Manipulate JSON data using AngularJS

Question

hope you well..

I have an API made using php, to grab from SQL and convert it to JSON which is working fine. My only problem was, i can't manipulate this php to fetch JSON like i wanted. The solution i believe just a placement in the $outp at my api.php .. i have the id as primary in SQL can i used it as index to display the JSON like i expected?

Thank YOu, any effort will be rewarded

Here is the output JSON :

{
talents: [
{
id: "1",
tag: "001",
name: "Jasmina",
images: "assets/images/Jasmina3.jpg",
images02: "assets/images/Jasmina4.jpg",
images03: "assets/images/Jasmina7.jpg",
skills: "Usher",
skills02: "Modeling",
indexing: "usher"
},
{
id: "2",
tag: "002",
name: "Bruna",
images: "assets/images/BrunaD17.jpg",
images02: "assets/images/BrunaD18.jpg",
images03: "assets/images/BrunaD10.jpg",
skills: "Usher",
skills02: "Modeling",
indexing: "usher"
}
]
}

Here is the Outcome i expected :

{
talents: [
1: {
tag: "001",
name: "Jasmina",
images: "assets/images/Jasmina3.jpg",
images02: "assets/images/Jasmina4.jpg",
images03: "assets/images/Jasmina7.jpg",
skills: "Usher",
skills02: "Modeling",
indexing: "usher"
},

2:{
tag: "002",
name: "Bruna",
images: "assets/images/BrunaD17.jpg",
images02: "assets/images/BrunaD18.jpg",
images03: "assets/images/BrunaD10.jpg",
skills: "Usher",
skills02: "Modeling",
indexing: "usher"
}
]
}

And take a peek at my API, please help me play around with that $outp. to get it.

<?php
//header("Access-Control-Allow-Origin: *");
header("Content-Type: application/json; charset=UTF-8");

$conn = new mysqli("localhost", "root", "", "application");

$result = $conn->query("SELECT * FROM talents");

$outp = "";
while($rs = $result->fetch_array(MYSQLI_ASSOC)) {
    if ($outp != "") {$outp .= ",";}
    $outp .= '{"id":"'  . $rs["id"] . '",';
    $outp .= '"tag":"'  . $rs["tag"] . '",';
    $outp .= '"name":"'   . $rs["name"]        . '",';
    $outp .= '"images":"'. $rs["images"]     . '",';
    $outp .= '"images02":"'. $rs["images02"]     . '",';
    $outp .= '"images03":"'. $rs["images03"]     . '",';
    $outp .= '"skills":"'. $rs["skills"]     . '",';
    $outp .= '"skills02":"'. $rs["skills02"]     . '",';
    $outp .= '"indexing":"'. $rs["indexing"] . '"}';
}
$outp ='{"talents":['.$outp.']}';
$conn->close();

echo($outp);
?>

felixmosh · Accepted Answer · 2017-11-12 06:55:40Z

2

Instead of trying to build the json by yourself use native json_encode. It solves all the edge cases that there are with JSONs.

All you need to prepare for it is an associative array, you can specify field by field (like i did in the example, it is recommended because that way you are specifying the exact fields to send to the user, it prevent sensitive data exposure) or just $outp[$rs["id"]] = $rs;

<?php
header("Content-Type: application/json; charset=UTF-8");

$conn = new mysqli("localhost", "root", "", "application");

$result = $conn->query("SELECT * FROM talents");

$outp = array();
while ($rs = $result->fetch_array(MYSQLI_ASSOC)) {
    $outp[$rs["id"]] = array(
        "tag" => $rs["tag"],
        "name" => $rs["name"],
        "images" => $rs["images"],
        "images02" => $rs["images02"],
        "images03" => $rs["images02"],
        "skills" => $rs["skills"],
        "skills02" => $rs["skills02"],
        "indexing" => $rs["indexing"]
    );
}
$outp["talents"] = array($outp);
$conn->close();

echo (json_encode($outp));
?>

answered Nov 12, 2017 at 6:55

felixmosh

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1 Comment

Andre Ramadhan Over a year ago

Working like charms. Thank you mate

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