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I have a $.POST call that return the name of a function that need to be ran but it wont execute the function and I don't know why.

Here is an example:

JS file:

$(function(){
     $.post('test.php',{event: 'add'},
          function(data){
               data.func(data.msg);
          },'json');

     function test(msg){
          alert(msg);
     }
});

PHP Ajax:

<?php
     switch($_POST['event']){
          case 'add':
               $output['func'] = 'test';
               $output['msg'] = 'This is add message';
               break;
          case 'delete':
               $output['func'] = 'test';
               $output['msg'] = 'This is delete message';
               break;
     }
     echo json_encode($output);
 ?>

The problem I am having is the ajax is returning the name of the function (test) but it will not run the function, how do I fix this?

share|improve this question
Probably needs eval.... – Pekka 웃 Nov 6 '10 at 12:26

2 Answers

up vote 6 down vote accepted

DO NOT USE EVAL.

Rather, make an object with the functions you want to be executable. For example:

var functionTable = {
    test: function (msg) {
        alert(msg);
    }
};

$.post('test.php', { event: 'add' }, function (data) {
    if (functionTable.hasOwnProperty(data.func)) {
        functionTable[data.func](data.msg);
    }
}, 'json');
share|improve this answer
Great, this is exactly the same as I have ;) +1 – Harmen Nov 6 '10 at 12:34
@Harmen, Yup, I saw that soon after I posted my answer. I prefer to use hasOwnProperty, however. (Who knows; someone may add a dangerous function to Object.prototype which gets called remotely... – strager Nov 6 '10 at 12:36
Eval() is only as "evil" as your usage of it. If he's generating that JSON in a way that's isolated from user generated data (which it appears he is), there's not much pragmatic reason to avoid eval(). It's particularly unnecessary to avoid it in this case, because $.post will already be using an eval() derivative to deserialize the JSON response in all but the most current browsers anyway. – Dave Ward Nov 6 '10 at 13:04
@Dave Ward, jQuery will use the JSON.parse() function to parse JSON. And still, you cannot be sure that the output is a JSON structure with a func and msg key – Harmen Nov 6 '10 at 13:07
@Harmen: Only in newer browsers that have implemented that. There are still an awful lot of browsers currently in use (anything below IE8, Firefox 3.5, Chrome 3, and Safari 5, I believe) that don't implement JSON.parse natively. In those browsers, jQuery uses new Function(json) to deserialize JSON, which is effectively an eval(). – Dave Ward Nov 6 '10 at 13:19
show 2 more comments
up vote 2 down vote

I think it's better to move your function into an object here, so you can see if it exists or not:

var possibleFunctions = {
  test: function(val){
    alert(val);
  }
};

$(function(){
     $.post('test.php',{event: 'add'},
          function(data){
              if(possibleFunctions[data.func]){
                  possibleFunctions[data.func](data.msg);
              }
          },'json');
      });
});
share|improve this answer

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