2 for each statements at same time? - php

Question

I recently have been trying to make a way to easily add more fields onto my form without having to go back and add more rows to my database structure. So, to begin working on this, I created a table where the structure is this:
OptionTitle
Option1
Option2
Option3
Option4
Option5
Option6
As you can see, it goes up to 6 options, and OptionTitle is the label name of the form. Then I made another table, one that reflects the users input of the previous table. This table is named usersoption
fid
OptionTitle
Option1
Ok, so FID reflects which form it is referencing to. This way, when displaying the submitted form, it'll pull information from this table where the FID is the same. OptionTitle is the label of the form, and Option1 is the option the user submitted.
Now, onto the form where it actually includes the options to select from. Here is a simplified version of how my code is included:
$query100 = $db->query("SELECT * FROM options WHERE fid='" . $id . "'");
while($row2 = mysqli_fetch_array($query100))
{
echo "
<div class=\"divform\" id=\"optiontitle\">
<label for=\"optiontitle\">$row2[optiontitle]:</label>
<select name=\"option1[]\" id=\"option1\">";
echo "<option value='$row6[option1]'>$row6[option1]</option>";
echo "<option value='$row6[option2]'>$row6[option2]</option>";
echo "<option value='$row6[option3]'>$row6[option3]</option>";
echo "<option value='$row6[option4]'>$row6[option4]</option>";
echo "<option value='$row6[option5]'>$row6[option5]</option>";
echo "<option value='$row6[option6]'>$row6[option6]</option>";
echo "
</select>
</div>
";
}
As you can see, the select name is option1[]. This is so I can have multiple select fields on the same form, and in return this will bring over the multiple difference select fields onto the submitted process. So now onto where my issue is, in the submission process. Here is what I have so far:
foreach($_POST['option1'] as $val){
$val = $db->escape_string($val);
$query30 = $db->query("INSERT `usersoption` SET `gid` = '".$id."', `fid` = '".$fid."', `optiontitle` = 'Where OptionTitle should go', `option1` = '$val'")or die( mysqli_error());
}
As you can see, I can successfully bring the option through a foreach statement. What I can't do, is bring in the OptionTitle. It seems almost unnecessary to bring in the OptionTitle, but it is necessary for the person reading the submitted form to know which option was being submitted. I'm not sure how to carry the OptionTitle over, it seems simple but all my attempts failed miserably. I did some research and one of the suggestions was to create a hidden input with the name and carry it over that way. Here is the addon that would be in the form:
<input type=\"hidden\" name=\"optiontitle[]\" value=\"test\">
This would be added on to the form and then carried over, but the issue is how do I bring it over? I would need to do a multiple foreach statement which does not work. For example, here was what I tried to bring over (it did not work):
foreach($_POST['option1'] as $val) && ($_POST['optiontitle'] as $val2)){
$val = $db->escape_string($val);
$val2 = $db->escape_string($val2);
$query30 = $db->query("INSERT `usersoption` SET `gid` = '".$id."', `fid` = '".$fid."', `optiontitle` = '$val2', `option1` = '$val'")or die( mysqli_error());
}

Answer 1

Have you tried giving your option array a key?
echo "<select name=\"option1[$row2[optiontitle]]\" id=\"option1\">";
Then change your foreach to:
foreach($_POST['option1'] as $title=>$val)

Answer 2

You can use key in foreach to access more array:
Try this code:
foreach($_POST['option1'] as $key=>$val){
$val = $db->escape_string($val);
$val2 = $db->escape_string(isset($_POST['optiontitle'][$key])?$_POST['optiontitle'][$key]:'');
$query30 = $db->query("INSERT `usersoption` SET `gid` = '".$id."', `fid` = '".$fid."', `optiontitle` = '$val2', `option1` = '$val'")or die( mysqli_error());
}

Answer 3

For the hidden input solution: Just do each query as you normally would, but add $_POST['optiontitle']:
foreach($_POST['option1'] as $val){
$val = $db->escape_string($val);
$query30 = $db->query("
INSERT `usersoption` SET
`gid` = '".$id."',
`fid` = '".$fid."',
`optiontitle` = '".$_POST['optiontitle']."',
`option1` = '$val'
")or die(mysqli_error());
}
By the way you should read up on prepared statements. These allow you to sanitise your data before inserting into the database. They are essential to good coding practise.