Question

我试图使用ajax从具有相同结构和名称的多个输入字段的表单填充我的mysqli数据库，但我非常困在这里。我已经尝试了几个解决方案，但都有相同的结果，只有我的最后两个输入字段被发送到我的数据库。

我的表单看起来像这样;

<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>

ajax部分;

$('#sendform').submit(function(e){
$.ajax({
    type: 'POST',
    url: 'inc/post.php', 
    data: $(this).serialize(),
    success: function(data){
        alert("Data Save: " + data);
    }
});
e.preventDefault();
});

我的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();

我在这里做错了什么？日Thnx！

Answer 1

试试这个：

<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>

和PHP页面

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

您可以使表单看起来像这样（请注意，member []现在是一个数组本身）

<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>

在PHP使用中

foreach ($_POST['member'] as $member) {
    $values[] = '(' . $member['shortname'] . ',' . $member['fullname'] . ')';
}

Answer 3

您在html中使用了成员[shortname]和成员[fullname]两次。因此，第二个字段集中的值将覆盖第一个字段集。

因此，我们只得到第二个字段集中的值。

您可以尝试使用以下代码：

<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>

我的php，进行以下更改：

$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 ..............

将具有相同名称的多个输入添加到mysql数据库

4 个答案: