2016-09-23 25 views
0

我嘗試收集從本地服務器(的Acquia開發桌面)使用此角的js代碼採集圖像(的Acquia開發桌面)

控制器的數據和圖像

var app = angular.module('App', []); 
app.controller('Ctrl', function($scope, $http) { 
    $scope.images = []; 
    $http({ 
     method : "GET", 
     url : 'http://docroot.com.dd:8083/catalogue/11/images/json' 
    }).then(function mySucces(response) { 
     $scope.images = response.data; 
    }, function myError(response) { 
     $scope.images = response.statusText; 
    }); 
}); 

的Json

[{"image":" <a href=\"http:\/\/docroot.com.dd:8083\/sites\/docroot.com.dd\/files\/catalogues\/2016-09\/images\/Pty%20Prs.compressedjpg_Page1.jpg\">Property Press.compressedjpg_Page1.jpg<\/a>"}] 

// i got out put like this : 

<a href=\"http:\/\/docroot.com.dd:8083\/sites\/docroot.com.dd\/files\/catalogues\/2016-09\/images\/Pty%20Prs.compressedjpg_Page1.jpg\">Property Press.compressedjpg_Page1.jpg<\/a> 

我需要收集的,而不是整個鏈路只圖像的URL,

+0

只是個人意見所採取的代碼,但我不知道該把HTML代碼在你的JSON是一個好主意 –

+0

@ C.Champagne IAM不確定,但我的JSON就像HTML代碼。 我們可以從它的首頁去除HTML嗎? – sarath

+0

你不能改變你的JSON嗎? –

回答

0

那麼,在一個JSON發送HTML元素似乎並不好,但是,無論如何,如果你不能改變它...

就我而言,我會解析HTML字符串與內置的XML解析器。

這裏是this answer

//XML parser 
 
var parseXml; 
 

 
if (typeof window.DOMParser != "undefined") { 
 
    parseXml = function(xmlStr) { 
 
     //should work with any recent browser 
 
     return (new window.DOMParser()).parseFromString(xmlStr, "text/xml"); 
 
    }; 
 
} else if (typeof window.ActiveXObject != "undefined" && 
 
     new window.ActiveXObject("Microsoft.XMLDOM")) { 
 
    //This part is intended to very old browsers 
 
    parseXml = function(xmlStr) { 
 
     var xmlDoc = new window.ActiveXObject("Microsoft.XMLDOM"); 
 
     xmlDoc.async = "false"; 
 
     xmlDoc.loadXML(xmlStr); 
 
     return xmlDoc; 
 
    }; 
 
} else { 
 
    throw new Error("No XML parser found"); 
 
} 
 

 
//Your code 
 

 
var jsonContent= [{"image":" <a href=\"http:\/\/docroot.com.dd:8083\/sites\/docroot.com.dd\/files\/catalogues\/2016-09\/images\/Pty%20Prs.compressedjpg_Page1.jpg\">Property Press.compressedjpg_Page1.jpg<\/a>"}]; 
 

 
var elem = jsonContent[0].image; 
 

 
var link = parseXml(elem); 
 

 
try { 
 
\t document.getElementById("out").innerHTML = link.documentElement.getAttribute("href"); 
 
} catch (e) { 
 
    alert(e); 
 
}
<span id="out" />