Asked  7 Months ago    Answers:  5   Viewed   32 times

I tried to request the weather from a web service supplying data in JSON format. My PHP request code, which did not succeed was:

$json = file_get_contents($url);
$data = json_decode($json, TRUE);
echo $data[0]->weather->weatherIconUrl[0]->value;    

This is some of the data that was returned. Some of the details have been truncated for brevity, but object integrity is retained:

{ "data": 
    { "current_condition": 
        [ { "cloudcover": "31",
            ... } ],  
        [ { "query": "Schruns, Austria",
            "type": "City" } ],
        [ { "date": "2010-10-27",
            "precipMM": "0.0",
            "tempMaxC": "3",
            "tempMaxF": "38",
            "tempMinC": "-13",
            "tempMinF": "9",
            "weatherCode": "113",
            "weatherDesc": [ {"value": "Sunny" } ],
            "weatherIconUrl": [ {"value": "" } ],
            "winddir16Point": "N",
            "winddirDegree": "356",
            "winddirection": "N",
            "windspeedKmph": "5",
            "windspeedMiles": "3" }, 
          { "date": "2010-10-28",
            ... },

          ... ]



This appears to work:

$url = ',austria&format=json&num_of_days=5&key=8f2d1ea151085304102710%22';
$content = file_get_contents($url);
$json = json_decode($content, true);

foreach($json['data']['weather'] as $item) {
    print $item['date'];
    print ' - ';
    print $item['weatherDesc'][0]['value'];
    print ' - ';
    print '<img src="' . $item['weatherIconUrl'][0]['value'] . '" border="0" alt="" />';
    print '<br>';

If you set the second parameter of json_decode to true, you get an array, so you cant use the -> syntax. I would also suggest you install the JSONview Firefox extension, so you can view generated json documents in a nice formatted tree view similiar to how Firefox displays XML structures. This makes things a lot easier.

Wednesday, March 31, 2021
answered 7 Months ago

What a HORRENDOUS debug session.. well there's good news.. I figured it out..

I started looking at it using AJAX and logging it with Firebug... and it turns out json_decode (or eval by the way) cannot handle &quot;, which is what PHPUnit sends back (Come on Sebastian!), so to fix it:

$json = str_replace('&quot;', '"', $json);

Now I thought they were the same.. maybe someone can enlighten me..

Wednesday, March 31, 2021
answered 7 Months ago

I answered a similar question here: Sencha seems to not like Rails' json. It is specific to Rails but the concept still applies.

Essentially James Pearce is correct. What you are returning needs to be wrapped in a tag and the callback function. This will insert the code on your page and run the script, which has the effect of calling the function you provide.

$response = "<script type='text/javascript'>";
$response .= $_GET['callback'] . "(" . json_encode($row) . ")";
$response .= "</script>";
Wednesday, March 31, 2021
answered 7 Months ago

You were close. In your JSON object, the URLs are one more level deep. You need access it as below:

foreach ($json2->attachments as $value) {
    echo $value->url;


Saturday, May 29, 2021
answered 5 Months ago

You need to put them inside a curly brace with a single quote:

$place_name = $response->places[0]->{'place name'};
echo $place_name;

Or as @scragar said in the comments, if you're not confortable accessing them thru objects, you can put a true flag on json_decode($response, true), so that you can access them as associative arrays instead.

Friday, August 6, 2021
answered 3 Months ago
Only authorized users can answer the question. Please sign in first, or register a free account.
Not the answer you're looking for? Browse other questions tagged :