[英]How to pass PHP values to js script in Wordpress using AJAX
[英]How to pass the values to php script using ajax?
我已经写了一个代码,使用ajax将输入字段的值传递给php脚本,但是它不起作用。 任何人都可以建议如何纠正此代码? 我想在php文件中显示通过ajax传递的值。
ex.php
<?php
$temp = $_POST['start_date'];
$name = $_POST['end_date'];
echo $temp.$name;
?>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<title>Untitled Document</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<script src="js/jquery-1.11.1.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$("input").change(function(){
var fname = $("#fname").val();
var lname = $("#lname").val();
$.ajax({
method: "POST",
url: "ex.php", // path to php file
data: { start_date: fname, end_date: lname } // send required data here
})
.done(function( msg ) {
});
});
});
</script>
</head>
<body>
<form action="#" method="post" name="form1">
<input type="text" name="fname" id="fname"/>
<input type="text" name="lname" id="lname"/>
</form>
</body>
</html>
try using below code :
<?php
$temp = $_POST['start_date'];
$name = $_POST['end_date'];
echo $temp.$name;
?>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<title>Untitled Document</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<script src="js/jquery-1.11.1.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$("input").change(function(){
var fname = $("#fname").val();
var lname = $("#lname").val();
var post = "&start_date=" + fname + "&end_date=" + lname;
$.ajax({
'url': ex.php,
'data': post,
'type': 'POST',
'success': function (data)
{
}
});
});
});
</script>
</head>
<body>
<form action="" method="post" name="form1">
<input type="text" name="fname" id="fname"/>
<input type="text" name="lname" id="lname"/>
</form>
</body>
</html>
为妆容添加元素
<span id="idOfSomeElementYouWantToUse"></span>
并在您的回调集中设置为文本。
.done(function( msg ) {
$("#idOfSomeElementYouWantToUse").text(msg);
});
将div添加到您的页面,然后使用html()
附加来自ajax请求的数据
<body>
<form action="#" method="post" name="form1">
<input type="text" name="fname" id="fname"/>
<input type="text" name="lname" id="lname"/>
</form>
<div class="ajax-result"></div>
</body>
js:
$("input").change(function(){
var fname = $("#fname").val();
var lname = $("#lname").val();
$.ajax({
method: "POST",
url: "other_file.php", // path to php file
data: { start_date: fname, end_date: lname }, // send required data here
success:function(data){
$(',ajax-result').htm(data);
}
});
注意:从我可以告诉您的内容来看,我强烈建议您使用逻辑将其创建为一个新的php文件(other_file.php)
从上面的评论中我可以理解,您想要向数据库中插入文本框值,而您不想在textChange事件上调用ajax,因为当时并不能显示所有值,建议添加提交按钮到表单,并在表单提交时调用ajax。
这是主文件:
ajax_ex.php (您应该根据需要命名)
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<title>Untitled Document</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>
<body>
<form id="frmInsert" method="post" name="form1">
<input type="text" name="fname" id="fname"/>
<input type="text" name="lname" id="lname"/>
<input type="submit" value="Insert" />
</form>
<div id="msg">
</div>
</body>
<!-- You should put javascript at bottom -->
<script src="jQuery-2.1.4.min.js"></script>
<script type="text/javascript">
$("#frmInsert").submit(function(e){
var fname = $("#fname").val();
var lname = $("#lname").val();
$.ajax({
method: "POST",
url: "insert.php", // path to php file
data: { start_date: fname, end_date: lname } // send required data here
})
.done(function( msg ) {
msg = $.trim(msg);
if (msg == 'true') {
$('#msg').html("Data is inserted successfully");
}else {
$('#msg').html("Data insertion failed");
}
});
e.preventDefault();
});
并且您不应该在ajax中调用相同的文件,我建议您创建将在ajax中调用的单个文件,该文件将处理所有后端处理。
insert.php
<?php
//Check if post data is available (You should also check for particular attribute if it is available)
if (!empty($_POST)) {
$temp = $_POST['start_date'];
$name = $_POST['end_date'];
//Insert Data into database
// Your code
$result = 'true'; //check for operation if it is success and store it in result
//Echo result so you can check if insert is success of not in your ajax callback.
echo $result;
}
?>
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