[SOLVED] display problem

[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.

But display.php code currently isn't working.

 

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);

 

 

Display.php:

 

$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!';
}
?>

 

I don't see whats wrong with the display.php page code,

but its not displaying the data.

Can anyone see a problem with it?

 

 

[DarkWater]

if(mysql_num_rows($result) == 1){

$row = mysql_fetch_assoc($sql);

//print out information

}else{

 

 

You aren't actually printing the data. >_>

[Lambneck]

what would the syntax look like then?

 

ive tried:

echo $row;

 

and

 

print $row;

 

with no luck  :'(

[DarkWater]

echo $row['COLUMN_NAME'];

 

mysql_fetch_assoc() returns an array with the keys named as each column, so if you had a column named "name", it'd be in $row['name'].

[Lambneck]

I tried the following:

 

$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);

echo $row['col_2'];
echo $row['col_3'];
echo $row['col_4'];
echo $row['col_5'];

}else{
echo 'That record ID does not exist!';
}
?>

 

However its still not displaying.

notice anything else missing?

 

thanks for all the help.

 

[phpSensei]

try

 

$id = mysql_real_escape_string(strip_tags($_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){
$row = mysql_fetch_array($sql);

echo $row[0];

}else{
echo 'That record ID does not exist!';
}
?>

[Lambneck]

same result... nothin

???

[phpSensei]

Your using mysql_fetch_array($sql)

 

change it to $result.

 

change

$row = mysql_fetch_array($sql);

to

$row = mysql_fetch_array($result);

[Lambneck]

sweet

its working!

 

Thanks phpSensei

Thanks BlackWater