javascript 通过ajax调用从php文件接收多个值
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/24715018/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Receive multiple value from php file via ajax call
提问by user3817516
Below is my ajax call code. I want to send one data in .php file via ajax call and want to get two values from .php file. This two values I want to set in different 'input' tags whose id are 'course_name' and 'course_credit'.
下面是我的ajax调用代码。我想通过 ajax 调用在 .php 文件中发送一个数据，并想从 .php 文件中获取两个值。我想在不同的“输入”标签中设置这两个值，它们的 id 是“course_name”和“course_credit”。
Here my ajax call return correct value(real value from DB table) of 'course_name' input tag.
But 'MY PROBLEM IS' the value of input tag whose id is 'course_credit' shows 'success'. How can I get the correct value(real value from DB table) of id 'course_credit' ?
这里我的 ajax 调用返回了“course_name”输入标签的正确值（来自 DB 表的实际值）。
但是“我的问题是”id 为“course_credit”的输入标签的值显示“成功”。如何获得 id 'course_credit' 的正确值（来自数据库表的实际值）？
I have a 'select' tag which id is 'c_select'
我有一个 'select' 标签，它的 id 是 'c_select'
HTML:
HTML：
<input type="text" name="course_name" id="course_name" value=""/>
<input type="text" name="course_credit" id="course_credit" value=""/>
AJAX :
阿贾克斯：
$('#c_select').change(function(){
$.ajax({
type:'post',
url:'get_course_info_db.php',
data: 'c_id='+ $(this).val(),
success: function(reply_data1,reply_data2){
$('#course_name').val(reply_data1);
$('#course_credit').val(reply_data2);
}
});
});
get_course_info_db.php
get_course_info_db.php
<?php
include('db_connection.php');
$c_id = $_POST['c_id'];
$result = mysql_query("SELECT * FROM course WHERE c_id = '$c_id'");
$all_course_data = mysql_fetch_array($result);
$c_name = $all_course_data['c_name'];
$c_credit = $all_course_data['c_credit'];
echo $c_name,$c_credit;
exit();
?>
回答by WisdmLabs
AJAX code:-
AJAX 代码：-
$('#c_select').change(function(){
$.ajax({
type:'post',
url:'get_course_info_db.php',
data: 'c_id='+ $(this).val(),
success: function(value){
var data = value.split(",");
$('#course_name').val(data[0]);
$('#course_credit').val(data[1]);
}
});
});
PHP code:-
PHP代码：-
<?php
include('db_connection.php');
$c_id = $_POST['c_id'];
$result = mysql_query("SELECT * FROM course WHERE c_id = '$c_id'");
$all_course_data = mysql_fetch_array($result);
$c_name = $all_course_data['c_name'];
$c_credit = $all_course_data['c_credit'];
echo $c_name.",".$c_credit;
exit();
?>
回答by ThomasP1988
success: function(reply_data1,reply_data2){
$('#course_name').val(reply_data1);
$('#course_credit').val(reply_data2);
}
second arguement is the status of http request, you have to encode the answer, i suggest you JSON in your php
第二个争论是http请求的状态，你必须编码答案，我建议你在你的php中使用JSON
$c_credit = $all_course_data['c_credit'];
echo json_encode(array('name' => $c_name,'credit' => $c_credit));
exit();
and in your javascript
并在您的 javascript 中
success: function(response,status){
var datas = JSON.parse(response);
$('#course_name').val(datas.name);
$('#course_credit').val(data.credit);
}
this is not tested, but this is the way to do it
这没有经过测试，但这是做到这一点的方法
回答by Raven
The successcallback is Function( PlainObject data, String textStatus, jqXHR jqXHR );http://api.jquery.com/jQuery.ajax/
的成功回调 函数（PlainObject数据，字符串textStatus，jqXHR jqXHR）; http://api.jquery.com/jQuery.ajax/
php:
php:
$data = array(
'name' => $c_name,
'credit' => $c_credit,
);
echo json_encode($data);
javascript:
javascript:
success: function(data) {
var result = $.parseJSON(data);
$('#course_name').val(result.name);
$('#course_credit').val(result.credit);
}
回答by Hexus
I'd suggest using JSONto encode the data you fetch from the database.
我建议使用JSON对您从数据库中获取的数据进行编码。
Try changing your ajax call as follows:
尝试按如下方式更改您的 ajax 调用：
$('#c_select').change(function(){
$.ajax({
type:'post',
url:'get_course_info_db.php',
data: 'c_id='+ $(this).val(),
dataType: 'json', // jQuery will expect JSON and decode it for you
success: function(reply_data){
$('#course_name').val(reply_data['c_name']);
$('#course_credit').val(reply_data['c_credit']);
}
});
});
And your PHP as follows:
而你的 PHP 如下：
include('db_connection.php');
// Escape your input to prevent SQL injection!
$c_id = mysql_real_escape_string($_POST['c_id']);
$result = mysql_query("SELECT * FROM course WHERE c_id = '$c_id'");
$all_course_data = mysql_fetch_array($result);
echo json_encode($all_course_data);
exit();
I haven't tested this but I imagine it'd work for you.
我还没有测试过这个，但我想它对你有用。
