dongyong6332
2016-10-26 09:54
浏览 64
已采纳

使用ajax从Javascript函数调用Php函数

Hello everyone I am new to php. I have been trying out this thing when a user enter a product name need to validate that the product is valid or not. For that purpose I have used onchange event when the text is entered.The onchange function will call the javascript function.From javascript function I am calling the php which is in the same file.So when I am entering the product name somehow the php function is not working.

Here is my code :

<?php
  include 'conf.php';//it contains the php database configuration
   session_start();

   $quantityRequired=0; 
   $productName_error="";
    if(is_ajax()){
      if(isset($_POST["productName"])){
        $productName=$_POST["productName"];
        $row=mysqli_query($conn,"SELECT * from OrderDetails where ProductName='".$productName."'");
        if($row)
        {
          $result=mysqli_fetch_assoc($row);
          $quantityRequired=$result["Quantity"];
        }   
        else
         {
          $productName_error="The product name is not valid or product does not exist";
          echo $productName_error;
         }  
      }
    } 
  function is_ajax() {
    $flag=(isset($_SERVER['HTTP_X_REQUESTED_WITH']) && strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) == 'xmlhttprequest'); 
    return $flag;
  }
?>
<html>
   <head>
       <title>Order Page </title>
       <script type = "text/javascript" 
         src = "https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
   </head>
   <body>
      <form action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>" method="POST">
         <label for="userName">Username</label><br>
         Product Name<input type="text" name="productName" id="productName" onchange="validateProduct()"><?php echo $productName_error?><br>
         Quantity Required<input type="text" name="quantityRequired" id="quantityRequired"><br>
         Availability<input type="text" name="availability">
         <p id="demo"></p>
      </form>
      <script>
        function validateProduct()
        {
            $.ajax({
             type: "POST"
         });

        }
      </script>
   </body>  
</html> 

so the code is when the user enters the product name.The function validate product is called.From validate product it will call the php which is in the same file. is_ajax() function is used to check whether it is the ajax request or not.

  • 写回答
  • 好问题 提建议
  • 关注问题
  • 收藏
  • 邀请回答

3条回答 默认 最新

  • doujiang3997 2016-10-26 10:15
    已采纳

    There may be other problems I haven't spotted, but the first thing that jumps out to me is that your server-side code runs conditionally:

    if(isset($_POST["productName"]))
    

    And that condition was never satisfied because you didn't send any values in the AJAX request:

    $.ajax({
        type: "POST"
    });
    

    Send the value(s) you're looking for:

    $.ajax({
        type: "POST",
        data: { productName: $('#productName').val() }
    });
    

    You may also need to specify a couple other options if they don't default correctly. Explicit code is generally better than implicit in many cases:

    $.ajax({
        url: 'yourUrl.php',
        type: "POST",
        dataType: 'html',
        data: { productName: $('#productName').val() }
    });
    

    In general you'll probably want to check the documentation for $.ajax() and see what you can and should tell it. You'll also want to take a look at your browser's debugging tools to see more specifically why and how it fails when testing these things.

    Speaking of things you should do, you should read this and this. Your code is wide open to SQL injection attacks at the moment, which basically means that you are executing as code anything your users send you.

    已采纳该答案
    评论
    解决 无用
    打赏 举报
  • drpkcwwav20524605 2016-10-26 10:09
    var data1 = "Something";
    $.ajax({
        url: "script.php",
        type: "POST",
        data: { data1: data1 }
    }).done(function(resp) {
        console.log( resp )
    }).fail(function(jqXHR, textStatus) {
        alert("Request failed: " + textStatus + " - Please try again.")
    })
    

    Here you have a script that will send the data1 variable across to the php script. The resp in the done portion is the return you send back from the php script.

    If you want to send more data just add it { data1: data1, data2: data2 } and so on.

    Just adjust to suit your needs.

    评论
    解决 无用
    打赏 举报
  • dsf1222 2016-10-26 10:30

    A PHP library for Ajax

    Jaxon is an open source PHP library for easily creating Ajax web applications. It allows into a web page to make direct Ajax calls to PHP classes that will in turn update its content, without reloading the entire page.

    Jaxon implements a complete set of PHP functions to define the contents and properties of the web page. Several plugins exist to extend its functionalities and provide integration with various PHP frameworks and CMS.

    How does Jaxon work

    Define and register your PHP classes with Jaxon.

    $jaxon->register(Jaxon::CALLABLE_OBJECT, new MyClass);
    

    Call your classes using the javascript code generated by Jaxon.

    <input type="button" onclick="JaxonMyClass.myMethod()" />
    

    check link https://www.jaxon-php.org/docs.html

    评论
    解决 无用
    打赏 举报

相关推荐 更多相似问题