Here is the error?

General PHP Help
#1

Hi everyone again,

Insert multi row

[php]$query = ‘INSERT INTO table_name (id, receiver_id, receiver_amount) VALUES (5, (’ .implode(’), (’, $_REQUEST[‘receiver_id’]). .implode(’)’, ‘(’, $_REQUEST[‘receiver_amount’]) . ‘))’;
$result = mysql_query($query)
or die(‘Could not execute INSERT query’);[/php]

Sample:

  id         receiver_id          amount
   5             3                   12
   5             1                    5
   5             5                    8

Thanks

#2

using implode isn’t the way to do it, not if you’re just putting a single number in the row.
[php]
$query = ‘INSERT INTO table_name (id, receiver_id, receiver_amount) VALUES (5, $_POST[‘receiver_id’], $_POST[‘receiver_amount’])’;
$result = mysql_query($query) or die(mysql_error());[/php]

I don’t know how this info is being gotten, so we really can’t tell you how to do multiple inserts, other than you’ll probably need a for loop.

#3

Thanks for the reply,

To me need multi-line

$id=5;

$receiver_id=implode(", ", $_REQUEST[‘receiver_id’]);
Output: 2, 5, 7

$receiver_amount=implode(", ", $_REQUEST[‘receiver_amount’]);
Output: 5, 8, 17

Table

  id         receiver_id          amount
   5             2                   5
   5             5                    8
   5             7                    17

id, Not AUTO_INCREMENT

Thanks

#4

Right, but the table you gave doesn’t support that, you’re showing us just 1 number in each column. If you’re wanting to store mutiples in one field, then you need to change the data type to text. Then you can do
[php]
$query = “INSERT INTO table_name (id, receiver_id, receiver_amount) VALUES (5, '$receiver_id, ‘$receiver_amount’)”;
$result = mysql_query($query) or die(mysql_error());[/php]

#5

If $_REQUEST[‘receiver_id’] and $_REQUEST[‘receiver_amount’] are always the same size and same order you can do something like this:

[php]
$id = 5;
$receiver_id = array(2, 5, 7);
$receiver_amount = array(5, 8, 17);
$values = array();

for($i = 0; $i < count($receiver_id); $i++) {
$values[] = “({$id}, {$receiver_id[$i]}, {$receiver_amount[$i]})”;
}
echo “INSERT INTO table_name (id, receiver_id, receiver_amount) VALUES " . implode(”, ", $values);
[/php]

It will output

INSERT INTO table_name (`id`, `receiver_id`, `receiver_amount`) VALUES (5, 2, 5), (5, 5, 8), (5, 7, 17)
#6

@m@tt, Php You a professor
I do not know how to thank you
Thank you very very very

The code works no problem,
but,

name=“receiver_id[]”
name=“receiver_amount[]”;

How do I get into array()?

$_REQUEST[‘receiver_id’] and $_REQUEST[‘receiver_amount’]

$receiver_id = array(2, 5, 7);
$receiver_amount = array(5, 8, 17);

Thank you again

#7

for($i=0; $i<count($_POST[‘receiver_id’]); $i++) {
$receiver_id[] = $_POST[‘receiver_id’][$i];
$receiver_amt[] = $_POST[‘receiver_amount’][$i];
}

#8

Thank you very much

Best Regards

#9

Hello again,

I have a smal problem,

It is not working without selecting first row.

1st row is selected : running
1st and 2nd rows are selected : running
1st , 2nd and 3rd rows are selected : running
1st, 2nd, 3rd and 4th rows are selected : running

1st row is not selected but the other rows are selected : not running

http://awesomescreenshot.com/0c8m4to53
NOTE: The rows are not standart. They may be 1 or more.

Thanks

#10

I’m not sure if I understand the problem. It sounds like you should remove the empty value from the select box so there must be an option selected?

#11

Hello again,

1. row selected no problem at. working

1. row not selected, there is a problem . does not work
Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ‘)’ at line 1

step7.php code:
[php]
<?php
while ($row = mysql_fetch_array($receiverler)) {
?>


<?=$row['receiver_markasi']?>
<?=$row['receiver_tipi']?>
<?=$row['receiver_ozelligi']?>
<?=$row['receiver_fiyati']?><?=$row['receiver_para_cinsi']?>






<?php
$sayi = 1;
while($sayi < $encok=$dairesayisi[‘secilen_daire_sayisi’]+1) {
echo ‘’.$sayi.’’;
$sayi++;
}
?>



<?php
}
?>
[/php]

step8.php code
[php]
$id = $_SESSION[‘lastid’];
for($i=0; $i<count($_POST[‘receiver_id’]); $i++) {
$receiver_id[] = $_POST[‘receiver_id’][$i];
$receiver_amount[] = $_POST[‘receiver_amount’][$i];
}
$values = array();
for($i=0; $i<count($receiver_id); $i++) {
$values[]="({$id}, {$receiver_id[$i]}, {$receiver_amount[$i]})";
}

      $receiver_amounts=mysql_query("INSERT INTO receiver_amounts (`id`, `receiver_id`, `receiver_amount`) VALUES " . implode(", ", $values));
      if (!$receiver_amounts)
     {
     die(mysql_error());
     }

[/php]

Again, thank you very, very much
Best Regards
Adem GENÇ

#12

fixed the problem “”

Thank you so much for your help and interest.

Best Regards
Adem GENÇ

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