[method]

Hi all. I am using the following php code to grab passed data of post request(<input type="hidden" name="copy_list" value="23434,34346,34234">) and write each value in to mysql db but i keep getting the following error:

Warning: reset() [function.reset]: Passed variable is not an array or object in
write.php on line 64

pointing at this line:

reset($id); // Set array pointer to first array element

I am trying to write 23434,34346,34234 from :

<input type="hidden" name="copy_list" value="23434,34346,34234">

and write them to mysql.(Note: number of values posted is not know)
I be happy if some one help me grab these data correctly and write them to mysql.Thanks

PHP Code:

<?php
$id = array();
$amountselected = 0;
$id = $_POST['copy_list'];

echo "$id";

if(!$id == 0)
{

// Find out how many items were selected and protect against no selection
If (is_array($id)) {
      $amountselected = count($id);
} Else {
      $amountselected = 0;
}
echo "Number value passed:";
echo $amountselected;
echo "<br>";


reset($id); // Set array pointer to first array element
      foreach ($id as $addrecid) {
            // Build SQL Insert query
        
            $insert = "INSERT INTO test ( SELECT * FROM $s WHERE friendID = '$addrecid')";
            
            echo $insert;
            //echo $addrecid;
            //echo "<br>";
            $addsuccess = mysql_db_query ($dbname, $insert);
            If ($addsuccess) {

                  $howmanynew++;
            }
      }

If ($amountselected == $howmanynew)
{
?>
    <div class="confirmBox">
        The selected videos have been copied.
    </div>
<?
}

}

HTML Code:

<form action="./write.php" method=post >
       
	<input name="go" type="hidden" value="yes">
        <input type="hidden" name="copy_list" value="23434,34346,34234">
    <tr>
      <td width="20%">PlayLists Name:
      </td>
      <td width="50%"><input type="text" name="ListName" size="20"> Max 100 Character</td>
    </tr>
    <tr>
      <td colspan=2>
      <p align="center"><input type="submit" value="Create This PlayList" name="B1">
      </td>
    </tr>
    </form>

[Nico]

PHP Code:

$id = array_map('intval', explode(',', $_POST['copy_list'])); 

... and you can drop the reset() line. foreach() will always start from the beginning of the array, regardless of the position of the pointer.

[mab]

Edit:Too late on the post button...

$_POST['copy_list'] will contain a string with one or more coma separated items. To get that into an array would require using the explode() function.

[method]

Thanks that solved the problem but now my insert stoped working after i added another insert before $id = array(); . could you tell me what i am doing wrong. i printed out the insert statment and it shows the write statments!!. The insert2 is working but insert inside foreach loop stoped working after i add insert1!!

PHP Code:

$user = "root";
$pw = "";
$db = "test";
$mysql_access = mysql_connect("localhost", $user, $pw);
mysql_select_db($db, $mysql_access);




            // Build SQL Insert query
            $insert2 = "INSERT INTO playlistnames (ID,User_ID,playlistname,playlistfoldername) VALUES ('$ID',1,'$playlistname','$playlistname2')";
            //echo $insert2;
            //echo "<br>";
            $addsuccess = mysql_db_query ($db, $insert2);
            If ($addsuccess) {
                
                  $howmanynew++;
            }


$id = array();
$amountselected = 0;
$id = $_POST['copy_list'];

....
.... 

[Nico]

PHP Code:

$id = array(); 

PHP Code:

$id = $_POST['copy_list']; 

... none of these lines are supposed to be there. Can you post your full code? I don't see any obvious errors here, except these two lines.

[method]

I post the full code here. I think problem is with db connection because after i added insert2 the insert1 stoped working and only insert2 working now!!

Full code:

PHP Code:

<?php

$playlistname=$_POST['ListName'];
$playlistname2=$_POST['ListName2'];
//echo '<br>';


$user = "root";
$pw = "";
$db = "test";
$mysql_access = mysql_connect("localhost", $user, $pw);
mysql_select_db($db, $mysql_access);


            // Build SQL Insert query
            $insert2 = "INSERT INTO playlistnames (ID,User_ID,playlistname,playlistfoldername) VALUES ('$ID',1,'$playlistname','$playlistname2')";
            //echo $insert2;
            //echo "<br>";
            $addsuccess = mysql_db_query ($db, $insert2);
            If ($addsuccess) {
                  
                  $howmanynew++;
            }



$id = array();
$amountselected = 0;
//$id = $_POST['copy_list'];
$id = array_map('intval', explode(',', $_POST['copy_list']));  

echo "$id";

if(!$id == 0)
{

// Find out how many items were selected and protect against no selection
If (is_array($id)) {
      $amountselected = count($id);
} Else {
      $amountselected = 0;
}
echo "Number value passed:";
echo $amountselected;
echo "<br>";


reset($id); // Set array pointer to first array element
      foreach ($id as $addrecid) {


            // Build SQL Insert query
          
            $insert1 = "INSERT INTO singers ( SELECT * FROM test2 WHERE friendID = '$addrecid')";
            
            echo $insert1;
            //echo $addrecid;
            //echo "<br>";
            $addsuccess = mysql_db_query ($dbname, $insert1);
            If ($addsuccess) {
                 
                  $howmanynew++;
            }
      }

If ($amountselected == $howmanynew)
{
?>
    <div class="confirmBox">
        The selected videos have been copied.
    </div>
<?
}

}


?>

[Nico]

Let MySQL tell you what the problem is:

PHP Code:

$addsuccess = mysql_db_query ($dbname, $insert1) OR die(mysql_error()); 

[method]

I added line for insert1and got this error :

No database selected

[Nico]

Well that should give you a good hint.

[Keith]

Why on earth are you using mysql_db_query()? How old is the tutorial you were reading?