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?
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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?
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 = `$data.circulating_supply/1000000`;
append(ul, span);
)
.catch(function(error)
console.log(error);
);
</script>
The problem is that somehow it works when I create a HTML file locally but not online :
https://codepen.io/anon/pen/LaqqpB
api javascript circulating-supply
New contributor
add a comment |
I have an issue trying to use javascript to get the circulating supply by requesting data from Tzscan API :
HTML code to request the circulating supply :
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width">
<title>JS Bin</title>
</head>
<body>
<h1>There Tezos Circulating : </h1>
<div id="getapi"></div>
</body>
</html>
Javascript Code to get the data from the API :
<script>
function createNode(element)
return document.createElement(element);
function append(parent, el)
return parent.appendChild(el);
const ul = document.getElementById('getapi');
const url = 'http://api6.tzscan.io/v3/supply';
fetch(url)
.then((resp) => resp.json())
.then(function(data)
let span = createNode('span');
span.innerHTML = `$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
New contributor
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
28 mins ago
add a comment |
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 = `$data.circulating_supply/1000000`;
append(ul, span);
)
.catch(function(error)
console.log(error);
);
</script>
The problem is that somehow it works when I create a HTML file locally but not online :
https://codepen.io/anon/pen/LaqqpB
api javascript circulating-supply
New contributor
I have an issue trying to use javascript to get the circulating supply by requesting data from Tzscan API :
HTML code to request the circulating supply :
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width">
<title>JS Bin</title>
</head>
<body>
<h1>There Tezos Circulating : </h1>
<div id="getapi"></div>
</body>
</html>
Javascript Code to get the data from the API :
<script>
function createNode(element)
return document.createElement(element);
function append(parent, el)
return parent.appendChild(el);
const ul = document.getElementById('getapi');
const url = 'http://api6.tzscan.io/v3/supply';
fetch(url)
.then((resp) => resp.json())
.then(function(data)
let span = createNode('span');
span.innerHTML = `$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
api javascript circulating-supply
New contributor
New contributor
edited 30 mins ago
Frank
1,117317
1,117317
New contributor
asked 5 hours ago
LexxorLexxor
83
83
New contributor
New contributor
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
28 mins ago
add a comment |
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
28 mins 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
28 mins 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
28 mins ago
add a comment |
2 Answers
2
active
oldest
votes
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);
add a comment |
OMG, it works perfectly.
Thank you a lot.
New contributor
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
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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);
add a comment |
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);
add a comment |
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);
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);
edited 1 hour ago
answered 4 hours ago
Richard AyotteRichard Ayotte
16416
16416
add a comment |
add a comment |
OMG, it works perfectly.
Thank you a lot.
New contributor
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
add a comment |
OMG, it works perfectly.
Thank you a lot.
New contributor
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
add a comment |
OMG, it works perfectly.
Thank you a lot.
New contributor
OMG, it works perfectly.
Thank you a lot.
New contributor
New contributor
answered 4 hours ago
LexxorLexxor
83
83
New contributor
New contributor
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
add a comment |
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
add a comment |
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.
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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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
28 mins ago