Simple PHP rest json duplicates data fields

Question

I'm iOS developer and for finishing prototype I need simple php code which could return data from particular table as JSON. Here is a code I developed:

<?php
    $connection = mysqli_connect($db_host, $db_user, $db_pass, $db_database) or die(mysql_error());

    $query = "select * from Events" or die("Error in the consult.." . mysqli_error($connection));
    $result = $connection->query($query);
    $events = array();

    while ($event = $result->fetch_array(MYSQLI_ASSOC)) {       

        $events[] = $event; 
    }

    echo json_encode($events);

    $result->free();
    $connection->close();
?>

The problem is that it duplicates data fields. Here is a result I receive:

[
   {
      "0":"1",
      "id":"1",
      "1":"Some title",
      "title":"Some title"
   }
]

In database I have one record only.

How to solve the following problem?

Shashikumar Misal · Accepted Answer · 2014-05-02 14:08:20Z

2

Use associative array, will definately work...!

while ($event = mysqli_fetch_assoc($result)) {

    $events[] = $event; 
}

And You used header('Content-Type: application/json'); - its good, but also try by removing it as well.

answered May 2, 2014 at 14:08

Shashikumar Misal

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1 Comment

Serge Over a year ago

Thanks. It works well. Since I use object oriented style I updated code in next way: (see edited topic)

Community · Accepted Answer · 2023-11-17 20:38:18Z

1

https://www.php.net/mysqli_fetch_array

"Fetch a result row as an associative, a numeric array, or both"

resulttype parameter: This optional parameter is a constant indicating what type of array should be produced from the current row data. The possible values for this parameter are the constants MYSQLI_ASSOC, MYSQLI_NUM, or MYSQLI_BOTH.

It would appear you are using BOTH. Switch to ASSOC or NUM. ASSOC is usually easier.

https://www.php.net/manual/en/mysqli-result.fetch-assoc.php

edited Nov 17, 2023 at 20:38

CommunityBot

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answered May 2, 2014 at 13:50

Jessica

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Comments

user1908308 · Accepted Answer · 2014-05-02 13:51:03Z

0

For JSON you would proably want to do:

while ($event = mysqli_fetch_array($result, MYSQL_ASSOC)) {

answered May 2, 2014 at 13:51

user1908308

Comments

Community · Accepted Answer · 2017-05-23 12:28:14Z

0

you can see this OT then you can use mysqli_fetch_assoc to fix the error, since the one you are using right now returns also an array, that is why you are getting duplicated data.

edited May 23, 2017 at 12:28

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answered May 2, 2014 at 13:59

Jorge Y. C. Rodriguez

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