Solved

Notice: Use of undefined constant picture - assumed 'picture' on line 24

Posted on 2003-11-08
3
28,071 Views
Last Modified: 2013-12-13
In the following code I get the error "Use of undefined constant picture - assumed 'picture' on line 24". how can I resolve this error?
<?php

include('include_fns.php');
include('header.php');

$conn = db_connect();


$pages_sql = 'select * from pages order by code';
$pages_result = mysql_query($pages_sql, $conn);

while ($pages = mysql_fetch_array($pages_result)) {

  $story_sql = "select * from stories
                where page = '".$pages['code']."'
                and published is not null
                order by published desc";
  $story_result = mysql_query($story_sql, $conn);
  if (mysql_num_rows($story_result)) {
    $story = mysql_fetch_array($story_result);
    print '<table border="0" width="400">';
    print '<tr>';
    print '<td rowspan="2" width="100">';
(line 24)    if ($story[picture])
      print "<img src=\"resize_image.php?image=$story[picture]\" />";
    print '</td>';
    print '<td>';
    print '<h3>'.$pages['description'].'</h3>';
    print $story['headline'];
    print '</td>';
    print '</tr>';
    print '<tr><td align="right">';
    print '<a href="page.php?page='.$pages['code'].'">';
    print '<font size="1">Read more '.$pages['code'].' ...</font>';
    print '</a>';
    print '</table>';
  }
}

include('footer.php');
?>
0
Comment
Question by:djbell
  • 2
3 Comments
 
LVL 6

Accepted Solution

by:
DoppyNL earned 125 total points
ID: 9712747
if ($story[picture])

picture is syntacticly a constant here:
it's not a variable, because there is no $ in front of it
it's not a string, because there aren't ' or " around it.

Since it should be a string here, you need to change your code to:

if ($story['picture'])
0
 
LVL 6

Expert Comment

by:DoppyNL
ID: 9716611
Why the b?
not working correctly?
Still having problems?

Could you motivate please?

tnx!
0
 
LVL 1

Expert Comment

by:erevia
ID: 11111517
Hi all

don't forget quotes :-)

just replace
  if ($story[picture])
with :
  if ($story['picture'])

bye
0

Featured Post

Courses: Start Training Online With Pros, Today

Brush up on the basics or master the advanced techniques required to earn essential industry certifications, with Courses. Enroll in a course and start learning today. Training topics range from Android App Dev to the Xen Virtualization Platform.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

Title # Comments Views Activity
Why my select dropdown does not work? 8 40
Insert won't work due to space in column name 5 21
mysqli insert query problems 4 22
Decrypt string by php 7 29
Things That Drive Us Nuts Have you noticed the use of the reCaptcha feature at EE and other web sites?  It wants you to read and retype something that looks like this.Insanity!  It's not EE's fault - that's just the way reCaptcha works.  But it is …
This article discusses four methods for overlaying images in a container on a web page
The viewer will learn how to look for a specific file type in a local or remote server directory using PHP.
This tutorial will teach you the core code needed to finalize the addition of a watermark to your image. The viewer will use a small PHP class to learn and create a watermark.

776 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question