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how to loop or display json data using php? – Photoshop

I want to display skills and image from JSON using PHP I’m trying to research but not working

Error Message:

Notice: Array to string conversion in C:xampphtdocstestview.php on line 27

Notice: Array to string conversion in C:xampphtdocstestview.php on line 32

data.json

{
    "UserData": [

    {
        "id": "name",
        "fname": "Joey",
        "lname": "Tulang",
        "age": "29",
        "gender": "Male",
        "skills": [ "HTML", "CSS", "JavaScript" ],
        "image": [ "html.jpg", "css.jpg", "js.jpg" ]
    },
    {
        "id": "2",
        "fname": "Angelica",
        "lname": "Balce",
        "age": "20",
        "gender": "Female",
        "skills": [ "C++", "Python", "PHP"],
        "image": [ "c++.jpg", "python.jpg", "php.jpg" ]
    },
    {
        "id": "3",
        "fname": "Mj",
        "lname": "King",
        "age": "21",
        "gender": "male",
        "skills": [ "Photoshop", "Figma", "Corel Draw"],
        "image": [ "photoshop.jpg", "figma.jpg", "cdraw.jpg" ]
    }       

    ]   
}

index.php

<!DOCTYPE html>
<html>
<head>
    <title>Users</title>
</head>
<body>

<?php

$jsondata = file_get_contents("data.json");
$json = json_decode($jsondata, true);
$output = "<ul>";

foreach ( $json['UserData'] as $display ) {
  $output .="<li>" .$display['fname']. " <a href='view.php?id=".$display['id']."'>View User Data</a></li> </br>";
}

$output .= "</ul>";

echo $output;

?>

</body>
</html>

view.php

<?php
$jsondata = file_get_contents("data.json");
$json = json_decode($jsondata, true);

$key = array_search((int) $_GET['id'], array_column($json['UserData'], 'id'));

// if $key false, was not found, presume it was...

$user = $json['UserData'][$key];
?>
<!DOCTYPE html>
<html>
<head>
    <title>View User Data</title>
</head>
<body>

    <!-- Display User Data Here -->

    <p>First Name: <?= $user['fname'] ?></p>
    <p>Last Name: <?= $user['lname'] ?></p>
    <p>Age: <?= $user['age'] ?></p>
    <p>Gender: <?= $user['gender'] ?></p>

    <p>Skills</p>
    <ul>
        <?= "<li>" .$user['skills']. "</li>" ?>
    </ul>

    <p>Image</p>

        <?= "<imge src='image/" .$user['image']. "'/>" ?>


</body>
</html>

view.php (Output)

https://ibb.co/fk4LPLG

Tags:

2

Answers


  1. Chosen as BEST ANSWER
    0

    Answer:

     foreach ($user['image'] as $value)
 {
    echo "<img src='".$value."'>";
 }

  2. 0

    because your $user['image'] is not string, that one is array.
    try print your $user['image'] and you will know.

    ps: use foreach instead your array.

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