Urgent :--Send with SG template an image inline

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1

Hello guys, I have a form on my website where the customer can upload an image.Then the site sends to him a template email with sendgrid, so I want to show this image uploaded in my email. How can I do that?

2

UP

3

https://www.wix.com/corvid/forum/community-discussion/include-images-documents-in-form-submission-email

4

I didn’t understand, can a moderator help me?

5

If you use the sendgrid npm
then in your backend pass the full image url:

export function myFunction(sender, recipient, message, imageUrl) {
    sgMail.setApiKey('API KEY');
    let msg = {
        to: recipient,
        from: sender,
        templateId: 'template-key',
        dynamic_template_data: {
         "message": message,
         "imageUrl": imageUrl
            },
        };
    sgMail.send(msg);
}

And on sendGrid create an html template with:

<img class="max-width" border="0" style="display:block;color:#000000;text-decoration:none;font-family:Helvetica, arial, sans-serif;font-size:16px;max-width:100% !important;width:100%;height:auto !important;" src={{imageUrl}} alt="" width="180">
6

P.S. full url is not the wix image source. To retrieve the full url, do:

let imageUrl = "https://static.wixstatic.com/media/" + $w("#image1").src.split("/")[3];
7

Don’t work

8

It worked for me. Maybe post your code here…

9

@jonatandor35 But how to get the url of an image uploaded?

10

How can I get the url from an image uploaded with the upload button in a database?

11

Is it a dynamic item page?
then:

$w.onReady( () => {
  $w("#dynamicDataset").onReady( () => {
  let imageSrc = $w("#dynamicDataset").getCurrentItem().image;
  let imageUrl = "https://static.wixstatic.com/media/" + imageSrc.split("/")[3];
  })
  })
12

@jonatandor35 No it isn’t a dynamic page

13

@fraspace5 so add details. What data do you have? What would you like to happen?

14

@jonatandor35 I have an upload button, with this button I send to a database the image, but always in the same page I send an email with that image

15

@fraspace5

let imageUrl;
//code..code...use your uploadButton code...then:
      $w("#myUploadButton").startUpload()
      .then( (uploadedFile) => {
        imageUrl =  "https://static.wixstatic.com/media/" + uploadedFile.url.split("/")[3];
      } )