在yii中有一种方法可以使参数无限制
例如,我有模块/管理员/
'urlManager'=>array(
'urlFormat'=>'path',
'showScriptName'=>false,
'caseSensitive'=>false,
'rules'=>array(
'<controller:\w+>/<id:\d+>'=>'<controller>/view',
'<controller:\w+>/<action:\w+>/<id:\d+>'=>'<controller>/<action>',
'<controller:\w+>/<action:\w+>'=>'<controller>/<action>',
'admin/<controller:\w+>/<action:\w+>/<id:\d+>' => 'admin/<controller>/<action>',
'admin/<controller:\w+>/<action:\w+>'=>'admin/<controller>/<action>',
),
),
Run Code Online (Sandbox Code Playgroud)
在管理模块中,我需要每个动作都有无限的参数,例如:
/admin/anycontroller/anyaction/anything
/admin/anycontroller/anyaction/anything/anything2
/admin/anycontroller/anyaction/anything/anything2/anything3
/admin/anycontroller/anyaction/anything/anything2/anything3/anything4
... and so on
Run Code Online (Sandbox Code Playgroud)
我应该在规则上逐一定义吗?或者有捷径可以做到这一点?
以及如何在控制器动作上捕获它?
我有这个脚本:
function downloadIt() {
var dataUri = "data:application/csv;charset=utf-8,Col1%2CCol2%2CCol3%0AVal1%2CVal2%2CVal3%0AVal11%2CVal22%2CVal33%0AVal111%2CVal222%2CVal333"
var filename = "somedata.csv"
$("<a download='" + filename + "' href='" + dataUri + "'></a>")[0].click();
}
Run Code Online (Sandbox Code Playgroud)
它可以在Chrome上运行,但不能在Firefox上运行,并且在控制台上没有任何错误。是什么原因以及如何解决?
使用Google AppEngine(Go)读取文件的正确方法是什么?
在Java中我读过context.getResourceAsStream,有没有相同的功能呢?
开发模式使用npm run dev,发布模式使用npm build
我怎么知道它当前是建立在开发模式上还是不在代码中,例如:
<script>
import {onMount} from 'svelte';
onMount(function(){
if(DEVMODE) { // --> what's the correct one?
console.log('this is x.svelte');
}
})
</script>
Run Code Online (Sandbox Code Playgroud) 如何获取当前 svelte 文件的所有属性?
例如这个Component1.svelte
<script>
let x = '';
let y = '';
let z = '';
onMount(function(){
console.log( what? );
// need to print x, y, and z which set by other code
// that using/importing this file
// without defining one by one, was there such property?
});
</script>
<span>{x}</span>
<div>{y}</div>
<p>{z}</p>
Run Code Online (Sandbox Code Playgroud)
使用者whatever.svelte
<script>
import Foo from `./Component1.svelte`
</script>
<Foo x="1" y="abc">test</Foo>
Run Code Online (Sandbox Code Playgroud) 我想调用带有可变长度参数的console.log函数
function debug_anything() {
var x = arguments;
var p = 'DEBUG from ' + (new Error).stack.split("\n")[2];
switch(x.length) {
case 0: console.log(p); break;
case 1: console.log(p,x[0]); break;
case 2: console.log(p,x[0],x[1]); break;
case 3: console.log(p,x[0],x[1],x[2]); break;
case 4: console.log(p,x[0],x[1],x[2],x[3]); break;
// so on..
}
}
Run Code Online (Sandbox Code Playgroud)
有没有(更短的)其他方式,请注意我不想要这个解决方案(因为x对象(参数或数组)的其他方法将被输出.
console.log(p,x);
Run Code Online (Sandbox Code Playgroud) 我有20个教室,每个教室有50个学生.
是否可以在一个查询语句中找到每个教室中的10名最佳学生?
表"klas":
int id
int cla_id
int stu_id
int final_score
Run Code Online (Sandbox Code Playgroud) 是否有NoSQL支持的数据库驱动程序列表Go?我只找到了SQL数据库列表:
https://code.google.com/p/go-wiki/wiki/SQLDrivers
Run Code Online (Sandbox Code Playgroud) 当tableStruct有gorm.Model里面时,如何通过id没有gorm自动添加来获取第一条记录deleted_at IS NULL?有这样的功能吗?
是否可以修改 json 序列化和反序列化,使结构如下:
type Foo struct {
A int64
B uint64
// and other stuff with int64 and uint64 and there's a lot of struct that are like this
}
x := Foo{A: 1234567890987654, B: 987654321012345678}
byt, err := json.Marshal(x)
fmt.Println(err)
fmt.Println(string(byt))
// 9223372036854775808 9223372036854775808
err = json.Unmarshal([]byte(`{"A":"12345678901234567", "B":"98765432101234567"}`), &x)
// ^ must be quoted since javascript can't represent those values properly (2^53)
// ^ json: cannot unmarshal string into Go struct field Foo.A of type int64
fmt.Println(err)
fmt.Printf("%#v\n", x) …Run Code Online (Sandbox Code Playgroud) go ×4
javascript ×3
svelte ×2
svelte-3 ×2
database ×1
firefox ×1
go-gorm ×1
jquery ×1
json ×1
nosql ×1
php ×1
postgresql ×1
sql ×1
sql-server ×1
yii ×1