[SOLVED] recognize pass

[Lambneck]

Hello,

I have a php code that displays a list of links.

when a link is chosen it is to display data from

mysql database based on the "submission_id"

specified by the link chosen. This data would

then be displayed on a new "display.php" page.

 

My question is how do i code the display.php

page in order for it to recognize the submission_id

being specified by the link chosen.

 

the code of the initial page:

 

    $result = mysql_query("SELECT submission_id, col_4 FROM $table ORDER BY submission_date DESC");
    if (!$result) {
        die("Query to show fields from table failed:".mysql_error());
    }

    while($row = mysql_fetch_array($result))
    {
      echo '<a href="Display.php?id='.$row['submission_id'].'">'.$row['col_4'].'</a>';
      echo "<br />";
    }
    mysql_free_result($result);

[dezkit]

    $result = mysql_query("SELECT submission_id, col_4 FROM $table ORDER BY submission_date DESC");
    if (!$result) {
        die("Query to show fields from table failed:".mysql_error());
    }

    while($row = mysql_fetch_array($result))
    {
      echo "<a href="Display.php?id=\".$row['submission_id'].'\">'.$row['col_4'].'</a>";
      echo "<br />";
    }
    mysql_free_result($result);

 

I didn't read what you wrote, but i found a mistake. `,=,

[Lambneck]

hello,

I appreciate the attempted help,

but I'd appreciate it a lot more if I could get an

answer to the question originally asked

instead of a wrong answer to an imagined question

 

 

[DarkWater]

Lol, Lambneck is right, your answer was wrong.

And just do $_GET['submission_id'] on the next page to get the submission ID. =D

[Lambneck]

thanks,

I'm working with the following code as of now,

but something isn't right cause it doesn't display.

 

$id = (int) $_GET['id']; // since the submission ID is named "id" in the query string!

$sql = "SELECT * FROM $table WHERE submission_id=$id";
$result = mysql_query($sql) or die("Error ". mysql_error(). " with query ". $sql);
if(mysql_num_rows($result) == 1){
$row = mysql_fetch_assoc($sql);
//print out information
}else{
echo 'That record ID does not exist!';
}
?>

[Lambneck]

ok it recognizes the pass; the submission id number is in the url

but on the page none of the data is displayed.  ???

anyone see anything wrong?

 

$id = (int) $_GET['id']; // since the submission ID is named "id" in the query string!

$sql = "SELECT * FROM $table WHERE submission_id=$id";
$result = mysql_query($sql) or die("Error ". mysql_error(). " with query ". $sql);
if(mysql_num_rows($result) == 1){
$row = mysql_fetch_assoc($sql);
//print out information
}else{
echo 'That record ID does not exist!';
}
?>