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[英]Javascript Files that enable sending and receiving json data in phonegap
[英]Receiving JSON object after sending data to server in javascript
我試圖將登錄單擊時的電子郵件和密碼傳遞給使用PHP的數據庫。 一旦測試了電子郵件和密碼組合,PHP就會發回一個json對象。 現在,我已經測試了php,它可以正常運行,並且將顯示具有正確“ success”或“ fail”值的json對象,但我的狀態是:請求已取消,或者“ success”從未返回,而是“失敗”。 我不知道這是怎么回事。
這是到目前為止的javascript:
//make sure the login is correct
function testLogin(email, password){
//make sure the boxes arent empty
if(email !== "" && password !=="")
{
//create a json object, and
var jsonObject = {"email":email, "password":password};
var finalObject = JSON.stringify(jsonObject);
//alert(finalObject);
//sending json object
$.ajax({
type: 'POST',
url: 'WebPHP/check_login.php',
data: finalObject,
dataType: 'json',
success: function(data)
{
if(data['result'] === "success"){
alet("good");
window.location.href = "AMessage.html";
} else{
alert("Invalid Email or Password");
}
}
});
} return false;
}
這是到目前為止我擁有的php:
<?php
require 'db_connect.php';
header('Content-type: application/json');
$json = file_get_contents('php://input');
$jsondata = json_decode($json);
$email = $jsondata->{'email'};
$password = $jsondata->{'password'}
//$email = "TestUser";
//$password = "TestPasss";
$sql1 = " SELECT *
FROM users
WHERE email = '$email' AND password = '$password';";
$result = mysqli_query($Thesisdb, $sql1) or die(mysqli_error($Thesisdb));
$rows = $result->num_rows;
$post_data = array();
if($rows == 1){
$post_data = array('result' => "success");
} else {
$post_data = array('result' => "fail");
}
echo json_encode($post_data);
mysqli_close($Thesisdb);
您的成功警報有一個錯字,在您的示例中,您使用alet
alert
:
if(data['result'] === "success"){
alert`("good");
window.location.href = "AMessage.html";
}
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