Question

• 制作街道静态图像,如:

https://maps.googleapis.com/maps/api/streetview?size=1080x400&position=%s&f=90&head=235&pitch=p%;ampa。

如果你访问,你就会看到一种形象,即“神,我们对此没有图像”。

是否有办法发现这一“音响”状态,以便我能够回到另一个形象?

Answer 1

this situation is already build in in the 3.0 version due the boolean test status === streetviewStatus.Ok, here is a snippet from my situation solving

if (status === google.maps.StreetViewStatus.OK) {
            var img = document.createElement("IMG");
            img.src =  http://maps.googleapis.com/maps/api/streetview?size=160x205&location= + lat + , + lng  + &sensor=false&key=AIzaSyC_OXsfB8-03ZXcslwOiN9EXSLZgwRy94s ;
            var oldImg = document.getElementById( streetViewImage );
            document.getElementById( streetViewContainerShow ).replaceChild(img, streetViewImage);
        } else {
            var img = document.createElement("IMG");
            img.src =  ../../images/ProfilnoProfilPicture.jpg ;
            img.height = 205;
            img.width = 160;
            var oldImg = document.getElementById( streetViewImage );
            document.getElementById( streetViewContainerShow ).replaceChild(img, streetViewImage);
        }

Answer 2

One quick solution would be to load the image file using xmlrpc and check that its md5sum is 30234b543d5438e0a0614bf07f1ebd25, or that its size is 1717 bytes (it s unlikely that another image can have exactly the same size), but that s not very robust since I have seen Google change the position of the text in the image. Though it s a very good start for a prototype.

You could go for image processing instead. Note that it s still not perfectly robust since Google could decide to change the looks of the image anytime. You ll have to decide whether it s worth it.

不管怎么说,我如何利用 j子这样做:

load the image and open a 2D context for direct pxiel access (see this question for how to do it)
analyse the image:
- sample groups of 2×2 pixels at random locations; I recommend at least 30 groups
- a group of 2×2 pixels is good if all the pixels have the same value and their R/G/B values do not differ by more than 10% (ie. they re grey)
- count the ratio of good pixel groups in the image
if there are more than 70% good pixel groups, then we are pretty sure this is the “no imagery” version: replace it with another image of your choice.

我不建议直接测试罗姆学生的价值的原因是,青年促进平等论坛抑制不同浏览器的行为可能略有不同。

Answer 3

到2016年,您可使用新的。

现在需要 status。实地了解是否发现了帕那马。

www.un.org/Depts/DGACM/index_spanish.htm Example requests:

{
   "status" : "ZERO_RESULTS"
}

www.un.org/Depts/DGACM/index_spanish.htm

{
   ...
   "status" : "OK"
}

Answer 4

您可使用SapnoramaByLocation功能(见。

试图这样做:

function handleMapClick()
{
 var ll= new google.maps.LatLng(latitude,longitude);
 sv.getPanoramaByLocation(ll, 50, processSVData);
}

function processSVData(data, status) {
 if (status==google.maps.StreetViewStatus.ZERO_RESULTS)
 {
   <DO SOMETHING>
 }
}

Answer 5

要求有一条眼光的街道观像,如果它有具体的档案规模,那就不是街头观。我是这样说的:

var url =  google street view url ;

var xhr = new XMLHttpRequest();
xhr.open( GET , url, true);
xhr.responseType =  blob ;

xhr.onload = function (e) {
  if (this.status == 200) {
    try {
      var image = new Blob([this.response], {type:  image/jpeg });

      if (image.size) {
        if (url.indexOf( 640x640 ) > -1 && image.size === 8410) {
          // Not street view
        }

        if (url.indexOf( 400x300 ) > -1 && image.size === 3946) {
           // Not street view
        }
      }
    } catch (err) {
      // IE 9 doesn t support blob
    }
  }
};

xhr.send();

Answer 6

另一种办法是装上图像,然后比较一些颜色。 go角的“无街道观”形象总是一样的。这里,你将如何比较2个图象:

var url = STREETVIEWURL
var img = new Image();
// Add some info to prevent cross origin tainting
img.src = url +  ?  + new Date().getTime();
img.setAttribute( crossOrigin ,   );
img.crossOrigin = "Anonymous";
img.onload = function() {
    var context = document.createElement( CANVAS ).getContext( 2d );
    context.drawImage(img, 0, 0);
    //load 2 pixels.  I chose the first one and the 5th row
    var data1 = context.getImageData(0, 0, 1, 1).data;
    var data2 = context.getImageData(0, 5, 1, 1).data;
    console.log(data1);
    // google unknown image is this pixel color [228,227,223,255]
    if(data1[0]==228 && data1[1]==227 && data1[2]==223 && data1[3]==255 && 
                     data2[0]==228 && data2[1]==227 && data2[2]==223 && data2[3]==255){
        console.log("NO StreetView Available");
    }else{
         console.log("StreetView is Available");
    }
};

Some potential issues: I ve seen some errors with CrossOrigin tainting. Also, if google changes the image returned this code will break.

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