asked on

Pass a variable through a hyperlink to a new page

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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idealws

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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foxymoron7

ASKER

Thanks for taking a look idealws. That change didn't work. Same error message. Any other thoughts?

SOLUTION

ijaaacek

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idealws

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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foxymoron7

ASKER

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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ASKER CERTIFIED SOLUTION

idealws

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SOLUTION

idealws

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foxymoron7

ASKER

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.

foxymoron7

ASKER

Thanks idealws and ijaaacek

idealws

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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idealws

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.

foxymoron7

ASKER

Thanks!