我有这个表单，用户可以在其中使用 JQuery 添加行：
<form id="frm_salary" role="form" action="" method="post">
<!--start of frm_salary-->
<div class="form-group">
<label for="ca_salary_date1">Date:</label>
<input type="date" class="form-control" name="ca_salary_date_filed" placeholder="Date" required/>
</div>
<div class="form-group">
<label for="ca_purpose1">Purpose:</label>
<input type="text" class="form-control" name="ca_salary_purpose" placeholder="Purpose" required/>
</div>
<table class="table table-bordered table-hover" id="tab_logic">
<thead>
<tr >
<th class="text-center">
Area
</th>
<th class="text-center">
Date/s
</th>
<th class="text-center">
Number of Local Hires
</th>
<th class="text-center">
Salary Per Local Hire Per Day
</th>
<th class="text-center">
Amount
</th>
</tr>
</thead>
<tbody>
<tr id='addr0'>
<td>
<input type="text" name='ca_salary_area[]' placeholder='Area' class="form-control" required/>
</td>
<td>
<input type="text" name='ca_salary_date[]' placeholder='Date/s' class="form-control" required/>
</td>
<td>
<input type="text" name='ca_salary_localhires[]' placeholder='Number of Local Hires' class="form-control" required/>
</td>
<td>
<input type="text" name='ca_salary_perday[]' placeholder='Salary Per Local Hire Per Day' class="form-control" required/>
</td>
<td>
<input type="text" name='ca_salary_amount[]' placeholder='Amount' class="form-control" />
</td>
</tr>
<tr id='addr1'></tr>
</tbody>
</table>
<div class="form-group">
<input type="hidden" class="form-control" name="ca_total_salary" >
</div>
<div class="text-center">
<input type="submit" name="submit_salary" class="btn btn-success" value="Submit" />
</div>
</form>
<!--end of frm_salary-->
我也有这个PHP代码，用于在数据库中插入其值。它正在工作。
<?php
$total_amount = 0;
if (isset($_POST['submit_salary'])) {
//calculations
for ($j = 0; $j < count($_POST['ca_salary_area']); $j++) {
$nos_hire1 = $_POST['ca_salary_localhires'][$j];
$nos_salaray1 = $_POST['ca_salary_perday'][$j];
$subtotal_salary_amount1 = $nos_hire1 * $nos_salaray1;
$total_amount += $subtotal_salary_amount1;
}
$form_data = array();
$form_data['date_filed'] = mysql_real_escape_string($_POST['ca_salary_date_filed']);
$form_data['purpose'] = mysql_real_escape_string($_POST['ca_salary_purpose']);
$form_data['total'] = $total_amount;
$tbl = "tblcasalaryform";
// retrieve the keys of the array (column titles)
$fields = array_keys($form_data);
// build the query
$sql_salary = "INSERT INTO " . $tbl . "
(`" . implode('`,`', $fields) . "`)
VALUES('" . implode("','", $form_data) . "')";
// run the query result resource
mysql_query($sql_salary);
$last_id_inserted = mysql_insert_id($bd);
echo $last_id_inserted;
for ($i = 0; $i < count($_POST['ca_salary_area']); $i++) {
$nos_hire = $_POST['ca_salary_localhires'][$i];
$nos_salaray = $_POST['ca_salary_perday'][$i];
$subtotal_salary_amount = $nos_hire * $nos_salaray;
$sql = "INSERT INTO `tblcasalaryformdetails` SET
`area` = '" . $_POST['ca_salary_area'][$i] . "',
`dates` = " . $_POST['ca_salary_date'][$i] . ",
`number_of_local_hires` = " . $_POST['ca_salary_localhires'][$i] . ",
`salary_per_local_hire_per_day` = " . $_POST['ca_salary_perday'][$i] . ",
`amount` = " . $subtotal_salary_amount . ",
`casalaryform_id` = " . $last_id_inserted . "";
mysql_query($sql);
}
echo '<p class="bg-success" style="padding:5px;border-radius:5px;text-align:center">Successfully Inserted</p>';
}
?>
我想为此使用 Ajax，我的问题是我如何从动态创建输入名称的表单表中传递/获取值？
是jQuery吗？
然后像这样：
var data = $( "#frm_salary" ).serialize();
$.post( "test.php", data )
但是由于jQuery.serialize()方括号存在一些问题，要将它们放在您的 _POST 美元中，您所要做的就是添加一点魔力：
var data = $( "#frm_salary" ).serialize().replace(/%5B%5D/g, '[]');
$.post( "test.php", data )
就这么简单！
如果你想从HTML表中动态提取数据，你可以使用JQuery的find（）方法。
例：
$('#tab_logic').click(function(){
var firstColumn= $(this).find('td:first').text();
var secondColumn= $(this).find('td:nth-child(2)').text()
var thirdColumn= $(this).find('td:nth-child(3)').text()
...
});
您可以在模态中显示提取的值，然后在数据库中发布值