Update mysql database fields with array php - php

Question

I'm trying to achieve a multiple update in one submit. I have a table with a number of rows and want to be able to update one field in each row just by tabbing to the next insert box.
My code is thus:-
//start a table
echo '
';
//start header of table
echo '<tr>
<td width="60" align="center"><strong>Lab Item ID</strong></td>
<td width="60" align="center"><strong>Test Suite Name</strong></td>
<td width="60" align="center"><strong>Test Name</strong></td>
<td width="50" align="center"><strong>Result</strong></td>
</tr>';
//loop through all results
while ($row=mysql_fetch_object($sql)) {
//print out table contents and add id into an array and email into an array
echo '<tr>
<td align="center"><input name="id[]" value='.$row->lab_item_id.' readonly> </td>
<td align="center">'.$row->test_suite_name.'</td>
<td align="center">'.$row->test_name.'</td>
<td><input name="test_result[]" type="text" value="'.$row->test_result.'"></td>
</tr>';
}
//submit the form
echo'<tr>
<td colspan="3" align="center"><input type="submit" name="Submit" value="Submit"></td>
</tr>
</table>
</form>';
//if form has been pressed proccess it
if($_POST["Submit"])
{
//get data from form
//$name = $_POST['name'];
//$_POST['check_number'] and $_POST['check_date'] are parallel arrays
foreach( $_POST['id'] as $id ) {
$tresult = trim($_POST['test_result']);
$query = "UPDATE tbl_lab_item SET test_result='$tresult' WHERE lab_item_id = '$id'";
//execute query
}
print_r($_POST);
var_dump($tresult);
//redirect user
$_SESSION['success'] = 'Updated';
//header("location:index.php");
}
?>
When I print the $_POST arrays, everything is populating fine, however the variable is Null. I know I can't do a foreach on multiple arrays (at least I don't think I can) so is there some other trick I'm missing please? I can't be far away as the $_Post print has the right data in it.
Incidentally, the whole thing is generated by a query, so I never know how many records I'll have to update.
I've been looking at this (and other) forums, but can't seem to get a solution. I thought I understood arrays, but now I'm beginning to wonder!
edit - it's the $tresult variable that isn't working.
Many thanks,
Jason
Edit Thursday 21st Feb (05:41 UK time)
Thanks for your input everybody. I've solved this one now, and your collective advice helped. The code that finally cracked it is:-
//get data from form
$id1 = $_POST['id'];
$test_result1 = $_POST['test_result'];
foreach ($id1 as $key => $value){
$query = "UPDATE tbl_lab_item SET test_result='$test_result1[$key]' WHERE lab_item_id=$value ";
//execute query
Working through which variables etc were populated and what they were populated with was the key. Back to first principles, isn't it?
Cheers all.
J

Answer 1

Actually, you might get it done by doing a simpler (basic) form of for loop:
//get data from form
//$name = $_POST['name'];
//$_POST['check_number'] and $_POST['check_date'] are parallel arrays
$numberOfData = count($_POST['id']);
for($index = 0; $index < $numberOfData; $index++)
{
$id = $_POST['id'][$index];
$tresult = trim($_POST['test_result'][$index]);
$query = "UPDATE tbl_lab_item SET test_result='$tresult' WHERE lab_item_id = '$id'";
//execute query
}
print_r($_POST);
I hope this helps.
Cheers

Answer 2

change the query like this :
$query = "UPDATE tbl_lab_item SET test_result=$tresult WHERE lab_item_id = $id";
By adding single quotes ' ' you tell it to read is as a String and not to take the value of the var.
Edit
Replace your foreach loop with the following and let me know :
$id1 = $_POST['id'];
$test_result1 = $_POST['test_result'];
foreach( $id1 as $key => $value ) {
$query = "UPDATE tbl_lab_item SET test_result='$test_result1[$key]' WHERE lab_item_id = '$key' ";
}

Answer 3

Problem is that you're telling PHP to build your input fields as arrays, but then treat it as a string later:
<td><input name="test_result[]" type="text" value="'.$row->test_result.'"></td>
^^--- array
$tresult = trim($_POST['test_result']);
^^^^^^^^^^^^^^^^^^^^^--- no array key, so you're assigning the entire array
$query = "UPDATE tbl_lab_item SET test_result='$tresult'
^^^^^^^^^^--- array in string context
trim() expects a string, but you pass in an array, so you get back a PHP NULL and a warning. That null then gets stuffed into your SQL statement, and there's your problem.