使用ajax将数组发布到PHP
我在使用AJAX将数组发送到PHP页面时遇到问题。我一直在使用this question作为指导，但无论出于什么原因，我仍然不能让它工作。从我使用print_r($_POST)可以看出的是，我发布了一个空数组，但是在HTML/Javascript页面上，我使用一个警告来查看数组已被填充。post之所以起作用，是因为它在post时将空值输入到MySQL数据库中，但我不明白它为什么要传递一个空数组。代码如下：
Javascript：
<script type="text/javascript">
var routeID = "testRoute";
var custID = "testCustID";
var stopnumber = "teststopnumber";
var customer = "testCustomer";
var lat = 10;
var lng = 20;
var timeStamp = "00:00:00";
var dataArray = new Array(7);
dataArray[0]= "routeID:" + routeID;
dataArray[1]= "custID:" + custID;
dataArray[2]= "stopnumber:" + stopnumber;
dataArray[3]= "customer:" + customer;
dataArray[4]= "latitude:" + lat;
dataArray[5]= "longitude:" + lng;
dataArray[6]= "timestamp:" + timeStamp;
var jsonString = JSON.stringify(dataArray);
function postData(){
$.ajax({
type: "POST",
url: "AddtoDatabase.php", //includes full webserver url
data: {data : jsonString},
cache: false,
success: function(){
alert("OK");
}
});
window.location = "AddtoDatabase.php"; //includes full webserver url
}
alert(JSON.stringify(dataArray))
</script>PHP：
<?php
print_r($_POST);
$routeID = $_POST['routeID'];
$custID = $_POST['custID'];
$stopnumber = $_POST['stopnumber'];
$customer = $_POST['customer'];
$latitude = $_POST['latitude'];
$longitude = $_POST['longitude'];
$timestamp = $_POST['timestamp'];
$mysqli= new mysqli("fdb5.biz.nf","username","password","database");
mysqli_select_db($mysqli,"database");
$sql = "INSERT INTO Locations (routeID, custID, stopnumber, customer, latitude, longitude, timestamp) VALUES " .
"('$routeID','$custID','$stopnumber','$customer','$latitude','$longitude','$timestamp')";
mysqli_query($mysqli, $sql);
$error = mysqli_error($mysqli);
echo $error;
?>print_r($_POST)仅在php页面上显示数组()，而javascript页面上的jsonString警告显示["routeID:testRoute", "custID:testCustID", "stopnumber:teststopnumber", "customer:testCustomer", "latitude:10", "longitude:20", "timestamp:00:00:00"]
有人看到我做错了什么吗？
回答 2
Stack Overflow用户
发布于 2013-04-17 00:15:21
注意：导致代码输出array()的主要原因是您在发送/处理异步(AJAX)请求之前重定向了客户端
基本上，将window.location = "AddtoDatabase.php";移到success回调，正如前面提到的那样。
第一个问题:不应该使用数组，而应该使用对象文字(php中的~= assoc数组)。
要执行此操作，请更改此位：
var dataArray = new Array(7);//<== NEVER do this again, btw
dataArray[0]= "routeID:" + routeID;
dataArray[1]= "custID:" + custID;
dataArray[2]= "stopnumber:" + stopnumber;
dataArray[3]= "customer:" + customer;
dataArray[4]= "latitude:" + lat;
dataArray[5]= "longitude:" + lng;
dataArray[6]= "timestamp:" + timeStamp; 改为这样写：
var dataObject = { routeID: routeID,
custID: custID,
stopnumber: stopnumber
customer: customer,
latitude: lat,
longitute: lng,
timestamp: timeStamp};没有比这更多的了。要完成此操作，只需像这样发送数据：
function postData()
{
$.ajax({ type: "POST",
url: "AddtoDatabase.php",
data: dataObject,//no need to call JSON.stringify etc... jQ does this for you
cache: false,
success: function(resopnse)
{//check response: it's always good to check server output when developing...
console.log(response);
alert('You will redirect in 10 seconds');
setTimeout(function()
{//just added timeout to give you some time to check console
window.location = 'AddtoDatabase.php';
},10000);
}
});其次，您的postData函数在发送AJAX请求之前重定向客户端！在调用$.ajax之后，您的代码中有一条window.location = "AddtoDatabase.php";语句。如果您希望在ajax调用之后重定向客户端，则必须将该表达式移动到第二个代码片段^中的success回调函数(我在其中记录response)。
当您更改了所有这些内容后，您的$_POST变量应该看起来是正确的。如果没有，打印出$_REQUEST对象，然后看看ajax调用的响应是什么。
最后，请注意，使用支持预准备语句的api (从而保护您免受大多数注入攻击)，这并不意味着将未经检查的POST/GET数据串接到查询中会比过去更安全……
底线:当您使用支持关键安全特性的API时，请使用这些特性。
为了绝对清楚和完整，这里也是PHP代码的一个稍微修改过的版本：
$routeID = $_POST['routeID'];
$custID = $_POST['custID'];
$stopnumber = $_POST['stopnumber'];
$customer = $_POST['customer'];
$latitude = $_POST['latitude'];
$longitude = $_POST['longitude'];
$timestamp = $_POST['timestamp'];
//you're connecting OO-style, why do you switch to procedural next?
//choose one, don't mix them, that makes for fugly code:
$mysqli = mysqli_connect('fdb5.biz.nf', 'username', 'password', 'database');//procedural
//or, more in tune with the times:
$mysqli= new mysqli("fdb5.biz.nf","username","password","database");//OO
mysqli_select_db($mysqli,"database");
//or
$mysqli->select_db('database');如果您愿意，请查看文档以查看我将从头到尾使用的所有方法的过程副本。我更喜欢OOP-API
//making a prepared statement:
$query = 'INSERT INTO Locations
(routeID, custID, stopnumber, customer, latitude, longitude, timestamp) VALUES
(?,?,?,?,?,?,?)';
if (!($stmt = $mysqli->prepare($query)))
{
echo $query.' failed to prepare';
exit();
}
$stmt->bind_param('s', $routeID);
$stmt->bind_param('s',$custID);
//and so on
$stmt->bind_param('d', $latitude);//will probably be a double
$stmt->execute();//query DB有关预准备语句的有用链接：
当涉及到在case: A PDO tutorial, too中获取data...
Stack Overflow用户
发布于 2013-04-17 00:19:43
您应该使用serialize。然后......
<script>
jQuery(document).ready(function($){
/* attach a submit handler to the form */
$("#submit").click( function(event) {
/* stop form from submitting normally */
event.preventDefault();
/*clear result div*/
$("#loginresponse").html('');
/* use serialize take everthing into array */
var frmdata = $("#formname").serialize();
$.ajax({
url: "/file.php",
type: "post",
dataType: "json",
data: frmdata,
success: function(data, textStatus){
if(data.redirect == 'true'){
$('#formresponse').html(data.message);
return true;
}else{
$('#formresponse').html(data.message);
return false;
}
},
error:function(){
$("#formresponse").html('error');
}
});
});
});
</script>而不是在php中使用post获取数据。
<?php
$routeID = $_POST['routeID'];
$custID = $_POST['custID'];
$stopnumber = $_POST['stopnumber'];
$customer = $_POST['customer'];
$latitude = $_POST['latitude'];
$longitude = $_POST['longitude'];
$timestamp = $_POST['timestamp'];
?>并使用json编码显示。这样你就可以显示错误了
<?php
if(true)
echo json_encode(array('redirect'=>'true', 'message'=>'form submitted'));
else
echo json_encode(array('redirect'=>'false', 'message'=>'form not submited'));
?>https://stackoverflow.com/questions/16041835
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