CODEWITHSUNDEEP
phpjsonapigoogle-apigoogle-api-php-client
user3786234
user3786234

Reputation: 115

google api search results displaying from json format to php/html

I want to display the actvivites posts results which i recieved through goole json api url on my website, therefore i used following codeings. but its not displaying results. please advice .

    <?php
   $jsonurl = "https://www.googleapis.com/plus/v1/activities?orderBy=recent&query=%23fifa&alt=json&key={API KEY }";
    $json = file_get_contents($jsonurl);

    $json_output = json_encode($json);
    //echo json_encode($json);
    echo '<pre>';
    //echo json_encode($json);
    print_r($json_output);
    echo '</pre>';
    ?>
    <div class="datagrid ">


    <?php
    $json=array();
    foreach ($json_output["data"] as $json_result) {
        $json[] = array(
            'value' => $json_result["title"],
            'value' => $json_result["url"]

        );

      ?>

    <div ><table>


    <tbody>

        <TR>
        <TD><h3><?php echo $json_result["title"];?></h3></TD>
         <TD><h3><?php echo $json_result["url"];?></h3></TD>
        </TR>
     </tbody>
    </table>
    </div>

      <?php

    }


    ?>

    </div>

google api source

Upvotes: 0

Views: 620

Answers (1)

Damon Hogan
Damon Hogan

Reputation: 552

I can point out a few things in your code. Some definitely need a little help.

You have these two lines.

$json = file_get_contents($jsonurl);

$json is already a json encoded string, I presume instead of the following line. You actually want to decode it into an array.

$json_output = json_encode($json);

Such as

$json_output = json_decode($json, true);

notice I used the true parameter to make sure it gets all the way decoded instead of an array of std class variables. but I suppose it would depend on what you want in your code.

Upvotes: 1

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