How to append to json_encode array

Question

i have a validation checker that checks the database to see if the "email and username" is being used and if so then echo the email and username as "false". And if its not in the database then echo to "true". I am returning the results of the users email and username being in the sql sever. So i want to append "$json1 or $json2" to the "$first or $second" like this

 {"Email":false, "Username":true};

PHP Code:

<?php

if(isset($_GET['submit'])){

    //Connect to database
    $link = mysqli_connect("localhost", "root", "");
    mysqli_select_db($link, "magicsever");

    if(mysqli_connect_error()){

        die ("Database connection error");
    }

    //Find the email in the database
    $query = "SELECT * FROM app_signup WHERE email = '".mysqli_real_escape_string($link, $_GET['email'])."'";
    $query2 = "SELECT * FROM app_signup WHERE username = '".mysqli_real_escape_string($link, $_GET['username'])."'";

    $result = mysqli_query($link, $query);
    $result2 = mysqli_query($link, $query2);

    $email_checker1 = json_encode(array("Email"=>false));
    $email_checker2 = json_encode(array("Email"=>true));

    $username_checker1 = json_encode(array("Username"=>false));
    $username_checker2 = json_encode(array("Username"=>true));

    if(mysqli_num_rows($result)>0){

        echo $email_checker1;


    }

}
?>

Chinito · Accepted Answer · 2017-05-18 06:32:28Z

1

make new array for email and username results

//Find the email in the database
$query = "SELECT * FROM app_signup WHERE email = '".mysqli_real_escape_string($link, $_GET['email'])."'";
$query2 = "SELECT * FROM app_signup WHERE username = '".mysqli_real_escape_string($link, $_GET['username'])."'";

$result = mysqli_query($link, $query);
$result2 = mysqli_query($link, $query2);

$dataResult = array();
$dataResult['Email'] = (mysqli_num_rows($result)>0)? false: true;
$dataResult['Username'] = (mysqli_num_rows($result2)>0)? false: true;

echo json_encode($dataResult );

edited May 18, 2017 at 6:32

answered May 18, 2017 at 6:23

Chinito

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14 Comments

Jon2713 Over a year ago

I got an error on the $dataResult = new array(); line.

Chinito Over a year ago

or just use $dataResult = [];

Chinito Over a year ago

or just remove the new in array. sorry. wrong syntax

Jon2713 Over a year ago

D: i am returning the results of the users email and username being in the sql sever. So the set booleans wont be the result

Chinito Over a year ago

but what if there's no result?

|

Bartosz Zasada · Accepted Answer · 2017-05-18 06:15:05Z

0

JSON is only a data format, a way to present a data structure. Until you have a data structure that is ready to be presented, you don't want to convert it to JSON.

What you should do is to build an array containing the validation errors:

// I'm assuming that the true/false means that there was an error with validating that value
// You can initialize the validation array with false values for all fields and change them to true in case of validation error
$errors = ['Email' => false, 'Username' => false];

// Then, change the values in case of an error
if (... /* invalid email */) {
    $errors[] = ['Email' => true];
}
if (... /* invalid username */) {
    $errors[] = ['Username' => true];
}

Then convert the array to JSON only when you display it:

echo json_encode($errors);

answered May 18, 2017 at 6:15

Bartosz Zasada

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3 Comments

Jon2713 Over a year ago

The true means "the email and username that was inputted is available to take" and false means " its has already been taken"

Bartosz Zasada Over a year ago

That doesn't change my answer in any way. But perhaps you'd like to write your code in a descriptive way, so that you don't have to explain what you just did with an extra comment. How about naming the keys in the array email_available and username_available?

Jon2713 Over a year ago

i see what you do, but i want them in the same curly brackets not like separated.

Amit Gaud · Accepted Answer · 2017-05-18 06:15:52Z

0

 $first = json_encode(array("Email"=>false));
    $second = json_encode(array("Email"=>true));

    $json1 = json_encode(array("Username"=>false));
    $json2 = json_encode(array("Username"=>true));

use like this

$first = json_encode(array("Email"=>false)+array("Username"=>true));
$second = json_encode(array("Email"=>true)+array("Username"=>false));

answered May 18, 2017 at 6:15

Amit Gaud

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1 Comment

Jon2713 Over a year ago

Close the way i structure the code is i check if the email is taken first and if it is, then check if the username is taken and then from the username side i append the message from username to the email in the first if statement. So your way kinda works

Community · Accepted Answer · 2017-05-23 12:34:42Z

0

If you really want merge them together go through this references

Good approach would be like

$resData1['email']=false;
$resData2['email']=true;

$resData1['Username']=false;
$resData2['Username']=true;

$json1 = json_encode($resData1);
$json2 = json_encode($resData2);

edited May 23, 2017 at 12:34

CommunityBot

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answered May 18, 2017 at 6:17

Kuru

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3 Comments

Jon2713 Over a year ago

Hmm the problem with that is i am getting the same result "false and false" or "true and true" back. Is it possible to change like append he false and true with different variable names?

Kuru Over a year ago

because the code is written like that.... you might need to use an if condition..like if email return as false set to false or give true. Something like that will works

Jon2713 Over a year ago

Yea i thought of that but i would have to create and if statement for different scenarios and i am trying to find a less crazier way of doing it.

LF-DevJourney · Accepted Answer · 2017-05-18 06:27:36Z

0

Why not do json_encode after the array append.

$first = json_encode(array("Email"=>false, "Username"=>false));

added.

$data['Email'] = false;
if('''')
{
  $data['Username'] = false;
}
$result = json_encode($data);

edited May 18, 2017 at 6:27

answered May 18, 2017 at 6:15

LF-DevJourney

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5 Comments

Jon2713 Over a year ago

This would work if the username was true and email is false. I dont want a simply echo where i type it, i want to append if either or happens.

Jon2713 Over a year ago

So you created a data varible and passed "email" to it and you set it to false? but the problem is i am returning the results of the users email and username being in the sql sever.

LF-DevJourney Over a year ago

You just store the json string to your sql server, it have no difference which variable store this value.

Jon2713 Over a year ago

its not being stored in the sql server. I am taking it info from the server then sending the json file to an app that reads the information and displays something to the user. Sorry if i seem "dumb" in programming D: i am still learning

LF-DevJourney Over a year ago

$variable just store the value, what the variable name is have no difference.

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