Question

我有两个数组

第一个数组

array(
    [0] => +970
    [1] => +971
    [2] => +972
)

和第二个数组

array(
    [0] => 465465454
    [1] => 321321355
    [2] => 987946546
)

我想像这样合并它们

array(
    [+970] => 465465454
    [+971] => 321321355
    [+972] => 987946546
)

我尝试array_merge，但这给了我一些我不想要的结果。

$busi_code  = $page1_data->business_code; //array
$busi_num  = $page1_data->business_number; //array

$business_phone_numbers = array_merge($busi_code, $busi_num);

echo '<pre>';
print_r($business_phone_numbers);
echo '</pre>';

结果是

[0] => +970
[1] => +971
[2] => +972
[3] => 465465454
[4] => 321321355
[5] => 987946546

所以请指导我如何达到我要求的结果。

Answer 1

您正在寻找array_combine，而不是array_merge：

使用keys数组中的值作为键，将values数组中的值作为相应的值，创建一个数组。

$business_phone_numbers = array_combine($busi_code, $busi_num);

请参阅https://eval.in/954799

Answer 2

这是$business_phone_numbers = array_combine($busi_code, $busi_num);功能的工作：

function groupByInterval(data, secondsPerGroup) {
    // First sort by the time stamps in ascending order
    data = data.sort( ([a], [b]) => a - b )
        // Then add the starting time stamps of the groups
        .map(([unix, price]) => ({ unixstart: unix - unix % secondsPerGroup, price }) );
    // Create an entry per group; initialise values as empty arrays (for prices)    
    const map = new Map(data.map( ({unixstart}) => [unixstart, []] ));
    return Array.from(
        // Collect the prices per group
        data.reduce( (acc, {unixstart, price}) => 
            acc.set(unixstart, acc.get(unixstart).concat(price)),
            map
        ).entries(),
        // Enrich the group values with formatted dates and price measures
        ([unixstart, prices]) => ({
            unixstart,
            start: new Date(unixstart * 1000).toJSON(), 
            end:   new Date((unixstart + secondsPerGroup) * 1000).toJSON(),
            minPrice: Math.min(...prices),
            maxPrice: Math.max(...prices),
            firstPrice: prices[0],
            lastPrice: prices[prices.length-1],
            countPrices: prices.length,
            sumPrices: prices.reduce( (a,b) => a+b, 0 )
        })
    );
}

// Sample input
const data = [[1518371928, 15440000],[1518371928, 15440000],[1518371928, 15440000],[1518371928, 15440000],[1518371928, 15440000],[1518294600, 15440000],[1518294600, 15440000],[1518294600, 15440000],[1518294600, 15440000],[1518294600, 15440000],[1518294600, 15440000],[1518294600, 15440000],[1518294600, 15440000],[1518294600, 15440000],[1518294600, 15440000],[1518294600, 15440000],[1518294600, 15440000],[1518294600, 15440000],[1518294600, 15440000],[1518294600, 15440000],[1518294600, 15440000],[1518294600, 15440000],[1518294600, 15440000],[1518294600, 15440000],[1518417005, 15430000],[1518417005, 15430000],[1518420449, 15465000],[1518422229, 15510000],[1518423849, 15535000],[1518425806, 15598000],[1518427146, 15615000],[1518428229, 15625000],[1518430648, 15635000]];

const result = groupByInterval(data, 60*60); // Group per hour (expressed in seconds)

console.log(result);

DOCS： http://php.net/manual/en/function.array-combine.php

Answer 3

您必须使用array_combine。

试试这个：

$a = array(
0 => +970,
1 => +971,
2 => +972);

$b = array(
0 => 465465454,
1 => 321321355,
2 => 987946546);

$r = array_combine($a,$b);
print_r($r);

多个阵列合并问题

3 个答案: