make link from a variable.

Question

tulipputih 0 Junior Poster

14 Years Ago

Hi all.. really need a help. this is looks easy peasy but I can't solve it :(

I want to make a link from lesson ID to a details of the selected lesson ID. <td> "<a href='details.php?lessonID=". $lessonID ."'> </a>" </td> when I run the coding the following error appear on that line.

Parse error: parse error, expecting `','' or `';'' in C:\xampp\htdocs\lina\resultlink.php on line 65.

How the details page should be constructed so that the lessonID can be passed to it.?

many thanks for your help.
-tulip

php

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Latest Post 14 Years Ago Latest Post by navi17

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djjjozsi 2 Junior Poster in Training

14 Years Ago

<?php

$lessonID =2;

?>
<td><a href='details.php?lessonID=<?php echo $lessonID; ?>'> </a></td>

diafol

14 Years Ago

<?php

$lessonID =2;

?>
<td><a href='details.php?lessonID=<?php echo $lessonID; ?>'> </a></td>

<?php
$lesson = 2;
echo "<td><a href='details.php?lessonID={$lessonID}'</a></td>";
?>

If you need to stripslash/escape the value, do it beforehand:

$lesson = stripslashes($other_var);
...

Rather than mess around with the "html output".

ShawnCplus 456 Code Monkey

14 Years Ago

Thanks all, for your input..
I have tried all but none is success..the same error still appear on the same line.
thanks

Then show all of the code, not just one line.

navi17 5 Junior Poster

14 Years Ago

it will print the data from the database.
i am sure.

tulipputih commented: very helpful +1

djjjozsi 2 Junior Poster in Training

14 Years Ago

it will print the data from the database.
i am sure.

i've seen your first version, and in that post i did not see the <?php echo $field... ; ?>
which needed to print the values.

if i saw it wrong then sorry about it.

tulipputih commented: good suggestion +1

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furqan219 -9 Junior Poster in Training · Answer 1 · 2009-06-09T17:23:57+00:00

try this

echo "<td bgcolor=#9B5DCA><a href='link_page.php?id={$lessonID'>".stripslashes(delete)."</a>
	   </td>";

tulipputih 0 Junior Poster · Answer 2 · 2009-06-10T03:04:41+00:00

Thanks all, for your input..
I have tried all but none is success..the same error still appear on the same line.

thanks

tulipputih 0 Junior Poster · Answer 3 · 2009-06-10T13:01:21+00:00

ShawnCplus,
many thanks for your kind help.
this is the full code..
I have the details page ( details.php) to be linked from the selected ID).

thanks,

<?php

mysql_connect("localhost","root","");

mysql_select_db("student") or die("Unable to select database"); 
?>

<form name="form" action="result.php" method="get">

Subject   <input type="text" name="q" />
Topic <input type="text" name="r" />
Ability <input type="text" name="s" />

  <input type="submit" name="Submit" value="Search" />
</form>

<?php
  // Get the search variable from URL
 $q = $_GET['q'] ;

  $r = $_GET['r'] ;

  $s = $_GET['s'] ;

// Build SQL Query  
 $myquery = "select * from lesson where subject like '%".$q."%'  AND learningArea like '%".$r."%'  AND  ability like '%".$s."%'";
//echo $myquery ;
 $result = mysql_query ($myquery); 


 echo "<table>"  ;
echo "<tr><th>Lesson ID</th>";
echo "<th>Subject</th>";
echo "<th>Learning Area</th>";
echo "<th>No of Students</th>";
echo "<th>Time Period</th>";
echo "<th>Class</th>";
echo "<th>Ability</th></tr>";

while ($row= mysql_fetch_array($result)) 
 {
$lessonID = $row["lessonID"];  
  $subject = $row["subject"];
  $learningArea= $row["learningArea"];
    $noofstudent= $row["noofstudents"];
    $minutes= $row["minutes"];
    $class= $row["class"];
    $ability= $row["ability"];

//display the row
   
echo "<tr>
<td><a href='details.php?lessonID=<?php echo $lessonID; ?>'> </a></td>
<td>$subject</td> 
<td>$learningArea</td> 
<td>$noofstudents</td> 
<td>$minutes</td>  
<td>$class</td>  
<td> $ability</td>  
</tr>" ;
}
echo "</table>";
  
?>

navi17 5 Junior Poster · Answer 4 · 2009-06-10T13:21:13+00:00

I think the problem is here

$myquery = "select * from lesson where subject like '%".$q."%' AND learningArea like '%".$r."%' AND ability like '%".$s."%'";

it should be:

$myquery = "select * from lesson where subject like '%'.$q.'%' AND learningArea like '%'.$r.'%' AND ability like '%'.$s.'%'";

Try this and let me know it is working or not?

tulipputih 0 Junior Poster · Answer 5 · 2009-06-10T14:13:16+00:00

Hi, I have tried the one that you suggested.
It doesn't work.
I don't have any problem to listed down the result..just to make a link to a detail page based on the lessonID chosen. Error on the same line
<td><a href='details.php?lessonID=<?php echo $lessonID; ?>'> </a></td>

furqan219 -9 Junior Poster in Training · Answer 6 · 2009-06-10T14:20:25+00:00

please use echo in start of
<td><a href='details.php?lessonID=<?php echo $lessonID; ?>'> </a></td>

this line and terminator at the end

djjjozsi 2 Junior Poster in Training · Answer 7 · 2009-06-10T14:23:48+00:00

echo "<tr>
<td><a href='details.php?lessonID=$lessonID'>Details</a></td>
<td>$subject</td>
<td>$learningArea</td>
<td>$noofstudents</td>
<td>$minutes</td>
<td>$class</td>
<td> $ability</td>
</tr>" ;

djjjozsi 2 Junior Poster in Training · Answer 8 · 2009-06-10T14:42:10+00:00

this is a typo error in your SQL query then:-O

..........
it should be:

$myquery = "select * from lesson where subject like '%'.$q.'%' AND learningArea like '%'.$r.'%' AND ability like '%'.$s.'%'";

........

and,
the SQL save version is this:

$q = isset($_GET['q'])? mysql_real_escape_string($_GET['q']):'';
$r = isset($_GET['r'])? mysql_real_escape_string($_GET['r']):'';
$s = isset($_GET['s'])? mysql_real_escape_string($_GET['s']):'';

// Build SQL Query
$myquery = "select * from lesson where subject like '%{$q}%' AND learningArea like '%{$r}%'
 AND ability like '%{$s}%'";

navi17 5 Junior Poster · Answer 9 · 2009-06-10T18:16:55+00:00

try this:

while ($row= mysql_fetch_array($result))
{
$lessonID = $row["lessonID"];
$subject = $row["subject"];
$learningArea= $row["learningArea"];
$noofstudent= $row["noofstudents"];
$minutes= $row["minutes"];
$class= $row["class"];
$ability= $row["ability"];

?>
//display the row

<tr>
<td><a href="details.php?lessonID=<?php echo $lessonID; ?>"> Details</a></td>
<td><?php echo $subject; ?></td>
<td><?php echo $learningArea; ?></td>
<td><?php echo $noofstudents; ?></td>
<td><?php echo $minutes; ?></td>
<td><?php echo $class; ?></td>
<td><?php echo $ability; ?> </td>
</tr>
<?php  } ?>
</table>

i am sure it works.

djjjozsi 2 Junior Poster in Training · Answer 10 · 2009-06-10T18:20:02+00:00

but its not printing the data from the database... you will see $minutes and other stuffs in the output...
.....
the shortest is the better...

tulipputih 0 Junior Poster · Answer 11 · 2009-06-10T23:47:57+00:00

I have tried that as well..
please anyone who can identify the error...

thanks.

tulipputih 0 Junior Poster · Answer 12 · 2009-06-10T23:54:36+00:00

try this:

while ($row= mysql_fetch_array($result))
{
$lessonID = $row["lessonID"];
$subject = $row["subject"];
$learningArea= $row["learningArea"];
$noofstudent= $row["noofstudents"];
$minutes= $row["minutes"];
$class= $row["class"];
$ability= $row["ability"];

?>
//display the row

<tr>
<td><a href="details.php?lessonID=<?php echo $lessonID; ?>"> Details</a></td>
<td><?php echo $subject; ?></td>
<td><?php echo $learningArea; ?></td>
<td><?php echo $noofstudents; ?></td>
<td><?php echo $minutes; ?></td>
<td><?php echo $class; ?></td>
<td><?php echo $ability; ?> </td>
</tr>
<?php  } ?>
</table>

i am sure it works.

Hi Navi17.
thanks a lot..can be linked to details page.
however, the Id number is not shown in the result..
all show the same string..details..but the lesson Id that sent to details page is correct (the lessonID)
just the Id is not displayed in the table.
any idea to correct it?

thanks

tulipputih 0 Junior Poster · Answer 13 · 2009-06-11T00:03:32+00:00

i've seen your first version, and in that post i did not see the <?php echo $field... ; ?>
which needed to print the values.
if i saw it wrong then sorry about it.

djjjozsi, it does work for me..I mean the link.
the code that I posted before also print data from the database..
The problem is to make a link to the details page.
now, Navi17 has solved that problem..however the ID is not printed here..but the value is passed to the details page... I need both..the lessonID printed & the value passed.
ANy solution. Mny thanks for your respond

tulipputih 0 Junior Poster · Answer 14 · 2009-06-11T00:09:27+00:00

Thanks all for your help..
NAvi17 & djjjozsi i have corrected it now..
credits for both of you :)
It works perfectly now.

djjjozsi 2 Junior Poster in Training · Answer 15 · 2009-06-11T00:16:02+00:00

you can print the $lessonID after the Datails word...

I don't like saying this but my version still created the same output like navi17's example,

decide which is shorten or which comes earier :)

echo "<tr>
      <td><a href='details.php?lessonID=$lessonID'>Details for $subject( $lessonID )</a></td>
      <td>$subject</td>
      <td>$learningArea</td>
      <td>$noofstudents</td>
      <td>$minutes</td>
      <td>$class</td>
      <td> $ability</td>
      </tr>" ;

Josh Connerty 20 Unverified User · Answer 16 · 2009-06-11T10:08:59+00:00

<?php

mysql_connect("localhost","root","");

mysql_select_db("student") or die("Unable to select database"); 
?>

<form name="form" action="result.php" method="get">

Subject   <input type="text" name="q" />
Topic <input type="text" name="r" />
Ability <input type="text" name="s" />

  <input type="submit" name="Submit" value="Search" />
</form>

<?php
  // Get the search variable from URL
 $q = $_GET['q'] ;

  $r = $_GET['r'] ;

  $s = $_GET['s'] ;

// Build SQL Query  
 $myquery = "select * from lesson where subject like '%".$q."%'  AND learningArea like '%".$r."%'  AND  ability like '%".$s."%'";
//echo $myquery ;
 $result = mysql_query ($myquery); 


 echo "<table>"  ;
echo "<tr><th>Lesson ID</th>";
echo "<th>Subject</th>";
echo "<th>Learning Area</th>";
echo "<th>No of Students</th>";
echo "<th>Time Period</th>";
echo "<th>Class</th>";
echo "<th>Ability</th></tr>";

while ($row= mysql_fetch_array($result)) 
 {
$lessonID = $row["lessonID"];  
  $subject = $row["subject"];
  $learningArea= $row["learningArea"];
    $noofstudent= $row["noofstudents"];
    $minutes= $row["minutes"];
    $class= $row["class"];
    $ability= $row["ability"];

//display the row
   
echo <<<EOD
<tr>
<td><a href="details.php?lessonID=$lessonID"> </a></td>
<td>$subject</td> 
<td>$learningArea</td> 
<td>$noofstudents</td> 
<td>$minutes</td>  
<td>$class</td>  
<td> $ability</td>  
</tr>
EOD;
}
echo "</table>";
  
?>

navi17 5 Junior Poster · Answer 17 · 2009-06-11T16:54:38+00:00

tulipputih if your problem is solved then mark this forum as solved.

make link from a variable.

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