Insert multiple form selections into one single table column (set) - php

Question

I have a simple form with the following:
<form action="#" method="post" name="form1" >
<input name="criteria1" size="64">
<input name="criteria2" size="64">
<select name="criteria3[]"multiple="multiple" >
<option value="5 ">5</option>
<option value="7">7</option>
<option value="10">10</option>
</select>
</form>
Now, if the user selects more then one option from criteria3, how do I insert that in a single database table field as well as insert the other two criteria in their own column?
I have tried many different ways and just get no where. My last attempt was as follows:
$values = array();
$type = explode(",", $_POST['criteria3']);
foreach ($type as $value) {
$values[] = sprintf('(%d)', $value);
}
$values = implode(',', $values);
if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "form1")) {
$insertSQL = sprintf("INSERT INTO table (name, date, number)
VALUES (%s, %s, '$values')",
GetSQLValueString($_POST['criteria1'], "text"),
GetSQLValueString($_POST['criteria2'], "text"));
Not really sure what is going wrong – have tried several iterations and am getting no where.
Here are the most recent errors:
Warning: explode() expects parameter 2 to be string, array given in /filename.php on line 84
Warning: Invalid argument supplied for foreach() in /filename.php on line 86

Answer 1

Try the following(If you want separate records corresponding to separate values for criteria3 for a single entity):
$criteria1 = $_POST['criteria1'];
$criteria2 = $_POST['criteria2'];
$arr=$_POST['criteria3'];
foreach($arr as $criteria3)
{
$sql="insert into table_name (criteria1, criteria2, criteria3) values('".$criteria1."', '".$criteria2."', '".$criteria3."')";
mysql_query($sql);
}
And if you want a single record for a single entity with multiple criteria3-values use the following code:
$criteria1 = $_POST['criteria1'];
$criteria2 = $_POST['criteria2'];
if(isset($_POST['criteria3'])) //true when at least on of the options is selected.
{
$criteria3=implode(',', $_POST['criteria3']);
}
else //i.e. when no option is selected.
{
$criteria3='';
}
$sql="insert into table_name (criteria1, criteria2, criteria3) values('".$criteria1."', '".$criteria2."', '".$criteria3."')";
mysql_query($sql);

Answer 2

Since I am the only one that will be inserting data - I found a crude work around until I can get the code that Rajesh provided to work - I think this is the right path if I can get it to work.Thanks to all for your input. If I could, I would give you all the up vote!

Answer 3

remove the explode function from $_POST['criteria3'], Because it is already array it not need of explode so you try directly
$type = $_POST['criteria3'];
foreach($type as $k => $v) {
// here do the INSERT query with value $v
}
or if you prefer this link
1).Getting data from a multiple select dropdown with PHP to insert into MySQL
hope this helpful for you :)
<form action="" method="post" name="form1" >
<input name="criteria1" size="64">
<input name="criteria2" size="64">
<select name="criteria3[]" multiple >
<option value="5 ">5</option>
<option value="7">7</option>
<option value="10">10</option>
</select>
<?php
if(isset($_POST['submit']))
{
$criteria1 = $_POST['criteria1'];
$criteria2 = $_POST['criteria2'];
if(isset($_POST['criteria3']))
{
$query = "insert into table_name (criteria1, criteria2, criteria3) values";
foreach($_POST['criteria3'] as $list)
{
$query .= "('$criteria1','$criteria2','$list')";
}
echo $query;
}
}
?>

Answer 4

Explode only works with a string.
What you get from your form for criteria3 is an array, so you should act accordingly:
$criteria1 = $_POST['criteria1'];
$criteria2 = $_POST['criteria2'];
$criteria3 = implode(",", $_POST['criteria3']);
Here we generate a string based on the content of the table returned by your form.
We seperate each element by a comma.
Then we create the SQL query.
$sql = sprintf("INSERT INTO table (name, date, number) VALUES (%s, %s, %s);", $criteria1, $criteria2, $criteria3);
Hope this helps.