【发布时间】:2023-04-09 22:03:01
【问题描述】:
我是 ajax 新手,有人帮助我,我想创建一个包含输入字段的表单。
每当我单击按钮时,我都会获取输入字段的值,并将其声明为 AJAX 中的数据,并将来自 ajax 的值传递给 PHP 脚本。它将显示一个表格。
我的问题是如何获取输入字段的值并将其声明为 AJAX 中的数据。单击该表后,将在 AJAX 脚本中成功声明该表,该脚本将显示一个表。
提前谢谢你
更新:
@J_D,这是我的表单的 html 代码:
<form method="post" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]); ?>">
<table cellpadding="15px">
<tr>
<td>Transmittal #</td>
<td><input type="text" class="form-control" style="padding-left:5px;" name="transmittal_number_inquiry" id="transmittal_number_inquiry" class="transmittal_number_inquiry" onKeyPress="return isNumberKey(event)" required></td>
</tr>
</table>
<div style="float:right; padding-right:110px; padding-top:10px;">
<a href="#modalTransmittalInquiry" class="btn btn-info" data-toggle="modal" id="btn-inquire-transmittal-number" data-backdrop="false" name="inquire-transmittal-number">Inquire</a>
<?php /*?><input type="submit" class="btn btn-info" data-toggle="modal" id="btn-inquire-transmittal-number" name="btn-inquire-transmittal-number" data-backdrop="false" value="Inquire"><?php */?>
<?php /*?><button type="submit" class="btn btn-info" data-toggle="modal" id="btn-inquire-transmittal-number" name="btn-inquire-transmittal-number" data-backdrop="false">Inquire</button><?php */?>
</div>
</form>
这是我的 AJAX 代码:
$(document).ready(function(){
$('.btn-inquire-traensmittal-number').click(function(){
$inputtextval = $('#transmittal_number_inquiry').val();
$.ajax({
type: 'POST',
url: getTransmittalNum.php,
data: {'transmittal_number_inquiry' : $inputtextval},
success: function(res){
}
});
});
});
这是 getTransmittalNum.php 代码
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "etransmittal";
$selectedTransmittal = $_GET['q'];
$con = mysqli_connect($servername,$username,$password,$dbname);
if(!$con){
die("Connection failed: " . mysqli_connect_error());
}
if(isset($_POST['inquire-transmittal-number'])){
$query = "SELECT en.transid, en.transdate, CONCAT(userlist.lname, ', ', userlist.fname, ' ', userlist.mname) AS sender_name,
userlist.`department`, en.document_number, doctype.document_type, doctype.document_description, vendor.`vendor_name`, en.`remarks`,
en.status_id, stat.status_name, en.total_amount
FROM tbl_encode_transmittal en
LEFT JOIN tbl_vendor vendor ON vendor.`vendor_id` = en.vendor_id
LEFT JOIN tbl_doctype doctype ON doctype.`doc_id` = en.doctype_id
LEFT JOIN tbl_userlist userlist ON userlist.userid = en.sender_id
LEFT JOIN tbl_userlist userlist1 ON userlist1.userid = en.`receiver_id`
LEFT JOIN tbl_status stat ON stat.status_id = en.status_id
WHERE en.`transid` = '{$_POST['transmittal_number_inquiry']}'";
$result = mysqli_query($con, $query);
$rows = array();
if($result){
while($row = mysqli_fetch_assoc($result)){
$rows[] = $row;
}
}
else{
echo 'MYSQL Error: ' . mysqli_error();
}
$json = json_encode($rows);
echo $json;
mysqli_close($con);
}
?>
【问题讨论】:
-
请发布您用于进行 ajax 调用的代码和表单的 html
-
你的尝试是什么?
-
@NorlihazmeyGhazali,将值传递给 PHP,然后显示包含从 Mysql 获取数据的表。
标签: javascript php jquery ajax