Question

我想将username和password传递给php脚本并检入数据库。在客户端，我使用以下脚本生成json object并将其发布到php文件。

var myobj = {};
myobj["usrname"]= $( "#customer option:selected" ).text();
myobj["usrpass"]= $("#psw").val();

var myjson = JSON.stringify(myobj);

$.ajax({
method: "POST",
url: "checkpass.php",
data: myjson
})
.done(function( msg ) {
    alert( msg );
});

在服务器端，当我在firebug中看到时，帖子将作为

传递

Parametersapplication / x-www-form-urlencoded不排序   { “usrname”：“XXXXXXX ...
  JSON

usrname       “XX”

usrpass       “去做就对了”   资源   { “usrname”： “XXX”， “usrpass”： “justdoit”}

然而，当我运行php脚本来检查查询时，它会返回错误

$usrname = $_POST['usrname'];
$usrpass = $_POST['usrpass'];

$sql = "select count(*) from glusers where EmpName='$usrname' and EmpPass='$usrpass'";
$result = $conn->query($sql);

if($result >0){
$output = 'Success';
} else
{
$output = 'fail';
}

我已经尝试了所有帖子，但无法让它发挥作用。

提前致谢。

此致

Answer 1

为了使ajax获得成功事件而回应并死亡该声明

Js档案

 var myobj = {};

        myobj["usrname"] = 'myUsername';

        myobj["usrpass"] = 'myPassword';



        $.ajax({
            type: "post",
            url: "url",
            dataType: "json",
            data: {post_data: myobj},
            contentType: "application/x-www-form-urlencoded",
            success: function (responseData) {
                console.log(responseData);
            },
            error: function (errorThrown) {
                console.log(errorThrown);
            }
        });

PHP操作文件

           /** if we print post we will get the following array * */
//print_r($_Post);
//die()
//Array
//(
//    [post_data] => Array
//        (
//            [usrname] => myUsername
//            [usrpass] => myPassword
//        )
//
//)

if (isset($_Post['post_data'])) {
    $myPost = $_Post['post_data'];
    $usrname = $myPost['usrname'];
    $usrpass = $myPost['usrpass'];

    $sql = "select count(*) from glusers where EmpName='$usrname' and EmpPass='$usrpass'";
    $result = $conn->query($sql);
    $num_row = $result->num_rows;

    if ($num_row > 0) {
        $output = 'Success';
    } else {
        $output = 'fail';
    }
    echo json_encode($output);
    die();
}

Answer 2

试试这个：在js文件中：

$(document).on("ready", function(){

            // Create an object using an object literal.
            var ourObj = {};

            // Create a string member called "data" and give it a string.
            // Also create an array of simple object literals for our object.
            ourObj.data = "Some Data Points";
            ourObj.arPoints = [{'x':1, 'y': 2},{'x': 2.3, 'y': 3.3},{'x': -1, 'y': -4}];


            var savedata = JSON.stringify(ourObj)
            $.ajax({
                type:"POST",
                url:"Users.php",
                data: {"points" : JSON.stringify(ourObj)},
               success: function(data) {
                    // Do something with data that came back. 
                    alert(data);
               }

            })
        });

在PHP文件中：

if (isset($_POST["points"])) {
$points = json_decode($_POST["points"]);


echo "Data is: " . $points->data . "<br>";
echo "Point 1: " . $points->arPoints[0]->x . ", " . $points->arPoints[0]->y;
}

Answer 3

试试这个：

var myobj = '{
usrname:'+$( "#customer option:selected" ).text()+',
usrpass:'+$("#psw").val()+'
}';

或

var myobj = {};
myobj.usrname= $( "#customer option:selected" ).text();
myobj.usrpass= $("#psw").val();

Answer 4

使用Json2库，如下所示，

var myobj = {};
myobj["usrname"]= $( "#customer option:selected" ).text();
myobj["usrpass"]= $("#psw").val();

var myjson = JSON2.stringify(myobj);

$.ajax({
 method: "POST",
 url: "checkpass.php",
 data: myjson
})
.done(function( msg ) {
  alert( msg );
});

Answer 5

实际上你的PHP代码是无效的，因为你传递了json日而不是名称值对，所以你不能从$ _POST [＆＃39;用户名＆＃39;]获得它。您需要获取整个发布数据并对其进行解码。

$ data = json_decode（file_get_contents（＆＃39; php：// input＆＃39;），true）;

现在$ data是用户名和密码的字典数组。在传递查询之前还要清理数据以避免sql注入。

php中的json对象未被读取

5 个答案: