Read JSON returned by PHP in Javascript

Question

I want to select some content from the database and return it to the javascript. There are several rows returned by the database. I tried this with JSON and also get a result, if I print it out. But if I want to convert the JSON string, there is always the error message below. (at the JSON.parse) So, I assume maybe an mistake while filling the array? Thanks in advance guys!

Javascript:

$.ajax({
  url: "./select_firsttracks.php",
  type: "post",
  success: function(resultset) {
    $("#erroroutput").html(resultset);

    var arr = JSON.parse("{" + resultset + "}"); // --> "Uncaught SyntaxError: Unexpected token {"            
  },
  error: function(output) {
    $("#erroroutput").html("fatal error while fetching tracks from db: " + output);
  }
});

PHP:

$storage = array();
while($row = $result->fetch_array())
{
    $storage[] = 
        array
        (
            "id" => $row["id"],
            "trackname" => $row["trackname"],
            "artist" => $row["artist"],
            "genre" => $row["genre"],
            "url" => $row["url"],
            "musicovideo" => $row["musicovideo"]
        );

    echo json_encode($storage);
}

Output on the console:

[{"id":"1","trackname":"yes","artist":"Lady Gaga","genre":"Pop","url":"ftp:\/development","musicovideo":"1"}][{"id":"1","trackname":"yes","artist":"Lady Gaga","genre":"Pop","url":"ftp:\/development","musicovideo":"1"},{"id":"2","trackname":"no","artist":"Prinz Pi","genre":"Rap","url":"ftp:\/development","musicovideo":"1"}]

Your resultset already is a valid JSON; adding braces makes it bad. — Amadan
– Amadan, Commented Nov 24, 2015 at 8:45
Don't need to concat { and } in JSON.parse("{" + resultset + "}"); Code: JSON.parse(resultset); or add dataType: 'json', in ajax configurations — Tushar
– Tushar, Commented Nov 24, 2015 at 8:45

madalinivascu · Accepted Answer · 2015-11-24 08:47:55Z

4

echo the json after the while

$storage = array();
while($row = $result->fetch_array())
{
    $storage[] = 
        array
        (
            "id" => $row["id"],
            "trackname" => $row["trackname"],
            "artist" => $row["artist"],
            "genre" => $row["genre"],
            "url" => $row["url"],
            "musicovideo" => $row["musicovideo"]
        );


}
echo json_encode($storage);

and change:

 var arr = JSON.parse(resultset);

answered Nov 24, 2015 at 8:47

madalinivascu

32.4k4 gold badges41 silver badges59 bronze badges

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

Comments

Pavlin · Accepted Answer · 2015-11-24 08:45:49Z

2

You're adding curly braces in front and behind your received JSON, here:

var arr = JSON.parse("{" + resultset + "}");

Phps json_encode returns perfectly valid JSON by itself. Try it without adding the braces:

var arr = JSON.parse(resultset);

answered Nov 24, 2015 at 8:45

Pavlin

5,5587 gold badges44 silver badges57 bronze badges

Comments

Alex Andrei · Accepted Answer · 2015-11-24 08:49:54Z

1

The resulting json string is not valid, you can check it with jsonlint

Modify your php code to echo outside the loop:

while($row = $result->fetch_array())
{
    $storage[] = 
        array
        (
            "id" => $row["id"],
            "trackname" => $row["trackname"],
            "artist" => $row["artist"],
            "genre" => $row["genre"],
            "url" => $row["url"],
            "musicovideo" => $row["musicovideo"]
        );


}
echo json_encode($storage);

And in javascript just use the output as a javascript object

success: function(resultset) {
    console.log(resultset)
    resultset.each(function(index,element){ console.log(index,element )})
  },

answered Nov 24, 2015 at 8:49

Alex Andrei

7,2763 gold badges30 silver badges44 bronze badges

Collectives™ on Stack Overflow

Read JSON returned by PHP in Javascript

3 Answers 3

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related