Question

我是PHP的新手，我遇到了一个问题，我找不到任何答案！这真让我烦恼...... 我想使用AJAX将数据发布到PHP脚本。我正在使用JS框架Prototype来进行AJAX通信。

这是JS代码：

new Ajax.Request("/ondemand/Radio.php", {
          method: 'POST',
          parameters: {programID: id}, 
          onSuccess: function(transport) {
                window.alert("Success, " + id);


          },
          onFailure: function() {
              window.alert("Communication problem");
          },
          onComplete: function() {
              window.alert("Complete");
          }
});

所有JS都在元素中......从框中选择选项时调用该函数

PHP代码：

<?php 
       //Selects all programs that have a podcast
       $QUERY_SELECT_ALL_PROGRAMS   = "SELECT DISTINCT d.defnr, d.name 
                       FROM definition d, podcast p 
                   WHERE p.program = d.defnr";
       //Select podcasts that belongs to a given program
       $QUERY_SELECT_PODCASTS_FOR_PROGRAM   = "SELECT p.title, p.refnr, p.filename
                           FROM podcast p
                       WHERE program =  ?";
       //Selects all podcasts
       $QUERY_SELECT_ALL_PODCASTS = "SELECT p.refnr, p.title, p.filename, p.filename 
                            FROM podcast";

        $BROADCAST_PATH = "";

        $programID = $_POST["programID"];


/* Returns true if DB connection to server and database is OK
 * Takes mysqli as parameter
 * Connect to the database using the MySQLi API in PHP 5.x
 * This is the prefered way*/
 function DBconnection($connection) {
   $result = false;
   //Refering to $con declared eralier
   //global $connection;
   //Check DB connection
 if ($connection->connect_error) { 
           die('Connect Error: '.$connection- >connect_error); }
else {
    //Refering to $DB_NAME declared earlier
    //Select DB
      global $DB_NAME;
       $DB_selected = $connection->select_db($DB_NAME);
      if (!$DB_selected) { die ('Can\'t use : ' . $connection->connect_error); }
       else { $result = true; }
  }
  return $result;
   }

   ?>


<?php
  echo "<form>";
  echo "<select>";
  //The MySQL connection object, must be created before connection
  $con = new mysqli($MYSQL_SERVER, $MYSQL_USER_NAME, $MYSQL_PASSWORD, $DB_NAME);
if (DBconnection($con)) {
    if ($stmt = $con->prepare($QUERY_SELECT_PODCASTS_FOR_PROGRAM)) {
        $stmt->bind_param("i", $program);
        //$stmt->bind_param("i", $program);
        //$program = $_POST["programs"];
        $program = $_POST["programID"];
        $stmt->execute();
        $stmt->bind_result($title, $refnr, $filepath);
    }
    if (is_null($_POST["programs"])) {
        echo "<option>Choose a program first...</option>";
        //echo "<option>".$file."</option>";
    }
    else {
        if (is_numeric($_POST["programs"])) {
            while($stmt->fetch()) {
                print_r($title);
                //$filepath holds the value of only the name of the broadcast without the entire path
                //40 is the starposition of the name
                $filename = substr($filepath, 40);
                echo "<option value=\"".$refnr."\" id=\"".$refnr."\" onclick=\"play('".$filename."')\">".utf8_encode($title).utf8_encode($filename)."</option>";
            }
        }
    }
     $con->close();
}
echo "</select>";
echo "</form>"; 
  ?>

这是我的问题...... $ _POST的值总是“Array（）”。当使用发布的常规表单时，一切正常，我得到了值，但是当使用AJAX（不仅是Prototype）时，我不会。

我想做的只是：用AJAX发布数据 - ＆gt;在sql查询中使用收到的数据 - ＆gt;根据sql wuery的结果制作HTML文件..

当我没有获得POST'ed variabel时，这有点困难

我还对发送的内容采取了措施，并且POSTDATA是正确的。拜托，我真的需要一些人的帮助...现在正在寻找几天的答案..

阅读此文章以便更好地理解.. Same problem

Answer 1

我不知道你是否在你的问题中粘贴了错误的代码，但$var = $_POST["programID"];似乎不在PHP块中，因此$ var将不会从POST数组中分配值。 ..

Answer 2

不知道如何评论所以我必须将此作为答案发布。如果从您的php脚本复制此代码：

$var = $_POST["programID"];
<p id="p">
<?php 
    echo "Program ID er " . utf8_encode($var);
?>
</p>

<?php and ?>

周围是否遗漏$var = $_POST["programID"];

编辑20101028：

好的，我看了一下你新发布的代码，找不到任何明显的东西，我对它做了一些修改：

 <?php

 $message = "<div id=\"php_debug\";
  style=\"border:1px solid red;
          background-color: black;
          color: white;\"
  >";// this is the debug message we will build and later print out where we need it

 //Selects all programs that have a podcast
 $QUERY_SELECT_ALL_PROGRAMS   = "SELECT DISTINCT d.defnr, d.name 
   FROM definition d, podcast p 
   WHERE p.program = d.defnr";
 //Select podcasts that belongs to a given program
 $QUERY_SELECT_PODCASTS_FOR_PROGRAM   = "SELECT p.title, p.refnr, p.filename
   FROM podcast p
   WHERE program =  ?";
 //Selects all podcasts
 $QUERY_SELECT_ALL_PODCASTS = "SELECT p.refnr, p.title, p.filename, p.filename 
   FROM podcast";

 $BROADCAST_PATH = "";
 $message .= "\$_POST["programID"] = ".$_POST["programID"]."<hr />";

 $programID = $_POST["programID"];
 $message .= "\$programID = ".$programID."<hr />";

 /* Returns true if DB connection to server and database is OK
 * Takes mysqli as parameter
 * Connect to the database using the MySQLi API in PHP 5.x
 * This is the prefered way*/
 function DBconnection($connection) {
   $message .= "I am in DBconnection($connection)<hr />";
   $result = false;
   //Refering to $con declared eralier
   //global $connection;
   //Check DB connection

   if ($connection->connect_error) { 
           die('Connect Error: '.$connection- >connect_error); }
   else{
     //Refering to $DB_NAME declared earlier
     //Select DB
     global $DB_NAME;
     $DB_selected = $connection->select_db($DB_NAME);
     if (!$DB_selected) { die ('Can\'t use : ' . $connection->connect_error); }
     else{ $result = true; }
   }
   return $result;
 }

 // Start generating form
 echo "<form>";
 echo "<select>";
 //The MySQL connection object, must be created before connection
 $con = new mysqli($MYSQL_SERVER, $MYSQL_USER_NAME, $MYSQL_PASSWORD, $DB_NAME);
 if (DBconnection($con)) {
   if ($stmt = $con->prepare($QUERY_SELECT_PODCASTS_FOR_PROGRAM)) {
   $message .= "\$program = ".$program."(\$program has never been seen before in this script)<hr />";
     //$stmt->bind_param("i", $program); This line was here originally, $program is null, has never been initialized before, $programID is the var where you store $_POST data, so I changed it to below
     $stmt->bind_param("i", $programID);
     $program = $_POST["programID"];//I left this here, becaause I dont know what you use $program for
     $message .= "\$program = ".$program."(\$program has been filled with \$_POST["programID"])<hr />";

     $stmt->execute();
     $stmt->bind_result($title, $refnr, $filepath);
   }
   $message .= "\$_POST["programs"] = ".$_POST["programs"]."(\$_POST["programs"] has never been sen before)<hr />";
   if(is_null($_POST["programs"])) {
   $message .= "\$_POST["programs"] was empty so I generated \"Choose program first...\" option"<hr />";
     echo "<option>Choose a program first...</option>";
     //echo "<option>".$file."</option>";
   }elseif(is_numeric($_POST["programs"])){
   $message .= "\$_POST["programs"] was numeric so I will start a loop here:<br /> START OF LOOP <br />";
     while($stmt->fetch()) {
       print_r($title);
       //$filepath holds the value of only the name of the broadcast without the entire path
       //40 is the starposition of the name
       $filename = substr($filepath, 40);
       $message .= "\$title = ".$title." and \$filename = ".$filename."<br />";
       echo "<option value=\"".$refnr."\" id=\"".$refnr."\" onclick=\"play('".$filename."')\">".utf8_encode($title).utf8_encode($filename)."</option>";
     }
   $message .= "END OF LOOP<hr />";
   }
   $con->close();
 }
 echo "</select>";
 echo "</form>";
 $message .= "</ div>";
 echo($message);//print out the debug message we were building through out the script 
 ?>

您可以使用上面发布的代码并告诉我们它打印出来的内容吗？

另外，我建议你在脚本开头处理所有的帖子数据：

$var1 = $_POST["var1"];
$var2 = $_POST["var2"];
$var3 = $_POST["var3"];
$var4 = $_POST["var4"];
$var5 = $_POST["var5"];

因为这样你可以省去很多困惑，以后只能使用简单的读取变量名。例如什么是$ _POST [“程序”]？是否与$ _POST [“programID”]相同？

总的来说，有很多特色，至少对我来说有变量名称：

$program
$programID
$programs
$_POST["program"]
$_POST["programs"]
$_POST["programID"]

使用哪些？哪些是正确的？

我倾向于通过以其目的开始变量名来解决这个问题：

$ id_program - ＆gt; db中单个程序或行的ID $ txt_program - ＆gt;程序描述或程序txt $ num_programs - ＆gt;程序数量，有多少程序（有多少个ID） $ arr_programs - ＆gt;程序数组，例如所有程序ID的数组

希望它有所帮助，让我知道。祝你好运：）

Answer 3

我可能会在这里遗漏一些东西，但你究竟想做什么？ Ajax.Request默认情况下不返回任何内容。可以使用responseText更新您的网页，但您的代码没有参考此内容。查看代码，您将不会在页面中看到任何更新内容（如果这是您期望的内容），但您应该从onSuccess回调中获取JavaScript警报。如果没有，请确认您收到的回调（或者如果没有错误）。

如果您希望使用AJAX更新页面中的值，则需要查看Ajax.Updater。需要说明的是，如果你想用Ajax调用的输出更新“p”的ID，你可以这样做：

new Ajax.Updater('p', '/ondemand/Radio.php', {
  parameters: { programID: id }
});

Answer 4

也许它正在烦恼并将请求作为GET发送？尝试：

print_r($_REQUEST);

$ _REQUEST var将包含$ _POST和$ _GET vars。

我也很难相信，但有可能代替method: 'POST'你应该使用method: 'post' - 大多数Prototype示例都使用小写。

Answer 5

如果您直接发布数据并且结果是您所期望的，那么生成的HTML就是您问题的根源。这一行：

 echo "<option value=\"".$refnr."\" id=\"".$refnr."\" onclick=\"play('".$filename."')\">".utf8_encode($title).utf8_encode($filename)."</option>";

是导致问题的原因。 "Array()"是数组连接到字符串时PHP所做的“toString（）”版本。 $refnr显然是一个数组，Prototype只是选择了这个值。

提示：使用printf('<option value="%d">%s</option>', ...)来拯救你的眼睛斜杠综合征的疼痛; - ）

使用AJAX（Prototype）时$ _POST为空

5 个答案: