Add multiple inputs with same name to mysql database

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!

@Jitendra Yadav Thnx, but when I use this code my result is 8 times the letter A. — fomfur
– fomfur, Commented Jun 6, 2014 at 5:17
You are doing wrong with form and foreach look answer with i have posted. :) — Jitendra Yadav
– Jitendra Yadav, Commented Jun 6, 2014 at 5:34

Jitendra Yadav · Accepted Answer · 2014-06-06 05:31:21Z

2

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();

answered Jun 6, 2014 at 5:31

Jitendra Yadav

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2 Comments

fomfur Over a year ago

This is exactly what I was looking for. I see the count and what I was doing wrong. Thank you very much!

fomfur Over a year ago

I have one question about this. How is it possible that the counter is stopping by '1'? When you add another fieldset is not showing up in the array?

Sherif · Accepted Answer · 2014-06-06 05:21:00Z

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'] . ')';
}

dipendra · Accepted Answer · 2014-06-06 05:37:14Z

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++;
}

sudhansu sekhar · Accepted Answer · 2015-06-01 06:22:43Z

                       <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 ..............

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