Displaying PHP results?

Question

I need a little help here with some php. just a little explanation: (im trying to display 5 results using this code

    $n_id = mysql_real_escape_string ($_GET['id']);
    $path = '';

    if(isset($n_id) && $n_id != "") {
        $sql = 'SELECT * FROM test2 WHERE id="' . $n_id . '"';
        $news = mysql_query($sql);

    if($result = mysql_fetch_array($news)) {
        $title = mysql_result($news,0,"title");
        $date = mysql_result($news,0,"date");
        echo '<b>' . $title . ' | ' . $date . '</b>
        <br/>
        <img src="images.php?id='. $n_id .'>';
    } else {
        header("Location: vendi.php");
    }
    echo '<br /><a href="/'.$path.'">Back to Archive</a>';
    }

It does display but i want that 1st result to be (image+title of the news and other results to be just title). Hope i wrote it clearly what i needed help with. Thank you

jrn.ak · Accepted Answer · 2011-05-18 18:58:57Z

1

Your SQL statement is only fetching a single row. This isn't a complete solution, but should get you closer:

if (isset($_GET['id']) && is_numeric($_GET['id'])) {
    $n_id = (int)$_GET['id'];
    $path = '';

    $count = 0;

    $sql = 'SELECT * FROM test2 WHERE id BETWEEN ' . $n_id ' AND ' . ($n_id + 5);
    $news = mysql_query($sql);

    while ($result = mysql_fetch_array($news)) {
        $title = $result['title'];
        $date = $result['date'];
        echo '<b>' . $title . ' | ' . $date . '</b>';
        if ($count < 1) {
            echo '<br/><img src="images.php?id='. $n_id .'>';
            $count++;
        }
    }

    if ($count == 0) { header("Location: vendi.php"); }

    echo '<br /><a href="/'.$path.'">Back to Archive</a>';
}

edited May 18, 2011 at 18:58

answered May 18, 2011 at 18:50

jrn.ak

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1 Comment

user692194 Over a year ago

THANKS YOUR ANSWER HELPED VERY MUCH TO DO THE JOB i modified my self but you were the head start thnx

simonbs · Accepted Answer · 2011-05-18 18:49:51Z

This should do it.

$n_id = mysql_real_escape_string ($_GET['id']);
    $path = '';

    if(isset($n_id) && $n_id != "") {
        $sql = 'SELECT * FROM test2 WHERE id="' . $n_id . '"';
        $news = mysql_query($sql);

        $first = TRUE;
        if($result = mysql_fetch_array($news)) {
            $title = mysql_result($news,0,"title");
            $date = mysql_result($news,0,"date");

            if($first == TRUE) {
                echo '<b>' . $title . ' | ' . $date . '</b>';
                $first = FALSE;
            }
            else {
                echo '<b>' . $title . ' | ' . $date . '</b>
                <br/>
                <img src="images.php?id='. $n_id .'>';
            }
        }
        else {
            header("Location: vendi.php");
        }
        echo '<br /><a href="/'.$path.'">Back to Archive</a>';
        }
    }

George Cummins · Accepted Answer · 2011-05-18 18:55:42Z

0

While looping through your results, you can check if you are at the first result:

$counter = 0;
while ($row = mysql_fetch_array($news)) {
    $title = mysql_result($news,0,"title");
    $date = mysql_result($news,0,"date");
    echo '<b>' . $title . ' | ' . $date . '</b>';
    if ( $counter == 0 ) {
        echo '<br /><img src="images.php?id='. $n_id .'>';
    }

}

edited May 18, 2011 at 18:55

answered May 18, 2011 at 18:48

George Cummins

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3 Comments

John Green Over a year ago

You're missing a close to your echo. I think he's going to have to switch up the SQL call too. Looks like he's doing a single row atm. That could get tetchier without more info.

jrn.ak Over a year ago

You're assigning a value here: if ( $counter = 0 ) { Should be ==

George Cummins Over a year ago

Fixed, and fixed. Thanks. As to the single row issue, he said he is trying to display five results.

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