小编for*_*une的帖子

我被困住了，无法解决这个错误，我认为这可能是我的数据库选项搞砸了

这是数据放入表格的地方

 <form name = "quoted" method="post" onsubmit="get_action(this);">
 <input id = "poster" type="text" name="poster" required="required" placeholder = "Credited Individual.">     <br>
 <textarea class = "actual_quote" name = "actual_quote" required="required" placeholder = "Write the question here!"></textarea><br><br><br>
 <div class = "checkboxes" required="required">
     <h3 style = "margin-top:-20px;">Please select one catagory that the quote falls into.</h3>
     <label for="x"><input type="radio" name="x" value="Inspirational" id = "inspirational.php" checked="checked" />    <span>Inspirational</span></label><br>
     <label for="x"><input type="radio" name="x" value="Funny" id = "funny.php" /> <span>Funny</span>    </label><br>
     <label for="x"><input type="radio" name="x" value="OutofContext" id = …

在页面中,两个变量在URL中传递.

在第二页中,初始化这些变量.提交时,新页面打开.问题出在这里,我无法恢复会话变量.

该脚本表示错误(未定义的索引).

提前感谢您的帮助.

这里的代码:

第一页:

<?php
$requete_cours="SELECT COURS.SIGLE AS SIGLE, ANNEE FROM COURS, MODULE WHERE           COURS.ID_MODULE = MODULE.ID_MODULE and ID_PERSONNE = $userid";
//$requete_cours;   
$res = mysqli_query($cxn, $requete_cours);
echo (mysqli_error ($cxn));
while($ligne = mysqli_fetch_array($res)){
echo '<a href="professeur_absences.php?classe='.$ligne['ANNEE'].'&amp;cours='.$ligne['SIGLE'].'">';
echo $ligne['ANNEE'].' - '.$ligne['SIGLE'].'</a>';
echo '<br/>';
}   
?>  

第二页:

$_SESSION['classe'] = $_GET['classe'];
$_SESSION['cours'] = $_GET['cours'];

第三页:

if(isset($_REQUEST['afficher'])){
    $semaine = $_REQUEST['semaine'];
    $classe = $_SESSION['classe'];
    $cours = $_SESSION['cours'];
    $_SESSION['semaine'] = $semaine;
    $requete_cours_id = "SELECT ID_COURS FROM COURS WHERE SIGLE = $cours";
    //echo …

所以，我想用单例创建一个客户端基本 URL。

这是我的 GuzzleClient.php，其中包含基本 URL

<?php

use GuzzleHttp\Client;
use GuzzleHttp\Exception\RequestException;

class GuzzleClient {
    public static function getClient()
    {
        static $client = null;
        if (null === $client) 
        {
            $client = new Client([
                'base_url' => 'http://localhost:8080/task_manager/v1/',
            ]);
        }

        return $client;
    }


    private function __construct() 
    {}
}

这是我应该放置基本网址的地方

require_once 'GuzzleClient.php';

class CardTypeAPIAccessor 
{

    private $client;

    public function __construct($client) 
    {
        $this->client = $client;
    }

    public function getCardTypes() {
        $cardTypes = array();

        try 
        {
            //this is where base URL should be
            $response = $client->get('admin/card/type',
                ['headers' …

我正在调用 Web 服务以获取 JSON 字符串响应，它包含不在原始字符串中的反斜杠。下面是我请求一个 JSON 字符串对象的代码，它是： {"name":"Name","id":1}

protected String doInBackground(String... uri) 
    {
        HttpClient httpclient = new DefaultHttpClient();
        HttpResponse response;
        String responseString = null;
        try {
            response = httpclient.execute(new HttpGet(uri[0]));
            StatusLine statusLine = response.getStatusLine();
            if(statusLine.getStatusCode() == HttpStatus.SC_OK){
                ByteArrayOutputStream out = new ByteArrayOutputStream();
                response.getEntity().writeTo(out);
                out.close();
                responseString = out.toString();
            } else{
                //Closes the connection.
                response.getEntity().getContent().close();
                throw new IOException(statusLine.getReasonPhrase());
            }
        } catch (ClientProtocolException e) {
            //TODO Handle problems..
        } catch (IOException e) {
            //TODO Handle problems..
        }
        return responseString;
    }

在执行后，我只是尝试将此响应字符串解析为 JSONObject。代码如下：

protected …

以前我用jQuery开发了所有项目,但最近开始使用Angular.我用Yeoman生成器创建了一个Angular样板,并从中扩展了我的代码.

我将jQuery代码的ajax表单翻译成以下Angular控制器.

'use strict';


angular.module('MyApp')
.controller('ShareformCtrl', function ($scope, $http) {
// create a blank object to hold our form information
// $scope will allow this to pass between controller and view
$scope.formData = {};

// process the form
$scope.processForm = function() {
    $http({
        method : 'POST',
        url : 'php/share_handling.php',
        data : $.param($scope.formData), // pass in data as strings
        headers : { 'Content-Type': 'application/x-www-form-urlencoded' } // set the headers so angular passing info as form data (not request payload)
    })
    .success(function(data) …

我刚下载了新发布的测试版Android Studio.尝试启动应用程序时,没有任何反应.

OS X控制台中显示以下消息:

"com.apple.launchd.peruser.501: (com.google.android.studio.58288[542]) 退出代码:1"

我正在运行OS X Mavericks并Java 1.7.0_51已安装.

任何帮助解决这个问题非常感谢:)