passing parameter from php to javascript

Question

I have written a code which sends a query to mysql database and then gets the results row by row using $row = mysql_fetch_assoc($result).

I need to send the data for each row to a javascript function.

Here is the piece of code I used:

while($row = mysql_fetch_assoc($result))
{

echo '<script type="text/javascript">'.'queryResults('. json_encode($row) . ');'.'</script>';

}

In my javascript function I have :

function queryResults(data){
var results = JSON.parse(data);}

However, I get error:

"SyntaxError: JSON.parse: unexpected character"

Where is the problem with my code?

It is literally impossible to day. Why don't you console.log(data) before calling JSON.parse() on it. — zallarak
– zallarak, Commented Feb 7, 2013 at 21:30
What is data? And do you really output tons of script tags? — Bergi
– Bergi, Commented Feb 7, 2013 at 21:31
You don't need JSON.parse here. Echoing JSON from PHP into JavaScript will just create an object literal in JavaScript. — Felix Kling
– Felix Kling, Commented Feb 7, 2013 at 21:33

Bergi · Accepted Answer · 2013-02-07 21:39:38Z

2

JSON directly inside a script tag is interpreted as a JavaScript object literal - you actually pass an object to the queryResults function. To use JSON.parse() (which you don't have to) you would need to wrap it inside JavaScript string delimiters and escape all quotes in the string returned by json_encode, so that JavaScript would see a valid string.

Btw, your sample function is missing the data parameter in its signature.

edited Feb 7, 2013 at 21:39

answered Feb 7, 2013 at 21:33

Bergi

671k162 gold badges1k silver badges1.5k bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

hAlE Over a year ago

Ok. great! But how can i use it as a json string? could you make an example of how to escape the quotes? I don't know how to do that.

Bergi Over a year ago

It would look like 'var obj = JSON.parse( "'.addslashes(json_encode($row)).'");'

Zevi Sternlicht · Accepted Answer · 2013-02-07 21:31:07Z

0

You need the field name of the table.

json_encode($row['fieldname'])

answered Feb 7, 2013 at 21:31

Zevi Sternlicht

5,39921 silver badges31 bronze badges

Comments

TKharaishvili · Accepted Answer · 2013-02-07 21:33:11Z

0

I think the problem is that you have no parameter specified in the function signature. This may solve the issue:

function queryResults(data){
var results = JSON.parse(data);
}

answered Feb 7, 2013 at 21:33

TKharaishvili

2,1391 gold badge21 silver badges32 bronze badges

Collectives™ on Stack Overflow

passing parameter from php to javascript

3 Answers 3

2 Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

2 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related