Angularjs 如何在我的控制器中获取url变量值
我的网址是: 我想删除所有if条件,并将“theme number变量”直接分配给$scope.banner变量。我试过$routeParams,但它没有显示任何价值。我是新生,对angularjs的工作原理有基本的了解。请帮帮我Angularjs 如何在我的控制器中获取url变量值,angularjs,Angularjs,我的网址是: 我想删除所有if条件,并将“theme number变量”直接分配给$scope.banner变量。我试过$routeParams,但它没有显示任何价值。我是新生,对angularjs的工作原理有基本的了解。请帮帮我 var myStr = $location.absUrl(); if (myStr.indexOf("theme11") != -1){ console.log ("THEME1
var myStr = $location.absUrl();
if (myStr.indexOf("theme11") != -1){
console.log ("THEME11 ");
$scope.banner = "theme11.jpg";
}else if(myStr.indexOf("theme2") != -1){
console.log ("THEME2");
$scope.banner = "theme2.jpg";
}else if(myStr.indexOf("theme3") != -1){
console.log ("THEME3");
$scope.banner = "theme3.jpg";
}else if(myStr.indexOf("theme4") != -1){
console.log ("THEME4");
$scope.banner = "theme4.jpg";
}
else{
$scope.banner = "default.jpg";
}
使用拆分
var themeName = $location.path().split("/")[3]
'http://192.168.1.4:8081/theme11#/createAd'.split("/")[3]
蛋氨酸含量
themeName = "theme11#"
然后
也许,试试这个:
var myStr = $location.absUrl();
var number = parseInt(myStr.replace((location.protocol+'//'), '').split('/')[1].replace('theme', ''));
$scope.banner = 'theme' + number + '.jpg";
jsfiddle:
非常感谢。。我浪费了一整天。。你救了我一个晚上谢谢你的时间
var myStr = $location.absUrl();
var number = parseInt(myStr.replace((location.protocol+'//'), '').split('/')[1].replace('theme', ''));
$scope.banner = 'theme' + number + '.jpg";