我需要在php中结合使用两个数组。
我所拥有的:
numbers = array(1, 2,3,4,5,6);
letters = array('q', 'w', 'e', 'r');
我需要什么:
1 q
2 w
3 e
4 r
5
6
重要提示:我永远不会知道“数字”或“字母”数组包含多少个项目(只是长度永远不相等),这会因情况而异。
到目前为止,使我最接近目标,但这将所有字母都提交给每个数字:
function array_combine2($arr1, $arr2) {
return array_combine(array_intersect_key($arr1, $arr2),
array_intersect_key($arr2, $arr1));
}
foreach(array_combine2($number,$array) as $n1 => $a1)
{
echo $n1 . $a1."<br>";
}
答案 0 :(得分:4)
您获得了max
并使用经典的for
循环。检查元素isset
是否使用空字符串。
$numbers = [1, 2,3,4,5,6];
$letters = ['q', 'w', 'e', 'r'];
$result = [];
for ($i = 0; $i < max(count($numbers), count($letters)); $i++) {
$result[] = [
"number" => isset( $numbers[$i] ) ? $numbers[$i] : "",
"letter" => isset( $letters[$i] ) ? $letters[$i] : "",
];
}
echo json_encode( $result );
这将导致:
[
{"number": 1,"letter": "q"},
{"number": 2,"letter": "w"},
{"number": 3,"letter": "e"},
{"number": 4,"letter": "r"},
{"number": 5,"letter": ""},
{"number": 6,"letter": ""}
]
更新:如果只想echo
个值,则可以:
$numbers = [1, 2,3,4,5,6];
$letters = ['q', 'w', 'e', 'r'];
for ($i = 0; $i < max(count($numbers), count($letters)); $i++) {
echo isset( $numbers[$i] ) ? $numbers[$i] : "";
echo " ";
echo isset( $letters[$i] ) ? $letters[$i] : "";
echo "<br />";
}
这将导致:
1 q
2 w
3 e
4 r
5
6
答案 1 :(得分:2)
$numbers = [1, 2,3,4,5,6];
$letters = ['q', 'w', 'e', 'r'];
$count = 0;
$combination = array();
foreach ($numbers as $num)
{
$val = '';
if (isset ($letters[$count]))
{
$val = $letters[$count];
}
$combination += array($num => $val);
$count ++;
}
foreach($combination as $n => $l)
{
echo $n . $l . "<br>";
}
答案 2 :(得分:2)
最短的解决方案
<?php
$numbers = array(1,2,3,4,5,6);
$letters = array('q','w','e','r');
foreach($numbers as $key => $num) {
echo $num .' ';
if (array_key_exists($key ,$letters)) // check if array element exists
echo $letters[$key].'<br>';
else echo "<br>";
}
说明。
首先,我们搜索数字数组或更长的数组,然后回显它的元素,同时检查字母数组中是否存在索引,如果存在,则使用数字回显该元素。