API Access HTML/JavascriptHow do I get a list of the transactions of an address using TzScan API v3?How do I get a list of all delegators of an address using TzScan API v3?API Request Cycle

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API Access HTML/Javascript


How do I get a list of the transactions of an address using TzScan API v3?How do I get a list of all delegators of an address using TzScan API v3?API Request Cycle













1















I have an issue trying to use javascript to get the circulating supply by requesting data from Tzscan API :



HTML code to request the circulating supply :



<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width">
<title>JS Bin</title>
</head>
<body>
<h1>There Tezos Circulating : </h1>
<div id="getapi"></div>
</body>
</html>


Javascript Code to get the data from the API :



<script>
function createNode(element)
return document.createElement(element);


function append(parent, el)
return parent.appendChild(el);


const ul = document.getElementById('getapi');
const url = 'http://api6.tzscan.io/v3/supply';
fetch(url)
.then((resp) => resp.json())
.then(function(data)
let span = createNode('span');
span.innerHTML = `$data.circulating_supply/1000000`;
append(ul, span);
)
.catch(function(error)
console.log(error);
);
</script>


The problem is that somehow it works when I create a HTML file locally but not online :



https://codepen.io/anon/pen/LaqqpB










share|improve this question









New contributor




Lexxor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • it'll help to also include the errors you are getting. I know we could probably look at your codepen but for the sake of the question, it's best include the errors in the question.

    – Frank
    1 hour ago















1















I have an issue trying to use javascript to get the circulating supply by requesting data from Tzscan API :



HTML code to request the circulating supply :



<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width">
<title>JS Bin</title>
</head>
<body>
<h1>There Tezos Circulating : </h1>
<div id="getapi"></div>
</body>
</html>


Javascript Code to get the data from the API :



<script>
function createNode(element)
return document.createElement(element);


function append(parent, el)
return parent.appendChild(el);


const ul = document.getElementById('getapi');
const url = 'http://api6.tzscan.io/v3/supply';
fetch(url)
.then((resp) => resp.json())
.then(function(data)
let span = createNode('span');
span.innerHTML = `$data.circulating_supply/1000000`;
append(ul, span);
)
.catch(function(error)
console.log(error);
);
</script>


The problem is that somehow it works when I create a HTML file locally but not online :



https://codepen.io/anon/pen/LaqqpB










share|improve this question









New contributor




Lexxor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • it'll help to also include the errors you are getting. I know we could probably look at your codepen but for the sake of the question, it's best include the errors in the question.

    – Frank
    1 hour ago













1












1








1








I have an issue trying to use javascript to get the circulating supply by requesting data from Tzscan API :



HTML code to request the circulating supply :



<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width">
<title>JS Bin</title>
</head>
<body>
<h1>There Tezos Circulating : </h1>
<div id="getapi"></div>
</body>
</html>


Javascript Code to get the data from the API :



<script>
function createNode(element)
return document.createElement(element);


function append(parent, el)
return parent.appendChild(el);


const ul = document.getElementById('getapi');
const url = 'http://api6.tzscan.io/v3/supply';
fetch(url)
.then((resp) => resp.json())
.then(function(data)
let span = createNode('span');
span.innerHTML = `$data.circulating_supply/1000000`;
append(ul, span);
)
.catch(function(error)
console.log(error);
);
</script>


The problem is that somehow it works when I create a HTML file locally but not online :



https://codepen.io/anon/pen/LaqqpB










share|improve this question









New contributor




Lexxor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












I have an issue trying to use javascript to get the circulating supply by requesting data from Tzscan API :



HTML code to request the circulating supply :



<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width">
<title>JS Bin</title>
</head>
<body>
<h1>There Tezos Circulating : </h1>
<div id="getapi"></div>
</body>
</html>


Javascript Code to get the data from the API :



<script>
function createNode(element)
return document.createElement(element);


function append(parent, el)
return parent.appendChild(el);


const ul = document.getElementById('getapi');
const url = 'http://api6.tzscan.io/v3/supply';
fetch(url)
.then((resp) => resp.json())
.then(function(data)
let span = createNode('span');
span.innerHTML = `$data.circulating_supply/1000000`;
append(ul, span);
)
.catch(function(error)
console.log(error);
);
</script>


The problem is that somehow it works when I create a HTML file locally but not online :



https://codepen.io/anon/pen/LaqqpB







api javascript circulating-supply






share|improve this question









New contributor




Lexxor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Lexxor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 1 hour ago









Frank

1,117317




1,117317






New contributor




Lexxor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 6 hours ago









LexxorLexxor

83




83




New contributor




Lexxor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Lexxor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Lexxor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • it'll help to also include the errors you are getting. I know we could probably look at your codepen but for the sake of the question, it's best include the errors in the question.

    – Frank
    1 hour ago

















  • it'll help to also include the errors you are getting. I know we could probably look at your codepen but for the sake of the question, it's best include the errors in the question.

    – Frank
    1 hour ago
















it'll help to also include the errors you are getting. I know we could probably look at your codepen but for the sake of the question, it's best include the errors in the question.

– Frank
1 hour ago





it'll help to also include the errors you are getting. I know we could probably look at your codepen but for the sake of the question, it's best include the errors in the question.

– Frank
1 hour ago










2 Answers
2






active

oldest

votes


















2














You've encountered a mixed content error. Requests can't be a mixture of http and https, they need to be all of the same, preferably https. Just change the const url = 'http://api6.tzscan.io/v3/supply'; to be https





const url = 'https://api6.tzscan.io/v3/supply';

fetch(url)
.then(resp => resp.json())
.then((dls) =>
let span = createNode('span');
span.innerHTML = dls;
append(ul, span);
)
.catch(console.log);





share|improve this answer
































    0














    OMG, it works perfectly.



    Thank you a lot.






    share|improve this answer








    New contributor




    Lexxor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.




















    • You're welcome. Better to mark the correct answer and just put comments under it instead of creating new answers to comment.

      – Richard Ayotte
      2 hours ago












    • Please don't add "thank you" as an answer. Instead, accept the answer that you found most helpful. - From Review

      – Tom
      1 hour ago










    Your Answer








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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    You've encountered a mixed content error. Requests can't be a mixture of http and https, they need to be all of the same, preferably https. Just change the const url = 'http://api6.tzscan.io/v3/supply'; to be https





    const url = 'https://api6.tzscan.io/v3/supply';

    fetch(url)
    .then(resp => resp.json())
    .then((dls) =>
    let span = createNode('span');
    span.innerHTML = dls;
    append(ul, span);
    )
    .catch(console.log);





    share|improve this answer





























      2














      You've encountered a mixed content error. Requests can't be a mixture of http and https, they need to be all of the same, preferably https. Just change the const url = 'http://api6.tzscan.io/v3/supply'; to be https





      const url = 'https://api6.tzscan.io/v3/supply';

      fetch(url)
      .then(resp => resp.json())
      .then((dls) =>
      let span = createNode('span');
      span.innerHTML = dls;
      append(ul, span);
      )
      .catch(console.log);





      share|improve this answer



























        2












        2








        2







        You've encountered a mixed content error. Requests can't be a mixture of http and https, they need to be all of the same, preferably https. Just change the const url = 'http://api6.tzscan.io/v3/supply'; to be https





        const url = 'https://api6.tzscan.io/v3/supply';

        fetch(url)
        .then(resp => resp.json())
        .then((dls) =>
        let span = createNode('span');
        span.innerHTML = dls;
        append(ul, span);
        )
        .catch(console.log);





        share|improve this answer















        You've encountered a mixed content error. Requests can't be a mixture of http and https, they need to be all of the same, preferably https. Just change the const url = 'http://api6.tzscan.io/v3/supply'; to be https





        const url = 'https://api6.tzscan.io/v3/supply';

        fetch(url)
        .then(resp => resp.json())
        .then((dls) =>
        let span = createNode('span');
        span.innerHTML = dls;
        append(ul, span);
        )
        .catch(console.log);






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 2 hours ago

























        answered 5 hours ago









        Richard AyotteRichard Ayotte

        16416




        16416





















            0














            OMG, it works perfectly.



            Thank you a lot.






            share|improve this answer








            New contributor




            Lexxor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.




















            • You're welcome. Better to mark the correct answer and just put comments under it instead of creating new answers to comment.

              – Richard Ayotte
              2 hours ago












            • Please don't add "thank you" as an answer. Instead, accept the answer that you found most helpful. - From Review

              – Tom
              1 hour ago















            0














            OMG, it works perfectly.



            Thank you a lot.






            share|improve this answer








            New contributor




            Lexxor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.




















            • You're welcome. Better to mark the correct answer and just put comments under it instead of creating new answers to comment.

              – Richard Ayotte
              2 hours ago












            • Please don't add "thank you" as an answer. Instead, accept the answer that you found most helpful. - From Review

              – Tom
              1 hour ago













            0












            0








            0







            OMG, it works perfectly.



            Thank you a lot.






            share|improve this answer








            New contributor




            Lexxor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.










            OMG, it works perfectly.



            Thank you a lot.







            share|improve this answer








            New contributor




            Lexxor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            share|improve this answer



            share|improve this answer






            New contributor




            Lexxor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            answered 4 hours ago









            LexxorLexxor

            83




            83




            New contributor




            Lexxor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.





            New contributor





            Lexxor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            Lexxor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.












            • You're welcome. Better to mark the correct answer and just put comments under it instead of creating new answers to comment.

              – Richard Ayotte
              2 hours ago












            • Please don't add "thank you" as an answer. Instead, accept the answer that you found most helpful. - From Review

              – Tom
              1 hour ago

















            • You're welcome. Better to mark the correct answer and just put comments under it instead of creating new answers to comment.

              – Richard Ayotte
              2 hours ago












            • Please don't add "thank you" as an answer. Instead, accept the answer that you found most helpful. - From Review

              – Tom
              1 hour ago
















            You're welcome. Better to mark the correct answer and just put comments under it instead of creating new answers to comment.

            – Richard Ayotte
            2 hours ago






            You're welcome. Better to mark the correct answer and just put comments under it instead of creating new answers to comment.

            – Richard Ayotte
            2 hours ago














            Please don't add "thank you" as an answer. Instead, accept the answer that you found most helpful. - From Review

            – Tom
            1 hour ago





            Please don't add "thank you" as an answer. Instead, accept the answer that you found most helpful. - From Review

            – Tom
            1 hour ago










            Lexxor is a new contributor. Be nice, and check out our Code of Conduct.









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            Lexxor is a new contributor. Be nice, and check out our Code of Conduct.











            Lexxor is a new contributor. Be nice, and check out our Code of Conduct.














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