pass a variable from PHP to JavaScript

Question

I am using a JSON string to pass a variable in JavaScript from PHP :

while( $row = mysql_fetch_array($result) )
        {
                $tmp = array('id'=>$row['id'], 'alert_type_id'=>$row['alert_type_id'], 'deviation'=>$row['deviation'], 'threshold_low'=>$row['threshold_low'], 'threshold_high'=>$row['threshold_high']) ;
                $settings[] = $tmp ;
        }

        echo '{"data":'.json_encode($settings).'}' ;

in Javascript, i am using the following snippet :

console.log( result ) ;                
var json = eval('('+ result +')') ;

and what appears in the Console is the following error :

1{"data":[{"id":"1","alert_type_id":"1","deviation":null,"threshold_low":"20","threshold_high":"80"}]}

SyntaxError: Expected token ')'

Could you please help me overcome this issue please ? Many Thanks.

score 2 · Answer 1 · answered Mar 10 '12 at 20:53

In your PHP, you probably want to use json_encode for encoding all of your data:

$result = array(
  'data' => json_encode($settings)
);

echo json_encode($result);

Second, in your javascript, eval is rarely (never) a good idea. From Douglas Crockford's style guide:

The eval function is the most misused feature of JavaScript. Avoid it.

Instead, you probably want to use JSON.parse() to rebuild the result returned by the server:

console.log(result) ;                
var resultObj = JSON.parse(result);
console.log(resultObj);

If your code is still broken, you might want to double-check your PHP to make sure it isn't spewing out any output beyond the json_encode statement.

score 1 · Answer 2 · answered Mar 10 '12 at 20:55

1

Something is outputting a "1" in front of your json string. Javascript is unable to resolve that properly during eval().

answered Mar 10 '12 at 20:55

davidethell

11,708
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score 0 · Accepted Answer · answered Mar 10 '12 at 20:51

0

Well that line of code isn't valid.

Write in the console:

'(1{"data":[{"id":"1","alert_type_id":"1","deviation":null,"threshold_low":"20","threshold_high":"80"}]})'

As this is what you're passing to the eval function. I should give you the same error.

If you want to save that string in a variable you need to adjust it a bit:

eval('var x =' + result);

Anyway It looks like you're doing something wrong, check again why do you need ti use the "evil" eval.

answered Mar 10 '12 at 20:51

gdoron

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isn't that the same as : '('+ result +')' i am sorry but i am not getting what's wrong – user690182 Mar 10 '12 at 20:54
@user690182. No it's not. it will just run the string "command" in side `()`. What are you trying to do with the `json`? – gdoron Mar 10 '12 at 20:56
thank you, i wrote what you just said in PHP : echo '('.'{"data":'.json_encode($settings).'}'.')' ; and everything is ok now :) thank you – user690182 Mar 10 '12 at 20:58
@user690182. I can't find where i said it... Any way you should accept an answer if you got one. **:=)** – gdoron Mar 10 '12 at 21:00
Don't use eval() use a JSON parser to get the data as stated in the other answers. – Guumaster Mar 10 '12 at 21:38
1

@Guumaster. Didn't you read the part which I recommended not to use `eval`??? – gdoron Mar 10 '12 at 21:40
sorry, I stopped reading after eval() ^_^. +1 to you – Guumaster Mar 11 '12 at 10:15

pass a variable from PHP to JavaScript

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