意外的令牌錯誤調用回調中的函數

Question

在qm150_submit $ .post中的ajax調用之後的回調中。 我想調用第二個名爲'send_email'的函數（它也有一個名爲'success_callback'的回調函數「

我正在這裏一個錯誤

function() {send_email(fromName,fromEmail,toEmail,subject,message,success_callback) }; 

錯誤：未捕獲的SyntaxError：意外的標記）

這裏是代碼：

function qm150_submit($title, $name, $email, $description, $send_email) { 

    $.post('<?PHP print API_SUBMIT; ?>', { "title": $title, "name": $name, "email": $email, "description": $description }, 
    function (data) {   // callback function after API_SUBMIT 

    // Send email with a link to their collection 
     if ($send_email) { 

     // parameters for the send_email() ajax function 

     var subject = "subject"; 
     var collection_id = data.collection_id; // data is json returned from the ajax above 
     var toEmail = $email 
     var message = "<?PHP print SHARE_COLLECTION;?>"+collection_id; 
     var fromEmail = "<?PHP print EMAIL_FROM_EMAIL; ?>"; 
     var fromName = "<?PHP print EMAIL_FROM_NAME; ?>"; 

     var success_callback = function (results) { 
      alert('send_email has returned with: '+results); 
     }; 

     alert('I am now calling the send_email'); 
     function() {send_email(fromName,fromEmail,toEmail,subject,message,success_callback) }; 

     } 
    }); 
     // missing a curly bracket ? no! note double indentation of the anonymous function (data) is a continuation of first statement 
} 

編輯：和爲SEND_EMAIL（）

function send_email(fromName,fromEmail,toEmail,subject,message,success_callback) { 
    alert('send_email called'); 
    $.ajax({ 
    type: 'post', 
    url: '<?PHP print API_SHARE_EMAIL;?>', 
    data: 'fromName=' + fromName + '&fromEmail=' + fromEmail + '&toEmail=' + toEmail + '&subject=' + subject + '&message=' + message, 
    dataType:'json', 
    success: success_callback 
    }); 
    alert('send_email finished'); 
    return true; 
} 

Answer 1

意外令牌function後(。

首先，你聲明一個匿名函數，而不用調用它。其次，匿名函數聲明不能​​是一個語句（或換句話說，函數聲明必須有一個名稱），這就是爲什麼(是意外（JavaScript期望函數名稱，而不是pamentalheses）。

只需直接調用send_email ...這已經是一個函數裏面，所以它不會「污染」的全局對象（沒有什麼用，污染也無妨） - 我認爲沒有必要爲一個匿名函數：

alert('I am now calling the send_email for real!'); 
send_email(fromName, fromEmail, toEmail, subject, message, success_callback); 

Answer 2

代碼如果您使用的JSLint檢查你的代碼，您將收到此錯誤：

Function statements cannot be placed in blocks. Use a function expression or move the statement to the top of the outer function.

您必須包裝這樣的匿名函數：(function{})();

固定碼：

function qm150_submit($title, $name, $email, $description, $send_email) { 

$.post('<?PHP print API_SUBMIT; ?>', { 
    "title": $title, 
    "name": $name, 
    "email": $email, 
    "description": $description 
}, function(data) { // callback function after API_SUBMIT 
    // Send email with a link to their collection 
    if ($send_email) { 

     // parameters for the send_email() ajax function 
     var subject = "subject"; 
     var collection_id = data.collection_id; // data is json returned from the ajax above 
     var toEmail = $email; 
     var message = "<?PHP print SHARE_COLLECTION;?>" + collection_id; 
     var fromEmail = "<?PHP print EMAIL_FROM_EMAIL; ?>"; 
     var fromName = "<?PHP print EMAIL_FROM_NAME; ?>"; 

     var success_callback = function(results) { 
      alert('send_email has returned with: ' + results); 
     }; 

     alert('I am now calling the send_email'); 

     (function() { 
      send_email(fromName, fromEmail, toEmail, subject, message, success_callback); 
     })(); 

    } 
}); 
}​ 

或者乾脆刪除匿名函數什麼也不做：

send_email(fromName, fromEmail, toEmail, subject, message, success_callback); 

instead of: 

(function() {send_email(fromName, fromEmail, toEmail, subject, message, success_callback); 
     })();