Question

我有一个项目，用户可以在线拥有自己的VirtualDiary。他/她注册登录和所有这些，并带到日期和文本区域的输入页面。我们的想法是，他们可以点击标题为NextPage和PreviousPage的页面顶部的两个按钮。这些将调用jquery ajax函数，该函数将反过来更改EntryID + 1或EntryID -1的值。这应该会改变页面上几乎所有内容的价值。但即使ajax调用记录成功，也没有任何反应。我对ajax很新，所以我可能做了一件非常愚蠢的事。提前致谢

PHP

<?php 
 session_start();

 //error_reporting(E_ERROR|E_WARNING);
 mysql_connect("localhost","root","") or die ("cannot");
 mysql_select_db("virtualdiary") or die ("db");

$JoinDateQuery = mysql_query("SELECT * FROM users WHERE UID = '".$_SESSION['UID']."' ");

if($JoinDateQuery === FALSE) {
die(mysql_error()); // TODO: better error handling
}

while($row = mysql_fetch_array($JoinDateQuery))
{
$JoinDate =  $row[4];
}


 $TodayDate = date("Y/m/d");
 $today = strtotime($TodayDate);
 $joinTime = strtotime($JoinDate);
 $datediff = $today - $joinTime;
 $_SESSION["EntryID"] = floor($datediff/(60*60*24));
 $EntryID = $_SESSION["EntryID"];
 $_SESSION['EntryDate'] = date('Y-m-d', strtotime($JoinDate. ' + '.$EntryID .'days'));
 $EntryDate = $_SESSION['EntryDate'];


$id = $_SESSION["UID"] ;








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

$deletion = "DELETE FROM entries WHERE UserID = '".$_SESSION['UID']."' and EntryID = '".$EntryID."' ";
mysql_query($deletion);
$submission = "INSERT INTO `virtualdiary`.`entries` (`Entry`, `UserID`,`EntryID`) VALUES ('". $entry . "',                    '".$_SESSION['UID']."', '".$EntryID."')";
mysql_query($submission);


}
$ThePost = 'SELECT * FROM entries WHERE UserID = "'.   $_SESSION['UID'] .'" and EntryID = "'.$EntryID.'"';
$result = mysql_query($ThePost);

if($result === FALSE) {
die(mysql_error());
}




 ?>

<html>

 <head>
 <link type="text/css" rel="stylesheet" href="Entry.css"/>
 <title>Home</title>
 <script src="http://code.jquery.com/jquery-latest.min.js"></script>
 <script>
  $(document).ready(function(){
 $('#NextDay').click(function(){
 $.ajax({
 type: "GET",
 url: "Home.php",
 data: "$EntryID = $EntryID + 1",
success: console.log("success")     

}); //ajax call
 });//on click next day
 });//document ready</script>

</head>
<body>
<section>
<button id="previousDay" class="day">Previous Day</button>
<button class = "day" id = "date">Joined: <?php echo $JoinDate; ?></br>
                 Entry Number: <?php echo $EntryID + 1; ?></br>
                 EntryDate: <?php echo $EntryDate ; ?>
                 </button>
<button class="day" id = "NextDay">Next Day</button>
<h1>Entry:  </h1>
<form method="post" action="Home.php">
<textarea name="entry" rows="24" cols="80">
<?php 
 while($row = mysql_fetch_array($result))
{
echo $row['Entry'];
}
?>
</textarea>
 </br>
 </br>
 <input name="submit" type="submit"/>    
 </form>

<a href="Calender.php"> <button id="calender" class = "day"><h1>Calender</h1></button></a>

 <button id="LogOut"><a href="LogOut.php">Log Out</a></button>
 </section>
 </body>

 </html>

顺便说一句EntryID返回4的结果（或者当天用户的默认值），所以很明显这个问题与ajax的Data:部分有关或者我我没有在正确的语境中使用ajax来实现我想要的目标。

编辑：我刚刚意识到$EntryID = $EntryID + 1必须在某处定义，但我不能在某处将其删除，因为这会改变我认为的第一个条目id实例。

Answer 1

来自http://api.jquery.com/jquery.ajax/

数据键入：PlainObject或String 要发送到服务器的数据。如果不是字符串，它将转换为查询字符串。它附加到GET请求的URL。请参阅processData选项以防止此自动处理。对象必须是键/值对。如果value是一个数组，jQuery会根据传统设置的值使用相同的键序列化多个值（如下所述）。

你正试图将“$ EntryID = EntryID + 1”改为“home.php”

“home.php $ EntryID = EntryID + 1”似乎不是正确的网址？

如果您想了解有关请求状态的更多信息，则应尝试使用错误回调集。

Answer 2

首先，我认为你应该制作一个AJAX＆＃34; POST＆＃34;要求，而不是＆＃34; GET＆＃34;请求。（只是行业标准，因为您将值发送到服务器）。

其次，我不确定，你在哪里获得了为该GET请求创建数据值的语法（真的很奇怪）！在我编辑代码的ajax调用中，现在应该是正确的创造和创造的方式发送可反序列化的数据。

修改忘记提及，可以使用$_REQUEST['xyx']或$_POST['xyz']检索从AJAX发送的值。如果您使用的是get，则可以使用$_GET['xyz']

<强> UNTESTED

//error_reporting(E_ERROR|E_WARNING); mysql_connect("localhost", "root", "") or die ("cannot"); mysql_select_db("virtualdiary") or die ("db"); $JoinDateQuery = mysql_query("SELECT * FROM users WHERE UID = '" . $_SESSION['UID'] . "' "); if ($JoinDateQuery === FALSE) { die(mysql_error()); // TODO: better error handling } while ($row = mysql_fetch_array($JoinDateQuery)) { $JoinDate = $row[4]; } $TodayDate = date("Y/m/d"); $today = strtotime($TodayDate); $joinTime = strtotime($JoinDate); $datediff = $today - $joinTime; $_SESSION["EntryID"] = floor($datediff / (60 * 60 * 24)); $EntryID = $_POST["EntryID"]; $_SESSION['EntryDate'] = date('Y-m-d', strtotime($JoinDate . ' + ' . $EntryID . 'days')); $EntryDate = $_SESSION['EntryDate']; $id = $_SESSION["UID"]; if (isset($_POST["entry"])) { $entry = $_POST["entry"]; $deletion = "DELETE FROM entries WHERE UserID = '" . $_SESSION['UID'] . "' and EntryID = '" . $EntryID . "' "; mysql_query($deletion); $submission = "INSERT INTO `virtualdiary`.`entries` (`Entry`, `UserID`,`EntryID`) VALUES ('" . $entry . "', '" . $_SESSION['UID'] . "', '" . $EntryID . "')"; mysql_query($submission); } $ThePost = 'SELECT * FROM entries WHERE UserID = "' . $_SESSION['UID'] . '" and EntryID = "' . $EntryID . '"'; $result = mysql_query($ThePost); if ($result === FALSE) { die(mysql_error()); } ?> <html> <head> <link type="text/css" rel="stylesheet" href="Entry.css"/> <title>Home</title> <script src="http://code.jquery.com/jquery-latest.min.js"></script> <script> $(document).ready(function () { $('#NextDay').click(function () { $.ajax({ type: "POST", url: "Home.php", data: { EntryID: '1' //not sure (this is quite conditional I guess) }, success: console.log("success") }); //ajax call });//on click next day });//document ready</script> </head> <body> <section> <button id="previousDay" class="day">Previous Day</button> <button class="day" id="date">Joined: <?php echo $JoinDate; ?></br> Entry Number: <?php echo $EntryID + 1; ?></br> EntryDate: <?php echo $EntryDate; ?> </button> <button class="day" id="NextDay">Next Day</button> <h1>Entry: </h1> <form method="post" action="Home.php"> <textarea name="entry" rows="24" cols="80"> <?php while ($row = mysql_fetch_array($result)) { echo $row['Entry']; } ?> </textarea> </br> </br> <input name="submit" type="submit"/> </form> <a href="Calender.php"> <button id="calender" class="day"><h1>Calender</h1></button> </a> <button id="LogOut"><a href="LogOut.php">Log Out</a></button> </section> </body> </html>

我显然有一个错误的Ajax调用

2 个答案: