post multiple fields with the same name into a database

[mrooks1984]

hello, i am hoping someone can help, i have multiple post results that need to go one by one into a table

 

heres the form i have:

	echo '<h2>Update Variation Values</h2>';
	echo '<div id ="content_form">';
	echo '<form method="post" action="">';
	echo '<input type="hidden" name="option_id" value="' . $option_id . '">';
	echo '<input type="hidden" name="variation_id" value="' . $variation_id . '">';
	echo '<input type="hidden" name="product_name" value="' . $product_name . '">';
	$sql = "SELECT * FROM store_option$variation_id WHERE product = '$product_name'";
	$res = mysql_query($sql) or die(mysql_error());
	while ($row = mysql_fetch_assoc($res)) {

		echo '<div id="content_title"><label for="title[]">Title:</label></div>';
		echo '<div id="content_box"><input name="title[]" type="text" id="title" value="' . $row['title'] . '" /></div>';
		echo '<div id="content_title"><label for="price[]">Price:</label></div>';
		echo '<div id="content_box"><input name="price[]" type="text" id="price[]" value="' . $row['price'] . '" /></div>';
	}
	echo '<div id="content_body"><input type="submit" name="post_update_variation_value" value="Save Values"></div></form>';
	echo '</div>';
}

heres my current code

	echo '<pre>',print_r($_POST,true);

foreach($_POST['title'] as $key => $value){
if($value !=''){

	$sql = "UPDATE store_option$variation_id SET title = '$value', price = '$price' WHERE product = '$product_name'";
	$res = mysql_query($sql) or die(mysql_error());
} else {
	echo "Option $key is empty<br />";
}
}

its currently just doing the last result and filling up the database with that result, many thanks all.

 

[jazzman1]

Does the table "store_option$variation_id", it has been updated correctly?

I'm seeing a syntax error in the code. 

[darkfreaks]

why do you have a string in an array loop

 

 

i read you could use the (array) type cast inside the foreach and it should be fine not sure if this is true or not to convert a string to array inside a foreach loop.

 

normally if it was a string you would be getting "invalid argument foreach" error.

 

also might change != to !==

 

 

[mrooks1984]

thanks for your replies, is it does update fine, but each field has the last result

[mrooks1984]

if i use this code, i get this message: Fatal error: Maximum execution time of 30 seconds exceeded in C:\xampp\htdocs\jubileeleather\site\admin\_class\store.php on line 654

 

	$option_id = mysql_real_escape_string($_POST['option_id']);
	$variation_id = mysql_real_escape_string($_POST['variation_id']);
	$product_name = mysql_real_escape_string($_POST['product_name']);

	foreach($_POST['title'] as $key1 => $value1){

		foreach($_POST['price'] as $key2 => $value2){

			foreach($_POST['option_value_id'] as $key3 => $value3){

				$sql = "UPDATE store_option$variation_id SET title = '$value1', price = '$value2' WHERE id = '$value3'";
				$res = mysql_query($sql) or die(mysql_error());
			}
		}
	}
	echo '<script>window.location="dashboard.php?page=store_variation"</script>';
}

[trq]

You need to use the $key to get the index of the price.

 

foreach ($_POST['title'] as $key => $value) {
  $title = $value;
  $price = $_POST['price'][$key];

  // execute your query
}

[mrooks1984]

thanks for your reply, i have tried this, but only the first field is changed in the database.

	function post_update_variation_value() {

	$option_id = mysql_real_escape_string($_POST['option_id']);
	$variation_id = mysql_real_escape_string($_POST['variation_id']);
	$product_name = mysql_real_escape_string($_POST['product_name']);

	foreach ($_POST['title'] as $key => $value) {
  		$title = $value;
  		$price = $_POST['price'][$key];

  		// execute your query
	$sql = "UPDATE store_option$variation_id SET title = '$title', price = '$price' WHERE id = '$key'";
	$res = mysql_query($sql) or die(mysql_error());
	}
	echo '<script>window.location="dashboard.php?page=store_variation"</script>';
}
}

[trq]

Are you sure that $key matches up with your ids? Your not using your ids in your initial form creation code.

[mrooks1984]

yea realised now, tried adding this to the code now all fine, thanks so much for your help

 

  $price = $_POST['price'][$key];

$id = $_POST['option_value_id'][$key];

 

  // execute your query

$sql = "UPDATE store_option$variation_id SET title = '$title', price = '$price' WHERE id = '$id'";

$res = mysql_query($sql) or die(mysql_error());

[trq]

I would use the $key when making your form, that way you know things are going to be held together as they should be:

 

echo '<div id="content_title"><label for="title[' . $row['id'] . ']">Title:</label></div>';
echo '<div id="content_box"><input name="title[' . $row['id'] . ']" type="text" id="title[' . $row['id'] . ']" value="' . $row['title'] . '" /></div>';
echo '<div id="content_title"><label for="price[' . $row['id'] . ']">Price:</label></div>';
echo '<div id="content_box"><input name="price[' . $row['id'] . ']" type="text" id="price[' . $row['id'] . ']" value="' . $row['price'] . '" /></div>';