通过ajax获取输入字段的值,然后将其传递给php
问题描述:
我是新来的ajax有人帮助我我想创建一个包含输入字段的表单。通过ajax获取输入字段的值,然后将其传递给php
每当我按一下按钮,我会得到输入字段的值,并将它宣布它在AJAX数据,并从阿贾克斯将它传递给PHP脚本的价值。它会显示一个表格。
我的问题是如何让输入字段的值,并宣布它为AJAX数据。点击表后将在AJAX脚本中声明成功,该表将显示一个表。
预先感谢您
UPDATE:
@J_D,这是我为我的表单的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>
这里是我的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){
}
});
});
});
这里的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);
}
?>
答
尝试下面的代码:
$(document).ready(function(){
$('#btn-inquire-transmittal-number').click(function(){
$inputtextval = $('#transmittal_number_inquiry').val();
$.ajax({
type: 'POST',
url: 'getTransmittalNum.php', // wrap code with quote
data: {'transmittal_number_inquiry' : $inputtextval},
dataType : 'json', // expecting result type json
success: function(res){
// once you got result,
// populate table here
}
});
});
});
PHP
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "etransmittal";
//$selectedTransmittal = $_GET['q']; //<---- u need this?????
$con = mysqli_connect($servername,$username,$password,$dbname);
if(!$con){
die("Connection failed: " . mysqli_connect_error());
}
if(isset($_POST['transmittal_number_inquiry'])){ // <-- check for existence
$query = "SELECT en.transid, en.transdate, CONCAT(userlist.lname, ', ', userlist.fname, ' ', userlist.mname) AS sender_name,
userlist.`department`, en.document_number, doctype.document_type, doctype.document_description, vendor.`vendor_name`, en.`remarks`,
en.status_id, stat.status_name, en.total_amount
FROM tbl_encode_transmittal en
LEFT JOIN tbl_vendor vendor ON vendor.`vendor_id` = en.vendor_id
LEFT JOIN tbl_doctype doctype ON doctype.`doc_id` = en.doctype_id
LEFT JOIN tbl_userlist userlist ON userlist.userid = en.sender_id
LEFT JOIN tbl_userlist userlist1 ON userlist1.userid = en.`receiver_id`
LEFT JOIN tbl_status stat ON stat.status_id = en.status_id
WHERE en.`transid` = '{$_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);
}
?>
请发表你使用,让您的AJAX调用的代码,并为您的形式 –
什么是你尝试的HTML? –
@NorlihazmeyGhazali,将值传递给PHP,然后显示包含Mysql的提取数据的表。 – pvegetah