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Pass a variable through a hyperlink to a new page

Posted on 2008-10-01
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Last Modified: 2010-04-21
Hi.  I'd like to pass a variable with/through a hyperlink to another page that will execute a mysql query using the passed variable.  The variable is a result from a query executed.   Page 1 works well (enough for me at least).  I get the following error when I click a hyperlink to go to Page 2:  
Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in ...projectsearch_grid.php on line 200.

Page 1 executes a query and displays the results in a simple table.  I want to make the results in the "Project" column hyperlinks that point toward Page 2.  Page 2 would then execute a second query and display the results using the selected project.  Page 1 displays results from a search to find a project(s) for whatever criteria.  Page 2 will display more details about the specific project that the user clicked on.

Page 1 query: SELECT Name, Address, City FROM Projects WHERE City LIKE '%city%'
Page 1 results:  Project 1
                          Project 2
                          Project 3
User clicks on "Project 3"

Page 2 query:  SELECT * FROM `Projects` WHERE `Project_ID`='Project 3'

My best but still probably horrible attempt is below.  Thanks in advance.
197   while($row = mysql_fetch_array($result))

198    {

199	echo '<tr><td valign="top" align="left">';

200	echo '<a href="project_detail.php"  target="_blank">'.$row['Project_Name'].'</a> <form

                     action="project_detail.php" method="post"> <input type="hidden" name="projnumber" value="<?php

                     $row['PROJECT_#']; ?>"></form>';

201	echo '</td><td valign="top" align="left">';

202	echo $row['Type_And_Team'];

203	echo '</td><td valign="top" align="left">';

204	echo $row['Address'];
 

	}

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Question by:foxymoron7
  • 6
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12 Comments
 
LVL 2

Expert Comment

by:idealws
ID: 22620237
At first glance looks like you just forgot to format your echo on line 200. Try the below it should solve your error problem.
// You have:

value="<?php $row['PROJECT_#']; ?>"
 

// Should be:

value="'.$row['PROJECT_#'].'"

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Author Comment

by:foxymoron7
ID: 22621133
Thanks for taking a look idealws.  That change didn't work.  Same error message.  Any other thoughts?
0
 

Assisted Solution

by:ijaaacek
ijaaacek earned 150 total points
ID: 22621852
What about this?

echo "<a href=\"project_detail.php\" target=\"_blank\">$row['Project_Name']</a><form action=\"project_detail.php\" method=\"post\"><input type=\"hidden\" name=\"projnumber\" value=\"$row['PROJECT_#']\"></form>";

0
 
LVL 2

Expert Comment

by:idealws
ID: 22626059
Modifying it this way should handle the error. A few changes to the code posted by ijaaacek
echo "<a href=\"project_detail.php\" target=\"_blank\">".$row['Project_Name']."</a><form action=\"project_detail.php\" method=\"post\"><input type=\"hidden\" name=\"projnumber\" value=\"".$row['PROJECT_#']."\"></form>";

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Author Comment

by:foxymoron7
ID: 22626233
Thanks to both of you.  That works to make the new window open.  Now I can't seem to get the variable in the "project_detail.php" page.  To test if I was passing the variable through, project_detail.php is only the code below.  The page returns "variable empty".

If you don't mind taking a look to see if you can help I would appreciate it.

Thanks again, idealws and ijaaacek.
<?php
 
 

$projnumber = $_POST['projnumber'];
 

If(!$projnumber)

	{echo "variable empty";}

	print $projnumber;
 

?>

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Accepted Solution

by:
idealws earned 350 total points
ID: 22626359
You can do it using get instead like so:
<?php

while($row = mysql_fetch_array($result))

   {

echo '<tr><td valign="top" align="left">';

echo "<a href=\"project_detail.php?projnumber=".$row['PROJECT_#']."\" target=\"_blank\">".$row['Project_Name']."</a>";

echo '</td><td valign="top" align="left">';

echo $row['Type_And_Team'];

echo '</td><td valign="top" align="left">';

echo $row['Address'];

 

}
 
 

?>
 
 

project_detail.php

<?php

 

 

$projnumber = $_GET['projnumber'];

 

If(!$projnumber)

	{echo "variable empty";}

	print $projnumber;

 

?>

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Assisted Solution

by:idealws
idealws earned 350 total points
ID: 22626411
Or like this for using post:
<?php

while($row = mysql_fetch_array($result))

   {

echo '<tr><td valign="top" align="left">';

echo "<form action=\"project_detail.php\" method=\"post\"><input type=\"hidden\" name=\"projnumber\" value=\"".$row['PROJECT_#']."\"><input type=\"submit\" value=\"".$row['Project_Name']."\"></form>";

echo '</td><td valign="top" align="left">';

echo $row['Type_And_Team'];

echo '</td><td valign="top" align="left">';

echo $row['Address'];

 

}
 

?>
 

project_detail.php

<?php

 

 

$projnumber = $_POST['projnumber'];

 

If(!$projnumber)

	{echo "variable empty";}

	print $projnumber;

 

?>

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0
 
LVL 1

Author Comment

by:foxymoron7
ID: 22626876


Thanks again idealws.  Both methods worked.  I'm going to go with the GET method.  Two follow up questions is you have time.  Why is it necessary to escape the quotes?  In theory, is it possible to use POST along with a hyperlink?  The GET method works for this specific example because the project # isn't sensitive information but I can imagine wanting to hide some info in the future but use a similar method.  In no way is this a criticism of your POST solution but I didn't like the look of the Submit button in the table of results.  Thanks for all of your help.
0
 
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Author Closing Comment

by:foxymoron7
ID: 31502219
Thanks  idealws and ijaaacek
0
 
LVL 2

Expert Comment

by:idealws
ID: 22626993
You can use some java to submit the form with a text link like so I was simply using a simple example of how you can do what your trying to do. Hope this helps:
<script language="JavaScript" type="text/javascript">

<!--

function submitNumber (selectednum)

{

  document.form1.projnumber.value = selectednum ;

  document.form1.submit() ;

}

-->

</script>
 

<?php

while($row = mysql_fetch_array($result))

   {

echo '<tr><td valign="top" align="left">';

echo "<form name=\"form1\" action=\"project_detail.php\" method=\"post\" target=\"_Blank\"><input type=\"hidden\" name=\"projnumber\"><a href=\"javascript:submitNumber('".$row['PROJECT_#']."')\">".$row['Project_Name']."</a></form>";

echo '</td><td valign="top" align="left">';

echo $row['Type_And_Team'];

echo '</td><td valign="top" align="left">';

echo $row['Address'];

 

}

 

?>

 

project_detail.php

<?php

 

 

$projnumber = $_POST['projnumber'];

 

If(!$projnumber)

	{echo "variable empty";}

	print $projnumber;

 

?>

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0
 
LVL 2

Expert Comment

by:idealws
ID: 22627013
By the way the above will only work for one form submittion. Im not the best with Java but it is a siomple solution to get your pointed in the right direction.
0
 
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Author Comment

by:foxymoron7
ID: 22627151
Thanks!
0

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