foreach only puts first check box in db

turpentyne · July 23, 2010

please please please, somebody help. I'm new at php, and I'm going nuts trying to figure out this one task. In one form, I need to enter the results of two sets of checkboxes into two tables. Nothing has worked, so I'm going back to basics and building from there.

I built this code below. It works but only puts the first row in, matching the first checkbox selected. Nothing else

Here's the code:

<?php

if (isset($_POST['submitted'])) {

$errors = array();

$item_id = ($_POST['item_id']);

$column2 = ($_POST['column2_id']);

if (empty($errors)) {

require ('databaseConnect.php');

foreach ($_POST['column2_id'] AS $key => $column2) {

$query = "INSERT INTO tablename(item_id, column2_id)

VALUES ('$item_id', '$column2')";

$result = mysql_query ($query);

if ($result) {

echo 'one plant has been added';

exit();

} else {

echo 'system error. No plant added';

echo '<p>' . mysql_error() . '<br><br>query:' . $query . '</p>';

exit();

}

mysql_close();

}

?>

<body>

<br>

<input type="checkbox" name="column2_id[]" value="3" > three <br>

</fieldset><br><br>

</form>

PravinS · July 23, 2010

Try like this for loop

$column2 = $_POST['column2_id'];

for ($i=0;$i<count($column2);$i++)
{
echo $column2[$i];
}

turpentyne · July 23, 2010

Ok, I tried that. Unfortunately, it's doing the same thing. Only the first row goes into the database. Here's what I wrote:

for ($i=0;$i<count($column2);$i++)

{

$query = "INSERT INTO tablename(item_id, column2_id)

VALUES ('$item_id', '$column2[$i]')";

$result = mysql_query ($query);

if ($result) {

echo 'your items have been added';

Pikachu2000 · July 23, 2010

Are the checkboxes checked when the form is submitted? They're only present in the $_POST array if they're actually checked.

turpentyne · July 23, 2010

hehe! I wish it was that simple. In the actual form, there are several checkboxes and at least half of them were checked.

Pikachu2000 · July 23, 2010

Alright then, to start debugging, see what the $_POST array actually contains. put echo '<pre>'; print_r($_POST); echo '</pre>'; at the head of the script and see if it contains the values you'd expect it to.

turpentyne · July 23, 2010

ok, I've done that. It's definitely knows what I've clicked. Still only the first entry goes into the database. Here's what it created.

Array

(

[item_id] => 20

[column2_id] => Array

(

[0] => 1

[1] => 5

[2] => 9

[3] => 10

)

[submitted] => TRUE

)

one item has been added

Pikachu2000 · July 23, 2010

Found it (I think). You have a curly bracket out of place. add one after $result = mysql_query ($query);, and remove one just prior to the closing ?> tag. Try that, and see if it needs any further tweaking.

turpentyne · July 23, 2010

uh-oh! Doing that, generated this error on reloading of the page:

Warning: Invalid argument supplied for foreach() in /filepath/htdocs/filename.php on line 15

Fatal error: Call to undefined function mysql_real_query() in /filepath/htdocs/filename.php on line 18

Pikachu2000 · July 23, 2010

Can you post the code as it is now, with the revisions?

turpentyne · July 23, 2010

Sure... here it is below: (by the way, how do you guys post this code in those windows, in the forum? are you just putting it in an html form text box?)

<html>

<title>title</title>

<?php

echo '<pre>';

print_r($_POST);

echo '</pre>';

if (isset($_POST['submitted']))

{

$errors = array();

$item_id = ($_POST['plant_id']);

$column2 = ($_POST['column2_id']);

if (empty($errors))

{

require ('databaseconnect.php');

for ($i=0;$i<count($medicine);$i++)

{

$query = "INSERT INTO tablename(item_id, column2_id)

VALUES ('$item_id', '$column2[$i]')";

$result = mysql_query ($query);

}

if ($result)

{

echo 'items have been added';

exit();

} else {

echo 'system error. Not added';

echo '<p>' . mysql_error() . '<br><br>query:' . $query . '</p>';

exit();

}

mysql_close();

}

?>

<body>

item id field:<br>

only enter numbers here<br>

//this will be a hidden passed variable in finished form

characteristics:<br>

<input type="checkbox" name="column2_id[]" value="3" > three <br>

<input type="checkbox" name="column2_id[]" value="7" > Seven <br>

</fieldset><br><br>

</form>

</body>

</html>

wildteen88 · July 23, 2010

You're original code you used is correct

<?php
if (isset($_POST['submitted'])) {
$errors = array();

  $item_id = ($_POST['item_id']);
  $column2 = ($_POST['column2_id']);


  if (empty($errors)) {
    require ('databaseConnect.php');

    foreach ($_POST['column2_id'] AS $key => $column2) {

      $query = "INSERT INTO tablename(item_id, column2_id)
      VALUES ('$item_id', '$column2')";

      $result = mysql_query ($query);

      if ($result) {
           echo 'one plant has been added';
           exit();

      } else {
      echo 'system error. No plant added';
      echo '<p>' . mysql_error() . '<br><br>query:' . $query . '</p>';
      exit();
      }
    mysql_close();
    } 
  }

} 

?>

<body>
<FORM action="test.php" method="post">

<input type="text" name="item_id" value="<?php if(isset($_POST['item_id'])) echo $_POST['item_id']; ?>" />
<br>
<input type="checkbox" name="column2_id[]" value="1" > one <br> 
<input type="checkbox" name="column2_id[]" value="2" > two <br> 
<input type="checkbox" name="column2_id[]" value="3" > three <br> 



</fieldset><br><br>
<input type="hidden" name="submitted" value="TRUE">
<input type="submit" />

</form>

The reason why your script is only adding one value into your database is because you're callingexit (on line 21) when your query successfully inserts a new row.

19.      if ($result) {
20.           echo 'one plant has been added';
21.           exit();
22.
23.      } else {

Delete exit(); and your script will execute as you intended.

(by the way, how do you guys post this code in those windows, in the forum? are you just putting it in an html form text box?)

Sure, wrap your code within code tags (

or

)

Pikachu2000 · July 23, 2010

To have the code in the code or php boxes, just post it within

 . . .

or

 . . .

bbcode tags.

I've just tested this, and it should work fine. There is no plant_id field in the form, there is item_id. See my comment in the code regarding that, also, to submit a form to itself, you can simply leave it as <form action="". . . There's no need to specify the script name.

<html>
<title>title</title>
<?php
if (isset($_POST['submitted'])) {
   $errors = array();
   $item_id = ($_POST['item_id']); // This was $_POST['plant_id'], which didn't exist in the form.
   $column2 = ($_POST['column2_id']);
   if (empty($errors)) {
      require ('databaseconnect.php');
      foreach ($column2 as $val ) {
         $query = "INSERT INTO tablename(item_id, column2_id) VALUES ('$item_id', '$val')";
         mysql_query($query) or die( mysql_error() );
         // echo $query . '<br />';
      }
      if ($result) {
         echo 'items have been added';
      }
   } else {
      echo 'system error. Not added';
      echo '<p>' . mysql_error() . '<br><br>query:' . $query . '</p>';
   }
   mysql_close();
}
?>

<body>


<FORM style="border: 1px dotted red; padding: 2px" action="" method="post">

item id field:<br>
only enter numbers here<br>
//this will be a hidden passed variable in finished form

<input type="text" name="item_id" value="<?php if(isset($_POST['item_id'])) echo $_POST['item_id']; ?>" />
<br><br>


characteristics:<br>
<input type="checkbox" name="column2_id[]" value="1" > one <br>
<input type="checkbox" name="column2_id[]" value="2" > two <br>
<input type="checkbox" name="column2_id[]" value="3" > three <br>
<input type="checkbox" name="column2_id[]" value="4" > four <br>
<input type="checkbox" name="column2_id[]" value="5" > five <br>
<input type="checkbox" name="column2_id[]" value="6" > six <br>
<input type="checkbox" name="column2_id[]" value="7" > Seven <br>

</fieldset><br><br>
<input type="hidden" name="submitted" value="TRUE">
<input type="submit" />

</form>

</body>

</html>

turpentyne · July 23, 2010

thanks on the info for how to post code here.

unfortunately, removing the exit(); didn't help. It still only posted one item, and gave me a huge list of errors about connecting to the database.

Pikachu2000 · July 23, 2010

Well, that's a start. At least we can narrow things down. Paste the errors here.

turpentyne · July 23, 2010

ah... I was changing the names of things 'to protect the innocent' and forgot one. that explains the plant_id. It is the same as item_id

turpentyne · July 23, 2010

hallelujah!!!

the script Pikachu2000 posted is working. Now I have to go through it to see the differences between that script and mine, so I can understand where I was going wrong.

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foreach only puts first check box in db

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