我正在尝试将多个值插入到一个表中。首先,我让用户选择instructors
. 根据初始选择,jquery 显示与其对应的输入显示的数量。因此,如果用户选择了两个讲师,则会出现两个输入字段。我想要两个将两个输入字段的值插入 mysql 数据库。现在,没有插入任何内容,我不确定原因。网站。
<script type="text/javascript">
$(document).ready(function () {
$('#btnAdd').click(function () {
var num = $('.clonedSection').length;
var newNum = new Number(num + 1);
var newSection = $('#pq_entry_' + num).clone().attr('id', 'pq_entry_' + newNum);
newSection.children(':first').children(':first').attr('id', 'inst_fname_' + newNum).attr('name', 'inst_fname_' + newNum).attr('placeholder', 'Instructor #' + newNum + ' First Name');
newSection.children(':nth-child(2)').children(':first').attr('id', 'inst_lname_' + newNum).attr('name', 'inst_lname_' + newNum);
newSection.insertAfter('#pq_entry_' + num).last();
$('#btnDel').prop('disabled', '');
if (newNum == 5) $('#btnAdd').prop('disabled', 'disabled');
});
$('#btnDel').click(function () {
var num = $('.clonedSection').length; // how many "duplicatable" input fields we currently have
$('#pq_entry_' + num).remove(); // remove the last element
// enable the "add" button
$('#btnAdd').prop('disabled', '');
// if only one element remains, disable the "remove" button
if (num - 1 == 1) $('#btnDel').prop('disabled', 'disabled');
});
$('#btnDel').prop('disabled', 'disabled');
});
</script>
</head>
<body>
<form action="index.php" method="post">
Instructor:
<ul id="pq_entry_1" class="clonedSection">
<li>
<input id="inst_fname_1" name="inst_fname_1" placeholder="Instructor #1 - First Name" type="text" />
</li>
<li>
<input id="inst_lname_1" name="inst_lname_1" placeholder="Last Name" type="text" />
</li>
</ul>
<input type='button' id='btnAdd' value='add another instructor' />
<input type='button' id='btnDel' value='delete instructor' />
</br>
<input value="SAVE" type="submit">
<?php
$db_con = new mysqli(DB_SERVER, DB_USERNAME, DB_PASSWORD, DB_DATABASE);
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$db_insert3 = "INSERT INTO instructor (`instructor_fname`, `instructor_lname`)";
$db_insert3 .= " VALUES ('" . $_POST['inst_fname_1'] ."', '" . $_POST['inst_lname_1'] ."')";
mysqli_query($db_con, $db_insert3);
?>
</form>