Help with Syntax problem

[Anon]

I'm using "PHP and MySQL Web Development" published by SAMS and there seems to be a typo, or two or a hundred, in the book. I need to use the Web Content Management script. I get an error when I try to submit a story. Can somebody please tell me what is wrong with this script:

<?php

// story_action.php
// add / modify story record

include "include_fns.php";

$conn = db_connect();

$time = time();

if ( ($html) && (dirname($html_type) == "text") ) {
$fp = fopen($html, "r");
$story_text = addslashes(fread($fp, filesize($html)));
fclose($fp);
}

if ($story) { // It's an update
$sql = "update stories
set headline = '$headline',
story_text = '$story_text',
page = '$page',
modified = $time
where id = $story";
}
else { // It's a new story
$sql = "insert into stories
(headline, story_text, page, writer, created, modified)
values
('$headline', '$story_text', '$page', '$auth_user', $time, $time)";
}

$result = mysql_query($sql, $conn);

if (!$result) {
print "There was a database error when executing <PRE>$sql</PRE>";
print mysql_error();
exit;
}

if ( ($picture) && ($picture != "none") ) {

if (!$story)
$story = mysql_insert_id();

$type = basename($picture_type);

switch ($type) {
case "jpeg":
case "pjpeg": $filename = "pictures/$story.jpg";
copy ($picture, $filename);
$sql = "update stories
set picture = '$filename'
where id = $story";
$result = mysql_query($sql, $conn);
break;
default: print "Invalid picture format: $picture_type";
}
}

header("Location: $destination");

?>

[Anon]

on the line:

if (!$story)

you need a { after it... like so:

if (!$story) {

if that doesn't solve the problem, please post the actual error you get.

[[deleted]]

cough

You could at least post the error.

[amcgrath]

No, that's ok... I think it's here: case "jpeg":
case "pjpeg": $filename = "pictures/$story.jpg";

there's no matching break...

[amcgrath]

Craig, that's actually legal, the if will only consider the following line and proceed. I personally don't like to do this, but ya can if you want.

[[deleted]]

Breaks are not mandatory.
A case statement begins executing on the first 'case' match. Then it runs untill the first 'break'. No break? then it continues to the end. That way you can use the same piece of code to be executed for several 'cases' without repeating the code.

as near as I can tell the code is ok. It even runs (minus the database commands) so PLEASE post the error message!!

[Anon]

I ended up modifiying the code a bit and taking out the portion for
uploading an html file and an image. I also modified the submited story
value to be "new" rather than "" if it was a new story. A value was always
being submited for "story" from the form even though it was blank so it was
always going into the update if statement. Now my problem is the submitted
data is now showing up in the database when I look at it in "MySQL Front".
But, I know the data is there because it outputs the data through the
browser. So, now there is no error but the data does not show up. The data
for "time" shows up but not the data actually submited through the form.
Here is my code:
<?php

include "include_fns.php";

if (isset($story) && (story != "new"))
$s = get_story_record($story);

$conn = db_connect();

$time = time();
if ($story=="new") {

$sql = "update stories
set headline = '$headline',
story_text = '$story_text',
page = '$page',
modified = $time
where id = $story";
}
else {
$sql = "insert into stories
(headline, story_text, page, writer, created, modified)
values
('$headline', '$story_text', '$page', '$auth_user', $time,
$time)";
}

$result = mysql_query($sql, $conn);

if (!$result) {
print "There was a database error when executing <PRE>$sql</PRE>";
print mysql_error();
exit;
}

header("Location: http://www.cncrenegade.com/");

?>

Thanks for your help. I really appreciate it.
~Steve

[Anon]

POST YOUR ERRORS. sorry to be rude, but you need to post your errors and try to specify where in your code the problem is. You can't expect people to wade through lines of code, we all have lives here.

[Anon]

Like I said in that post. I'm not getting an error. The data is going into the database and I know it is there because it outputs to the browser, but when I look at the table in MySQL-Front the data does show up even though it is outputting.

[amcgrath]

I stand corrected, thanks vince. (Though I still don't like doing it that way).

[[deleted]]

There's no accounting for taste :-)

[amcgrath]

True, I'm still kinda bothered by this: case "jpeg":
case "pjpeg": $filename = "pictures/$story.jpg";

What if the value is pjpeg? Wouldn't the first case statement force the pjpeg case to being read?

[[deleted]]

Either you made a mistake in explaining it or I'm missing something in your question.

if the value is pjpeg the case starting at
case 'pjpeg': will be executed until the first next break or the end of the switch statement.