Mysql_fetch_array print within a string. Possible?

Question

I'm kinda confused right now with this function I'm trying to edit. I would like to add the mysql_fetch_array to the feedback string. I'm pretty sure it's possible but can somebody point out the problem for me? The piece of code I'm trying to execute is in comments. And where's PRINTHERE, that where I want the mysql_fetch_array to spit out the results. Thx a lot!

foreach($_FILES as $k => $v){ 

    $img_type = "";

    ### $htmo .= "$k => $v<hr />";  ### print_r($_FILES);

    if( !$_FILES[$k]['error'] && preg_match("#^image/#i", $_FILES[$k]['type']) && $_FILES[$k]['size'] < $max_image_size ){

        $img_type = ($_FILES[$k]['type'] == "image/jpeg") ? ".jpg" : $img_type ;
        $img_type = ($_FILES[$k]['type'] == "image/gif") ? ".gif" : $img_type ;
        $img_type = ($_FILES[$k]['type'] == "image/png") ? ".png" : $img_type ;

        $img_rname = $_FILES[$k]['name'];
        $img_path = $upload_dir.$img_rname;

        copy( $_FILES[$k]['tmp_name'], $img_path ); 
        if($enable_thumbnails) make_thumbnails($upload_dir, $img_rname);
        /*
        mysql_connect("localhost", "", "") or die(mysql_error());
        mysql_select_db("");        
        $result = mysql_query("SELECT * FROM gallery_main ORDER BY datetime DESC");
        while($row = mysql_fetch_array($result))
          {
            $name=$row['name'];
            echo "<option value='$name'>$name</option>";
          }
         */ 
        $feedback .="<style>.stap1{display:none;}.stap2{display:block;}</style>
        <input name='src' style='width:400px;' type='hidden' value=\"$img_rname\"><br />
        <input name='thumbsrc' style='width:400px;' type='hidden' value=\"thumb_$img_rname\"><br />
        <select name='galleryname' id='galleryname'>PRINTHERE";
    }
}

Are you the Andrew Ng of Stanford machine learning fame?

hackartist
– hackartist

2012-02-19 00:00:30 +00:00
Commented Feb 19, 2012 at 0:00 — hackartist
– hackartist, Commented Feb 19, 2012 at 0:00
No, Im not. Just another guy with same name. hehehe

Andrew Ng
– Andrew Ng

2012-02-19 00:13:00 +00:00
Commented Feb 19, 2012 at 0:13 — Andrew Ng
– Andrew Ng, Commented Feb 19, 2012 at 0:13

Josh · Accepted Answer · 2012-02-19 00:26:30Z

1

Instead of echoing, store the results into a variable, like you did with $feedback.

$options = '';

while($row = mysql_fetch_array($result)) {
    $name=$row['name'];
    $options .= "<option value='$name'>$name</option>";
}

Then, append to $feedback

$feedback .= "<style>.stap1{display:none;}.stap2{display:block;}</style>
        <input name='src' style='width:400px;' type='hidden' value=\"http://www.djpassa.com/gallery/$img_rname\"><br />
        <input name='thumbsrc' style='width:400px;' type='hidden' value=\"http://www.djpassa.com/gallery/thumb_$img_rname\"><br />
        <select name='galleryname' id='galleryname'>" . $options;

Also, you need to select a database.

mysql_select_db("database_name");

edited Feb 19, 2012 at 0:26

answered Feb 19, 2012 at 0:04

Josh

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10 Comments

Andrew Ng Over a year ago

Hmm somehow it's still not working i've added the whole php file to this address: codepad.org/Y505gazf

Josh Over a year ago

Assuming your query is correct and retrieves at least 1 result, you need to select a database.

Andrew Ng Over a year ago

I've removed the info for security reasons. Even with the correct login info it wont parse the list out. Any more clues???

Josh Over a year ago

Can you confirm any content is concatenated to $feedback? If there is none, then there may be an issue with your SQL credentials or query. Also, add </select> to close off the select tag, but this shouldn't make a difference.

Andrew Ng Over a year ago

hi i've updated the code. This is what I have now: codepad.org/GpQwqJw4 it does spit out $lol which is lol in plain text. But not the $options

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