvertisersHome.php on line 24

It also doesnt print the results. I presume that goes hand in hand with the above error.

heeeeelp!!

bodzan

$result=mysqli_query($select,$link);

Sooz719

doing that only created another error, it now comes up with the following :

Warning: mysqli_query() expects parameter 1 to be mysqli, string given in C:\Program Files\Apache Group\Apache2\htdocs\AdvertisersHome.php on line 11

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in C:\Program Files\Apache Group\Apache2\htdocs\AdvertisersHome.php on line 24

bodzan

I just read in manual that the first parameter should be a link to a database not query like I thought... It is mysqli_query and not mysql_query after all, heh...

Then my second guess would be to put single quotes around a string in your query...

 $select="SELECT JobTitle, Salary, Location, Description FROM Jobs WHERE Username = '$stored_username'";