SyntaxError: JSON.parse: unexpected character when parsing output from PHP

Question

My php returns something as such:

  if(!$result = $db->query($sql)){
    echo $db->error;
      die('There was an error running the query [' . $db->error . ']');
  }
  echo 'Total results: ' . $result->num_rows . "\n";

  while($row = $result->fetch_assoc()){
      echo json_encode($row);
      echo "\n";
  }

In my javascript, I want to put the output on the page if it's a json object or just log it to the console if it's something else:

var divOutput = function(output) {
  try {
    // var x = output.trim();
    // x = JSON.parse(JSON.stringify(x));
    var x = JSON.parse(output);
    console.log(x);
    $("#dbOut").html(JSON.stringify(x, null, '\t'));
  } catch (e) {
    console.log(output);
    console.log(e);
  }
}

var getPlayerByID = function() {
  var myNumber = document.getElementById("PlayerInput").value;
  $.ajax({
    url : "db_funcs.php",
    data : {
      action : 'getPlayerByID',
      a : myNumber,
    },
    type : 'post',
    success : function(output) {
      divOutput(output);
    }

However it's throwing the JSON parse error when I query the database. How can I do this? }) }

Have you tested your JSON at jsonlint.com

Jay Blanchard
– Jay Blanchard

2015-12-01 17:31:03 +00:00
Commented Dec 1, 2015 at 17:31 — Jay Blanchard
– Jay Blanchard, Commented Dec 1, 2015 at 17:31

Marc B · Accepted Answer · 2015-12-01 17:42:25Z

2

JSON must be output as a SINGLE monolithic json string. You're building multiple SEPARATE json strings, which is illegal syntax.

Consider JSON to be pretty much equivalent to the right-hand side of a javascript variable assignment:

var foo = this_part_is_json;

e.g. you need to have

var foo = [[stuff from row 1], [stuff from row2], etc...];

but are producing

var foo = [stuff from row1][stuff from row2];
                           ^---syntax error would occur here

You need

$arr = array();
while($row = fetch from db) {
   $arr[] = $row;
}
echo json_encode($arr);

answered Dec 1, 2015 at 17:42

Marc B

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Comments

Niki van Stein · Accepted Answer · 2015-12-01 17:40:07Z

1

It throws the error because you are not only echoing json.

In the line

echo 'Total results: ' . $result->num_rows . "\n"

You already echo something which is not json encoded. With this code it should work:

//echo 'Total results: ' . $result->num_rows . "\n";  <---- this is not json
$arr = [];
while($row = $result->fetch_assoc()){
     $arr[] = $row; //save each row in array
    //echo "\n";
}
echo json_encode($arr); //encode all data as json

If you want the number of rows also in the json output add it in the $arr with a key like

$arr["num_rows"] = $result->num_rows;

edited Dec 1, 2015 at 17:40

answered Dec 1, 2015 at 17:33

Niki van Stein

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2 Comments

Austin Gayler Over a year ago

ah, ok. I thought the try/catch statement would handle that.

Niki van Stein Over a year ago

Nope, it sees the whole output as one thing and therefor it crashes. You can mark one of the answers given such that other users know how the problem was solved.

Niranjan N Raju · Accepted Answer · 2015-12-01 17:32:56Z

1

This is wrong

while($row = $result->fetch_assoc()){
  echo json_encode($row);
  echo "\n";
}

Try like this

$arr = array();
while($row = $result->fetch_assoc()){
  $arr[] = ($row);
}

echo json_encode($arr);

Use json_encode() only once, use arrays to store values in while()

answered Dec 1, 2015 at 17:32

Niranjan N Raju

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