JSON.parse: unexpected character at line 1 column 1 of the JSON data (php)

Question

I am unable to access json data as it always fails and gives error as SyntaxError: JSON.parse: unexpected character at line 1 column 1 of the JSON data search.php outputs the json data but scripts.js outputs json.parse error script.js

// execute when the DOM is fully loaded
$(function() {

    console.log("1");

    $("#q").typeahead({
        autoselect: true,
        highlight: true,
        minLength: 1
    },
    {
        source: search,
        templates: {
            empty: "no places found yet",
            suggestion: _.template("<p><%- subname %></p>")
        }
    });

    // re-center map after place is selected from drop-down
    $("#q").on("typeahead:selected", function(eventObject, suggestion, name) {


   });


});
function search(query, cb)
{
    // get places matching query (asynchronously)
    var parameters = {
        sub: query
    };
    $.getJSON("search.php", parameters)
    .done(function(data, textStatus, jqXHR) {
        cb(data);
    })
    .fail(function(jqXHR, textStatus, errorThrown) {
        console.log(errorThrown.toString());
    });
}

search.php

<?php

    require(__DIR__ . "/../includes/config.php");
    $subjects = [];
    $sub = $_GET["sub"]."%";
    $sql = "SELECT * from subjects where subname LIKE '$sub'";
    echo $sql;
    if($rows = mysqli_query($con,$sql))
    {
        $row = mysqli_fetch_array($rows);
        while($row){
            $subjects[] = [

            'subcode' =>$row["subcode"],
            'subname' => $row["subname"],
            'reg' => $row["reg"],
            'courseid' =>$row["courseid"],
            'branchid'  => $row["branchid"],
            'yrsem' => $row["yrsem"]
            ];
            $row = mysqli_fetch_array($rows);
        }
    }
    // output places as JSON (pretty-printed for debugging convenience)
    header("Content-type: application/json");
    print(json_encode($subjects, JSON_PRETTY_PRINT));

?>

Have you tried opening your Network browser tab and reviewing results from search.php? It looks like it is not valid JSON, at least since you have echo $sql; at line 6. — Yeldar Kurmangaliyev
– Yeldar Kurmangaliyev, Commented Feb 20, 2017 at 10:08
Another important issue: your code is vulnerable to SQL-injections. Do not use string concatenation for building SQL queries. — Yeldar Kurmangaliyev
– Yeldar Kurmangaliyev, Commented Feb 20, 2017 at 10:09

Anand Deep Singh · Accepted Answer · 2017-02-20 10:22:56Z

5

Solution for your problem is JSON.stringify

You need to convert your data output into string i.e.,

 var yourDataStr = JSON.stringify(yourdata)

and then validate your data with

JSON.parse(yourDataStr)

Then you can get your result. To validate, you can pass your data in below function :-

function validateJSON(str) {
    try {
       JSON.parse(str);
    } catch (e) {
     return false;
    }
  return true;
}

answered Feb 20, 2017 at 10:22

Anand Deep Singh

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1 Comment

AshrafNmd Over a year ago

can you please tell me where should I write this. When I write it in .done() then it gives error as missing ) argument list

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