Question

我有三个数组作为array1 array2 array3，我希望这个组合像array4一样。

array1
(
[1869] => 1
[1871] => 1
[0] => 2
[1807] => 1
[1875] => 1
[1811] => 1
[1877] => 1
[1878] => 1
[1879] => 1
[1880] => 1
[1886] => 1
[1850] => 2
[1618] => 3
[1679] => 1
)
array2
(
[1619] => 1
[1625] => 1
)

array3
(
[1111] => 1
[2222] => 1
)

需要输出如：

array4
(
 [1869] => 1
 [1871] => 1
 [0] => 2
 [1807] => 1
 [1875] => 1
 [1811] => 1
 [1877] => 1
 [1878] => 1
 [1879] => 1
 [1880] => 1
 [1886] => 1
 [1850] => 2
 [1618] => 3
 [1679] => 1
 [1619] => 1
 [1625] => 1
 [1111] => 1
 [2222] => 1
 )

我希望这个array1和array2像array3一样组合。请帮助我一些。谢谢。

Answer 1

如果每个数组都在一个变量上，你可以这样做，

String winHandleBefore = driver.getWindowHandle();

    //All tab's id (security + toolsQA) and storing it.
    Set<String> all = driver.getWindowHandles();

    //Get all the tab id's including security and iterate
    for(String winHandle : driver.getWindowHandles())
    {
       if(winHandle.equals(winHandleBefore))
       {
         //Condition satisfied. Switching to the security tab and closing it
         driver.switchTo().window(winHandle);
         driver.close();

       }
    }
    //Now the security tab will be closed we need to switch to the toolsQA whose id is in "all"
    //Removing the security tab id from all.
    all.remove(winHandleBefore);
    //Now we have only one toolsqa id in "all"
    for(String winHandle : all)
    {

         driver.switchTo().window(winHandle);
         //Now in toolsQA

这样你也不会松开钥匙。查看手册here。

Answer 2

或者

$array1 = array('1869' => 1,'1871' => 1,'0' => 2,'1807' => 1,'1875' => 1,'1811' => 1,'1877' => 1,'1878' => 1,'1879' => 1,'1880' => 1,'1886' => 1,'1850' => 2,'1618' => 3,'1679' => 1);
$array2 = array('1619' => 1,'1625' => 1);
$array3 = array('1111' => 1,'2222' => 1);

$newArray = array_replace( $array1,$array2,$array3);

但我认为只使用+是首选。

print_r($newArray);

数组与所有相结合，因为它就像相同的键和值

2 个答案: