JQUERY/AJAX請求不通過並衝突

Question

我在頁面中包含聯繫表單。在同一頁面中，我有一個腳本根據下拉的值獲取價格。現在，當我嘗試提交聯繫消息時，我與價格腳本發生衝突。基本上它試圖運行它，我不知道爲什麼。提交時的聯繫表單從來沒有工作...我只是得到一個新的網頁打開網址..？message = blablabla 任何想法是什麼問題？

我正在Laravel 4.2上工作，所以你看到的路線重定向到我的PHP函數。 這裏是JSfiddle這裏是PHP代碼：

public function postSendMessage() { 
    echo "<span class=\"alert alert-success\" >Your message has been received. Thanks!</span><br><br>"; 
} 

Answer 1

取消點擊這樣的形式不會提交

$("button#send").click(function(evt){ 
    evt.preventDefault(); 

---

新的錯誤，形式有接觸的ID，而不是類

data: $('form.contact').serialize(), 

需求是

data: $('form#contact').serialize(), 

Answer 2

這是我的相同情形

//For your drpbox use this code 
    $(document).on("change", "#yorDropBoxId", function(){   
     dropBoxValue=$("#yorDropBoxId").val(); 
     var request = $.ajax({//http://api.jquery.com/jQuery.ajax/ 
         url: "samePagePHPcript.php", 
         type: "POST", 
         data: { 
           ObjEvn:"dropBoxEvent", 
           dropBoxValue: dropBoxValue //You will use $myVar=$_POST["dropBoxValue"] to retrieve the information from javascript        
         }, 
         dataType: "json" 
      }); 
      request.done(function(dataset){ 
      //If you want to retrieve information from PHP sent by JSON. 
       for (var index in dataset){ 
        JsResponse=dataset[index].phpResponse; 
       } 

       if(JsResponse test someting){ 
       "do dometing" 
       control the beheaivor of your HTML elements 
       } 
      }); 
      request.fail(function(jqXHR, textStatus) { 
        alert("Request failed: " + textStatus); 
      }); 

    }); 



    //To submit your form use this code. You must use Prevent default if you are using a button or using a <a> link tag to trigger the evenrmrnt 
    $(document).on("click", "#btn_sendForm", function(e){ 
     e.preventDefault();  
     var dt={ 
       ObjEvn:"FormEvent", 
       input1:$("#txt_input1").val(), 
       input2: $("#txt_input2").val(), 
       input3: $("#txt_input3").val() 
      }; 
     var request = $.ajax({//http://api.jquery.com/jQuery.ajax/ 
          url: "samePagePHPcript.php", 
          type: "POST", 
          data: dt, 
          dataType: "json" 
       }); 
      request.done(function(dataset){ 
       //If you want to retrieve information from PHP send by JSON. 
        for (var index in dataset){ 
         JsResponse=dataset[index].phpResponse; 
        } 

        if(JsResponse test someting){ 
        "do dometing" 
        control the beheaivor of your HTML elements 
        } 
      }); 
      request.fail(function(jqXHR, textStatus) { 
         alert("Request failed: " + textStatus); 
      }); 
    }); 



    //In the samePagePHPcript.php you can do this:You will return your information from PHP using json like this 

    $event = $_POST["ObjEvn"]; 

    if(event==="FormEvent"){//Event to insert in your form 
    $arrToJSON = array(
      "phpResponse"=>"data you want to send to javascript", 
      "asYouWant"=>"<div class=\".class1\">more data</div>"  
      ); 
    echo json_encode(array($arrToJSON)); 

    } 
    elseif(event==="dropBoxEvent"){//Event to your dropbox - if you want 
    $arrToJSON = array(
      "phpResponse"=>"data you want to send to javascript", 
      "asYouWant"=>"<div class=\".class1\">more data</div>"  
      ); 
    echo json_encode(array($arrToJSON)); 


    }