我是PHP新手，我想在多维数组中找到0并替换为内部数组中缺少的数字。如果内部数组有两个以上的0，它将被忽略并转到下一个。
$list = array("First"=>array(0,1,2,3,0,5,6,7,8,9),
"Second"=>array(0,1,2,3,4,5,6,7,8,9),
"Third"=>array(0,1,2,3,4,5,0,0,8,9),
"Fourth"=>array(0,1,2,3,4,5,6,7,8,0),
"Fifth"=>array(0,1,2,3,4,5,0,7,8,9),
"Sixth"=>array(0,0,0,3,4,5,6,0,0,0),
"Seventh"=>array(0,1,2,3,0,0,6,7,8,9),
"Eighth"=>array(0,1,2,3,4,5,0,7,8,9),
"Ninth"=>array(0,1,2,3,4,0,6,7,8,9),
"Tenth"=>array(0,0,2,3,4,5,6,7,8,9));
$countZero = 0;
foreach($list as $lvl) {
foreach($lvl as $ind => $val) {
if($countZero = array_count_values($lvl[$val] === 0))
$list[$ind][$val] = 45 - array_sum($ind);
echo $count;
}
}
我希望所有内部数组，有两个0，只有一个，所有数字都按顺序排列，即
"First"=>array(0,1,2,3,4,5,6,7,8,9);
请帮帮我。
我尝试了下面的代码，试图找到0。
$counts = 0;
$newArr = array();
foreach($list as $lvl) {
if(is_array($lvl)) {
for($i = 0; $i < count($lvl) - 1; $i++) {
if(($lvl[$i] == 0) < 2){
$counts++;
$newArr[$i] = 45 - array_sum($lvl);
}
}
}
}
print_r($newArr);
这是一个使用array_walk:的解决方案
array_walk($list,
function(&$numbers) {
$zeroIndex = 0;
foreach($numbers as $i => $number) {
if( $number === 0 ) {
if( $zeroIndex > 0 ) {
return;
}
$zeroIndex = $i;
}
}
$numbers[$zeroIndex] = $zeroIndex;
});
- 你不需要数所有的零。你只需要检查是否有少于3个零
- 我正在保存零（
$zeroIndex = $i）的索引（位置） - 我假设第一个数字总是零（
$zeroIndex = 0） - 第二个零的索引大于零。如果我在最后找到的零的索引大于零（
if( $zeroIndex > 0 )）时找到一个零，这意味着有两个以上的零
事实上，以下是我所做和工作的内容。
$list = array(array(1,2,3,0,5,6,7,8,9),
array(1,2,3,4,5,6,7,8,9),
array(1,2,3,4,5,0,0,8,9),
array(1,2,3,4,5,6,7,8,0),
array(1,2,3,4,5,0,7,8,9),
array(0,0,3,4,5,6,0,0,0),
array(1,2,3,0,0,6,7,8,9),
array(1,2,3,4,5,0,7,8,9),
array(1,2,3,4,0,6,7,8,9));
for($l = 0; $l < count($list); $l++)
{
$total = 0;
$countZ = 0;
for($i=0; $i < 9; $i++)
{
if($list[$l][$i] == 0)
{
$countZ++;
$indexZero = $i;
}
$total += $list[$l][$i];
if($countZ > 1) {
break;
}
}
$list[$l][$indexZero] = 45 - $total;
}
print_r($list);
TY所有人。