CODEWITHSUNDEEP
phparraysjsonapiforeach
user5455037
user5455037

Reputation:

PHP search in json array

I want to get the value of all "name" keys which "category" value is not "Beilagen" or "Aktion".

Example of the wanted ouput using the json data below:

Zartweizen mit Gemüse

Kaiserschmarrn mit Apfelmus

Gebackene Tintenfischringe mit Knoblauchdip

The solution may be a foreach loop, but I can't really figure out how to make specific searches with that.

This is my quick fix solution, but as you can see in the json example below, the number of relevant data varies and it's not always 4 names I have to get. It can be more or less than that.

$name = "1. ".$updateArrayMensa[0]["name"].chr(10)."2. ".$updateArrayMensa[1]["name"].chr(10)."3. ".$updateArrayMensa[2]["name"].chr(10)."4. ".$updateArrayMensa[3]["name"];

Json data example:

[
   {
      "id":1542115,
      "name":"Zartweizen mit Gemüse",
      "category":"Tagesgericht 1",
      "prices":{
         "students":1.0,
         "employees":1.9,
         "pupils":null,
         "others":2.4
      },
      "notes":[
         "veganes Gericht"
      ]
   },
   {
      "id":1542116,
      "name":"Kaiserschmarrn mit Apfelmus",
      "category":"Tagesgericht 4",
      "prices":{
         "students":2.4,
         "employees":2.95,
         "pupils":null,
         "others":3.45
      },
      "notes":[
         "mit Antioxidationsmittel",
         "fleischloses Gericht"
      ]
   },
   {
      "id":1542117,
      "name":"Gebackene Tintenfischringe mit Knoblauchdip",
      "category":"Aktionsessen 3",
      "prices":{
         "students":2.4,
         "employees":2.95,
         "pupils":null,
         "others":3.45
      },
      "notes":[
         "mit einer Zuckerart und Süßungsmitteln",
         "mit Farbstoff",
         "mit Fleisch"
      ]
   },
   {
      "id":1542128,
      "name":"Ananaskompott",
      "category":"Beilagen",
      "prices":{
         "students":null,
         "employees":null,
         "pupils":null,
         "others":null
      },
      "notes":[
         "veganes Gericht"
      ]
   },
   {
      "id":1542129,
      "name":"Weiße Schokolade-Himbeer-Cookie",
      "category":"Aktion",
      "prices":{
         "students":null,
         "employees":null,
         "pupils":null,
         "others":null
      },
      "notes":[
         "fleischloses Gericht"
      ]
   }
]

Source of this json data: http://www.openmensa.org/api/v2/canteens/138/days/2015-10-16/meals

Upvotes: 3

Views: 7287

Answers (4)

Kristoffer Bohmann
Kristoffer Bohmann

Reputation: 4094

Using a foreach loop you can add matching rows to a $query array that will hold any matching rows:

$json = file_get_contents("9880387.json");
$json = json_decode($json);

# print_r($json);

foreach ($json as $v) {
    if( !in_array($v->category,["Beilagen","Aktion"]) ) {
        $query[] = $v->name;
    }
}
print_r($query);

Outputs an array with these values:

Zartweizen mit Gemüse
Kaiserschmarrn mit Apfelmus
Gebackene Tintenfischringe mit Knoblauchdip

Upvotes: 1

Tom
Tom

Reputation: 691

<?php

  $json = json_decode( file_get_contents( 'http://www.openmensa.org/api/v2/canteens/138/days/2015-10-16/meals' ) );
  $drop = array( 'Beilagen', 'Aktion' );
  $name = array();

  foreach( $json as $obj ) {
    if ( !in_array( $obj->category, $drop ) ) {
      $name[] = $obj->name;
    }
  }

  echo implode( '<br>', $name );

?>

Output:

Zartweizen mit Gemüse
Kaiserschmarrn mit Apfelmus
Gebackene Tintenfischringe mit Knoblauchdip

Upvotes: 1

Roman
Roman

Reputation: 41

You can convert json array to simple php array with function json_decode:

$items = json_decode("you_json_string");

$names = array();

foreach($items as $item) {
   if($item->category == 'Beilagen' || $item->category == 'Action')
       continue;

   $names[] = $item->name;
}
print_r($names);

Upvotes: 4

xjx424
xjx424

Reputation: 183

Have you tried json_decode? Convert to a php array then use your foreach loop. http://php.net/manual/en/function.json-decode.php

Upvotes: 0

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