PHP：从嵌套数组（json）创建HTML表但获取结果和“为foreach（）提供的无效参数”

Question

I am trying to create a HTML table from an array that is obtained through api/json.

It's a nested array (the actual data is nested within the session)

Array
(
[sessid] => -hnkGi-k1rgwhnymkcScR0bom-RRKURn2S1pgMZoBX4
[status] => 0
[players] => Array
    (
        [0] => Array
            (
                [first_name] => Chris
                [last_name] => Clausing
                [pdga_number] => 31635
                [rating] => 930
                [year] => 2014
                [class] => A
                [gender] => M
                [bracket] => Open
                [country] => NL
                [state_prov] => 
                [tournaments] => 11
                [rating_rounds_used] => 36
                [points] => 998
                [prize] => 0.00
            )

        [1] => Array
            (
                [first_name] => Uwe Walter
                [last_name] => Schlueter
                [pdga_number] => 37788
                [rating] => 902
                [year] => 2014
                [class] => A
                [gender] => M
                [bracket] => Master
                [country] => NL
                [state_prov] => 
                [tournaments] => 12
                [rating_rounds_used] => 33
                [points] => 970
                [prize] => 0.00
            )

        [2] => Array
            (
                [first_name] => Mark
                [last_name] => Steenhuis
                [pdga_number] => 50574
                [rating] => 859
                [year] => 2014
                [class] => A
                [gender] => M
                [bracket] => Master
                [country] => NL
                [state_prov] => 
                [tournaments] => 12
                [rating_rounds_used] => 28
                [points] => 678
                [prize] => 0.00
            )
      )
)

And the warning mesages:

Warning: Invalid argument supplied for foreach() in /home/surrealm/dvh/player_statistics.php on line 103

Warning: Invalid argument supplied for foreach() in /home/surrealm/dvh/player_statistics.php on line 103

See http://vliegende-hollander.com/player_statistics.php?year=2014&class=A&gender=M&country=NL

As you can see i can extract the desired data, and create the HTML table from it. But, I can not get rid of the two error messages.

If i take away one of the foreach() i only get first character of first value only (in this case, the first character of the sessid).

The actual bit of PHP i'm using to create the table is:

<?

if (is_array($player)) {
 foreach($player as $key){
  foreach($key as $k){             // <- this is line 103
   echo "<tr>";
    echo "<td>".$k['pdga_number']."</td>";
    echo "<td>".$k['first_name']." ".$k['last_name']."</td>";
    echo "<td>".$k['country']."</td>";
    echo "<td>".$k['rating']."</td>";
    echo "</tr>";
    }
   }
  }
 ?>

Can anyone help me clean up this code, so i get rid of the two warnings?

I'm not a full-time programmer, so it's probably just my not fully understanding the foreach() function. I usually extract and display data using MySQL, and while() loops instead.

Answer 1

The error states that the $key variable that you passed into the for each isn't an array. For this to work the player array would have to have arrays within it (i.e. be a multidimensional array), so the second foreach can iterate through it.

If your player array doesn't have arrays within it, then you only need one foreach to loop through and then you can access the data using the index (i.e. $key['pdga_number'] instead of $k['pdga_number']).

For me to fully help though, I would need to know the contents of the $player array.

EDIT

I've just seen the example given, try doing a second if(is_array($key) before calling the second foreach, i.e.

if (is_array($player)) {
    foreach($player as $key){
        if(is_array($key)){
            foreach($key as $k){             // <- this is line 103
                echo "<tr>";
                echo "<td>".$k['pdga_number']."</td>";
                echo "<td>".$k['first_name']." ".$k['last_name']."</td>";
                echo "<td>".$k['country']."</td>";
                echo "<td>".$k['rating']."</td>";
                echo "</tr>";
            }
        }
    }
}