Add multiple inputs with same name to mysql database - php

Question

I'm trying to fill my mysqli database from a form with multiple input fields with the same structure and names using ajax, but I'm pretty stuck here. I have tried several solutions but all with te same result, only my last two input fields are send to my database.
My form looks like this;
<form id="sendform" method="post" action="#">
<fieldset>
<input type="text" value="JD01" id="shortname" name="member[shortname]" readonly />
<input type="text" value="John Doe" id="fullname" name="member[fullname]" readonly />
</fieldset>
<fieldset>
<input type="text" value="JD02" id="shortname" name="member[shortname]" readonly />
<input type="text" value="Jane Doe" id="fullname" name="member[fullname]" readonly />
</fieldset>
<input type="submit" value="Send" />
</form>
The ajax part;
$('#sendform').submit(function(e){
$.ajax({
type: 'POST',
url: 'inc/post.php',
data: $(this).serialize(),
success: function(data){
alert("Data Save: " + data);
}
});
e.preventDefault();
});
And my php;
define('HOST', 'localhost');
define('USER', 'root');
define('PASS', 'root');
define('DBNAME', 'test');
$db = new mysqli(HOST, USER, PASS, DBNAME);
$values = array();
foreach ($_POST['member'] as $member) {
$values[] = '(' . $member['shortname'] . ',' . $member['fullname'] . ')';
}
$sql = "INSERT INTO sendform (shortname, fullname) VALUES " . implode(',', $values);
$result = $db->query($sql);
if($result) {
echo "Yep";
}
$db->close();
What am I doing wrong here? thnx!

Answer 1

Try this:
<form id="sendform" method="post" action="#">
<fieldset>
<input type="text" value="JD01" id="shortname" name="member[shortname][]" readonly />
<input type="text" value="John Doe" id="fullname" name="member[fullname][]" readonly />
</fieldset>
<fieldset>
<input type="text" value="JD02" id="shortname" name="member[shortname][]" readonly />
<input type="text" value="Jane Doe" id="fullname" name="member[fullname][]" readonly />
</fieldset>
<input type="submit" value="Send" />
</form>
And PHP Page:
define('HOST', 'localhost');
define('USER', 'root');
define('PASS', 'root');
define('DBNAME', 'test');
$db = new mysqli(HOST, USER, PASS, DBNAME);
$values = array();
for($i=0 ;$i < count($_POST['member']); $i++) {
$values[] = '("' . $_POST['member']['shortname'][$i] . '","' . $_POST['member']['fullname'][$i] . '")';
}
$sql = "INSERT INTO sendform (shortname, fullname) VALUES " . implode(',', $values);
$result = $db->query($sql);
if($result) {
echo "Yep";
}
$db->close();

Answer 2

You can make the form look like this (note that member[] is now an array itself)
<form id="sendform" method="post" action="">
<fieldset>
<input type="text" value="JD01" id="shortname" name="member[1][shortname]" readonly />
<input type="text" value="John Doe" id="fullname" name="member[1][fullname]" readonly />
</fieldset>
<fieldset>
<input type="text" value="JD02" id="shortname" name="member[2][shortname]" readonly />
<input type="text" value="Jane Doe" id="fullname" name="member[2][fullname]" readonly />
</fieldset>
<input type="submit" value="Send" />
</form>
in your PHP use
foreach ($_POST['member'] as $member) {
$values[] = '(' . $member['shortname'] . ',' . $member['fullname'] . ')';
}

Answer 3

You are using member[shortname] and member[fullname] two times in html. So values from second fieldset are overriding the first one.
Hence we are getting only the values the are put in second fieldset.
You can try using following code:
<form id="sendform" method="post" action="#">
<fieldset>
<input type="text" value="JD01" id="shortname" name="member['shortname'][0]" readonly />
<input type="text" value="John Doe" id="fullname" name="member['fullname'][0]" readonly />
</fieldset>
<fieldset>
<input type="text" value="JD02" id="shortname" name="member['shortname'][1]" readonly />
<input type="text" value="Jane Doe" id="fullname" name="member['fullname'][1]" readonly />
</fieldset>
<input type="submit" value="Send" />
</form>
And my php, make the following changes:
$values = array();
$i = 0;
foreach ($_POST['member'] as $member) {
$values[] = '(' . $member['shortname'][$i] . ',' . $member['fullname'][$i] . ')';
$i++;
}

Answer 4

<td class="text-left"><input type="text" class="form-control input-lg" name="expense[date][]" required placeholder="Enter Date (yy-mm-dd)"></td>
<td class="text-left"><input type="text" class="form-control input-lg" name="expense[msg][]" required placeholder="Enter Expenses Here"></td>
<td class="text-left"><input type="text" class="form-control input-lg" name="expense[vendor][]" required placeholder="Vendor Name Here"></td>
<td class="text-left"><input type="text" class="form-control input-lg" name="expense[amount][]" required placeholder="Enter Amount Here"></td>
<td class="text-left"><input type="text" class="form-control input-lg" name="expense[prove][]" required placeholder="Prove"></td>
PHP Code
<?php
if(isset($_POST['submit'])){
//$date = mysql_real_escape_string($_POST['date']);
//$date = strtotime($date);
// $date = date('Y-m-d', $date);
//$msg = mysql_real_escape_string($_POST['msg']);
//$vendor = mysql_real_escape_string($_POST['vendor']);
//$amount = mysql_real_escape_string($_POST['amount']);
//$prove = mysql_real_escape_string($_POST['prove']);
$values = array();
for($i=0 ;$i < count($_POST['expense']); $i++) {
$values[] = '("' . $_POST['expense']['date'][$i] . '","' . $_POST['expense']['msg'][$i] . '","' . $_POST['expense']['vendor'][$i] . '","' . $_POST['expense']['amount'][$i] . '","' . $_POST['expense']['prove'][$i] . '")';
}
$sql = ("INSERT INTO expenses (date, msg, vendor, amount, prove) VALUES " . implode(',', $values));
echo "<script>alert('Your expenses data inserted successfully!')</script>";
echo "<script>window.open('expenses.php','_self')</script>";
exit();
}
?>
show me the error Notice: Undefiened offset:2 in ..............