PHP array string key add number at the end

Question

Good day fellas!

I have this block of code where i create a JSON string and i have a dynamic number of data so what i need is also a dynamic name for a key..

if (mysql_num_rows($result) > 0) {
    $response["members"] = array();
    $x = 0;
    $members = array();
    while ($row = mysql_fetch_array($result)) {
        $members = array();
        $members["member" + (string)$x] = array();
        $member["member_id"] = $row["member_id"];
        $member["firstname"] = $row["firstname"];
        $member["mi"] = $row["mi"];
        $member["lastname"] = $row["lastname"];
        $member["email"] = $row["email"];
        $member["username"] = $row["username"];
        $member["password"] = $row["password"];
        $member["guild_id"] = $row["guild_id"];
        $member["guild_name"] = $row["guild_name"];
        $member["guild_code"] = $row["guild_code"];
        array_push($members["member" + (string)$x], $member);
        $x++;
    }
    array_push($response["members"], $members);

    echo json_encode($response);;
}

In JQuery this type of method works.. my question is, is there any way i can achieve it using php? If you know i will be very happy to know...

@Siege21x a much better solution given by me. check and try once — Death-is-the-real-truth
– Death-is-the-real-truth, Commented Apr 7, 2017 at 8:12
@AlivetoDie sure man. I'll make sure to try that out! Thanks alot! — sshanzel
– sshanzel, Commented Apr 7, 2017 at 9:01
Was able to have it man. I voted for your answer. Thanks a lot! — sshanzel
– sshanzel, Commented Apr 10, 2017 at 2:41

Death-is-the-real-truth · Accepted Answer · 2017-04-07 08:12:04Z

1

1.Suggestion:-stop using deprecated+removed(php5 +php7) version of mysql_*.Move towards mysqli_* OR PDO along with prepared statements(Prevent Sql Injection)

2.Why creating unnecessary array and multiple push.No need to do that.Do like below:-

$response["members"] = array();//put outside
if (mysql_num_rows($result) > 0) {
    $x = 0;
    while ($row = mysql_fetch_array($result)) {
        $response["members"][$x]["member_id"] = $row["member_id"]; //assign value directly to the resultant array
        $response["members"][$x]["firstname"] = $row["firstname"];
        $response["members"][$x]["mi"] = $row["mi"];
        $response["members"][$x]["lastname"] = $row["lastname"];
        $response["members"][$x]["email"] = $row["email"];
        $response["members"][$x]["username"] = $row["username"];
        $response["members"][$x]["password"] = $row["password"];
        $response["members"][$x]["guild_id"] = $row["guild_id"];
        $response["members"][$x]["guild_name"] = $row["guild_name"];
        $response["members"][$x]["guild_code"] = $row["guild_code"];
        $x++;
    }
}

if(count($response["members"])>0){ //check finally that array is not empty
 echo json_encode($response); //echo json encoded array data
}

answered Apr 7, 2017 at 8:12

Death-is-the-real-truth

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1 Comment

sshanzel Over a year ago

This is great! Thanks for the idea in this kind of saving the data within arrays. I'll read the difference between mysql and mysqli for me to understand what you are talking about :D Thanks a lot!

ewcz · Accepted Answer · 2017-04-07 07:45:19Z

1

if you want to achieve that the members array contains individual records with keys as member0, member1, etc., then you could proceed as:

if (mysql_num_rows($result) > 0) {
    $x = 0;
    $members = array();
    while ($row = mysql_fetch_array($result)) {
        $member = array();
        $member["member_id"] = $row["member_id"];
        $member["firstname"] = $row["firstname"];
        $member["mi"] = $row["mi"];
        $member["lastname"] = $row["lastname"];
        $member["email"] = $row["email"];
        $member["username"] = $row["username"];
        $member["password"] = $row["password"];
        $member["guild_id"] = $row["guild_id"];
        $member["guild_name"] = $row["guild_name"];
        $member["guild_code"] = $row["guild_code"];
        $members["member$x"] = $member;
        $x++;
    }
    $response["members"] = $members;

    echo json_encode($response);
}

answered Apr 7, 2017 at 7:45

ewcz

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1 Comment

sshanzel Over a year ago

Thanks for the answer! I'll make sure to try this out and declare this the answer if it will work! Thanks! :D

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