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Not found error


cat250

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One more problem...  I'm sorry if I bother u.

 

I have:

echo "Image: ".$json['Poster'].""; and if image exist I get in return URL... but if there's no image get a error:

Notice: Undefined index: Poster in C:\xampp\htdocs\file.php on line 48

 

How can I show a message error for this?

 

Maybe I have too many questions but it's first time when I use PHP and don't understand too much.

Help is appreciated, again. Thanks!

 

I readed about if, else... still can't solved.

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It's working as it should, you just have your messages flipped.

 

Change echo "Hi" with echo "Bye" and vice versa.

 

The 'else' part is what's supposed to happen IF json['Poster'] IS set.

 

To get a better understanding of this, instead of using "Hi and Bye" as your test messages, try this:

 

if (!isset($json['Poster'])){
   echo "Poster is NOT set!";
}
else {
  echo "Poster IS set!";
}

 

Hope that helps :)

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It's working as it should, you just have your messages flipped.

 

Change echo "Hi" with echo "Bye" and vice versa.

 

The 'else' part is what's supposed to happen IF json['Poster'] IS set.

 

To get a better understanding of this, instead of using "Hi and Bye" as your test messages, try this:

 

if (!isset($json['Poster'])){
   echo "Poster is NOT set!";
}
else {
  echo "Poster IS set!";
}

 

Hope that helps :)

 

Ok, now i get "Poster IS set!"... but that isn't true. This is what i wanted to say above.

 

I try to explain again: I have in mysql 2 entries, one with poster and one without.

Script return "Poster IS set!" in both cases. :)

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mysql_connect("localhost", "root", "") or die(mysql_error());

mysql_select_db("movies") or die(mysql_error());

$f = mysql_query("SELECT * FROM movies") or die(mysql_error());

while($inf = mysql_fetch_assoc($f)) {

$two= $inf['id'];

$link = "http://www.imdbapi.com/?i=$two";

$thr = file_get_contents($link);

$json = json_decode($thr,true);

echo "<img src='".$json['Poster']."' />"; 

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Perhaps Poster is set, but is empty.

 

mysql_connect("localhost", "root", "") or die(mysql_error());
mysql_select_db("movies") or die(mysql_error());
$f = mysql_query("SELECT * FROM movies") or die(mysql_error());
while($inf = mysql_fetch_assoc($f)) {
$two= $inf['id'];
$link = "http://www.imdbapi.com/?i=$two";
$thr = file_get_contents($link);
$json = json_decode($thr,true);
if(!empty($json['Poster'])) { echo "<img src='".$json['Poster']."' />";  }

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