Converting a string stored in JSON array to variable name in PHP

Question

I have created a JSON file which I will use as a datasource for global configurations of the app.

Excerpt from json file

 //Have not put the complete json file. No error in the file
 {
 "loginType":[
    {
        "name":"Facebook",
        "url":"#",
        "method":"",
        "label":"Continue with Facebook",
        "type":"social",
        "class":"",
        "icon":"",
        "callBack_url" : "fbLoginUrl",
        "providerButton":"<div class='fb-login-button' data-max-rows='1' 
        data-size='large' data-button-type='continue_with' data-use- 
        continue-as='true'></div>"        
    },
    {
        "name":"Twitter",
        "url":"#",
        "method":"logInWithTwitter()",
        "label":"Continue with Twitter",
        "type":"social",
        "class":"",
        "icon":"",
        "callBack_url" : "twitterLoginUrl",
        "providerButton" :""
      }
    ]
}

The callBack_url key in the json file has a variable with a similar name which has a url as its value e.g $twitterLoginUrl = "https://some.site.com/twitter_login?param1"

$jsonData_signIn =json_decode
(file_get_contents(/path/to/oauth2_provider.json)); 
 $oauth2Provider = jsonData_signIn->loginType;
       foreach($oauth2Provider as $type){
         if($type->type == 'local' ){
           echo "<a href=\"{$type->callBack_url}\">{$type->label}</a>";
         }
    }

For the above, as output for the link I get eg <a href="$fbLoginURL">Continue with facebook</a>

echo "<a href=\"{${$type->callBack_url}}\">{$type->label}</a>";

The reason I am not storing the complete URI is I will generate some parameters dynamically.

Do not do that. Do not use variable variables. Anywhere you use variable variables you should be using associative arrays instead. — Sammitch
– Sammitch, Commented Jun 4, 2018 at 20:20
... making a variable name out of an url no less... Yikes ;) — IncredibleHat
– IncredibleHat, Commented Jun 4, 2018 at 20:28
I put an actual answer to your question below but I 100% agree with @Sammitch in that you shouldn't be using variable variables. Just because PHP lets you be sloppy doesn't mean you should. Think about what happens if I put 'mysql_password' into the JSON... — Joseph Yancey
– Joseph Yancey, Commented Jun 4, 2018 at 20:35
Thanks Sammitch, I will use associative arrays. I wanted one file which for global configs.Associative arrays should do the job. — Blacksmith
– Blacksmith, Commented Jun 5, 2018 at 6:40
@IncredibleHat the url is stored in a variable. The json file contains the name of the variable. The solution should convert the string contained in the value of that key to a variable. — Blacksmith
– Blacksmith, Commented Jun 5, 2018 at 6:55

Joseph Yancey · Accepted Answer · 2018-06-04 20:27:32Z

2

Take a look at the manual for variable variables: http://php.net/manual/en/language.variables.variable.php

You basically just wrap the string variable name in ${ } to make it behave like an actual variable.

$fbLoginUrl = 'https://www.facebook.com/v2.10/dialog/oauth?client_id=xxxxxxx&state=xxxxxxx&response_type=code&sdk=php-sdk-5.6.2&redirect_uri=some.site.com/fbLogin.php&scope=public_profile';
$json = '{"callBack_url" : "fbLoginUrl"}';
$decoded = json_decode($json);
echo ${$decoded->callBack_url};

answered Jun 4, 2018 at 20:27

Joseph Yancey

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1 Comment

Blacksmith Over a year ago

Thanks. It works. The error I was getting was due to a blank field in my json file.

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