How to check whether dropbox item is selected?
I have a mysql database with two tables:
table1 (id, name, emailid)
table2 (id, email)
emailid
is a relationship totable2.id
I'm trying to make an html form that lists the contents of table1 with a dropdown box for the user to select the email field. The email field is populated with <options>
from table2
.
My question is: how do I check what value is already selected for emailid
and make that the already selected item in the dropbox when the form loads with selected="selected"
?
I've gotten one solution to work already but I don't think it is the proper way to do it and was looking for the best practices method.
The way I am currently doing it will only work with one item pulled from table1
but not if there are more than 1.
Here is my code so far:
This is where I get the data to make the form:
<?php
$sql = "SELECT table1.id as table1id, table1.name, table2.id as table2id
FROM table1
INNER JOIN table2 ON table1.emailid = table2.id
WHERE table1.id = {$id}";
$result = mysqli_query($link, $sql);
$row = mysqli_fetch_array($result);
$items1 = array('table1id' => $row['table1id']
'name' => $row['table1.name']
'table2id' =>开发者_开发问答; $row['table2id']);
$sql = "SELECT id, email FROM table2";
$result = mysqli_query($link, $sql);
$row = mysqli_fetch_array($result);
while ($row = mysqli_fetch_array($result)) {
$items2[] = array('id' => $row['id'],
'email' => $row['email'],
'selected' => ($row['id'] == $items1['table2id']) ? ' selected="selected" ' : '');
}
?>
This is my HTML:
<tr>
<td><?php echo $items1['name']; ?></td>
<td><input type="hidden" name="table1id" value="<?php echo $items1['table1id']; ?>" />
<?php echo $items1['table1id']; ?></td>
<td>
<select name="email">
<?php foreach ($items2 as $item2): ?>
<option value="<?php echo $item2['id']; ?>"<?php echo $item2['selected']; ?>>
<?php echo $item2['email']; ?></option>
<?php endforeach; ?>
</select>
</td>
</tr>
This is how the form is updated:
<?php
$id = mysql_real_escape_string($_POST['table1id']);
$email = mysql_real_escape_string($_POST['email']);
$sql = "UPDATE table1 SET emailid = {$email} WHERE id = {$id}";
?>
EDIT Killed that other stuff now that you have code up.
// Lets make our table2 the primary so we can organize our values in the $items array.
$sql = "SELECT table1.id as table1id, table1.name, table2.id as table2id, table2.email
FROM table2
LEFT JOIN table1 ON table2.emailid = table1.id";
// Do you need this WHERE? I'm thinking not.
// WHERE table1.id = {$id}";
$items = array();
while($row = mysqli_fetch_array($result)) {
// This will get set every time through the loop, but this is ok.
// You can check to see if its set and then set or not set.
$items[$row['table1id']]['name'] = $row['name'];
// We will keep appending email values here to the table1id array.
$items[$row['table1id']]['emails'][] = $row['email'];
}
Lets do the HTML markup now that we have all the items in a nice array for us.
<?php foreach($items as $id => $item): ?>
<tr>
<td><?php echo $item['name']; ?></td>
// Do you really need this hidden field?
<td><input type="hidden" name="table1id" value="<?php echo $id; ?>" /><?php echo $id; ?></td>
<td>
// Updated this to be unique for each id iteration.
<select name="email_<?php echo $id; ?>">
// Might want to consider a is_array() check on this value.
<?php foreach ($item['email'] as $email): ?>
<?php $selected = $email == $_POST['email_'. $id] ? ' selected="selected"' : ''; ?>
<option value="<?php echo $email; ?>"<?php echo $selected; ?>>
<?php echo $email; ?></option>
<?php endforeach; ?>
</select>
</td>
</tr>
<?php endforeach; ?>
I have not tested this so it may not work as a copy and paste. But I think this is can get you going correctly now.
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