[英]How to remove all the odd indexes (eg: a[1],a[3]..) value from the array
[英]how to read every odd value from below array list:
我的下拉值顯示為:
The
Are
company
we
belongsto
yes
America
can
我運行這段代碼,我在控制台中得到了上面的結果:
let arr = [];
let appswitcherlist = await profileidentity.profile_profile_Appswitcherdropdownlist.each((ele, index) => {
if (ele != undefined) {
ele.getText().then((text) => {
arr.push(text);
});
}
});
console.log(arr); // check what you get here.
const oddItems = arr.filter((item, index) => index%2 == 0)
console.log(oddItems); //check what you get here. You should bind this to drop-down.
我想閱讀每一個奇數行如下:
The
company
belongsto
America
您可以執行以下操作。 使用數組上的過濾器方法的地方。
let arr = ['The', 'Are', 'company', 'we', 'belongsto', 'yes', 'America', 'can'];
const oddItems = arr.filter((item, index) => index%2 == 0)
console.log(oddItems);
只需將(index % 2 !== 0)
添加到您現有的條件中,如下所示。
var appswitcherlist = await profileidentity.profile_profile_Appswitcherdropdownlist.each((ele, index) => {
if (ele != undefined && (index % 2 !== 0)) {
ele.getText().then((text) => {
console.log(text);
});
}
});
let listData = [
'The',
'Are',
'company',
'we',
'belongsto',
'yes',
'America',
'can',
]
let i = 0;
for(let row of listData){
i++;
if(i%2){
console.log(row);
}
}
const arr = ['The', 'Are', 'company', 'we', 'belongsto', 'yes', 'America', 'can'];
let oddItems = [];
for(a in arr){
a%2 === 0 ? oddItems.push(arr[a]) : null ;
}
console.log(oddItems);
Protractor 引入await之后真的很簡單,不需要再復雜了
首先在配置文件中禁用 selenium 控制流,只需在配置文件中添加這一行:
SELENIUM_PROMISE_MANAGER: false
https://www.protractortest.org/#/control-flow
現在在測試使用:
it('Some test name', async function() {
let arr=[]
let appswitcherlist=await profileidentity.profile_profile_Appswitcherdropdownlist.getText()
const oddItems=appswitcherlist.filter((item, index)= > index % 2 == 0)
console.log(oddItems)
})
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