1 次元配列と 2 次元配列の走査について質問する
Array
(
[8900] => A
[8898] ] => D
[8733] => B
[8799] => C
[8933] => A
[8827] = > C
[8854] => C
[8786] => A
[8738] => B
[8880] => B
[8835] => A
[8750] => D
[8760] => D
[8853] => A
[8863] => A
[8806] => D
[8766] => A
[8864] => D
[8896] => A
[8874] => A
[8828] => C
[8732] => D
[ 8925] => B
[8779] => B
[8895] => B
[8843] => B
[8873] ] => B
[8744] => C
[8731] => D
[8856] => B
[8969] = > ARAAY
(
[0] = & GT; A
## [1] = & GT; B# [3] = & GT # )
## [9051] => 配列 ( ) [1] => B [2] = > C )## [9059] => 配列
(
) [0] => A
[ 2] = & gt; c
[3] = & gt; d
##)[9032] = & gt;
([0] = & GT; A
## [1] = & GT; B
# [2] = & GT; C## [3] = > D
)
# [8953] => 配列 ( ) [0] => A[1 ] = & gt; b [2] = & gt; c
## [3] = & GT; d
#) ## [8960] => 配列(
[0] => A
[1] => B
[3] => D
)
[9110] =>配列
(
[0] => A
[1] => B
[2] => C
[3] => D
)
[9084] =>配列
(
[0] => A
[1] => B
[2] => C
)
[9044] =>配列
(
[0] => A
[1] => B
[2] => C
[3] => D
)
[8961] =>配列
(
[0] => A
[1] => B
[3] => D
)
[9023] =>配列
(
[0] => A
[1] => B
[2] => C
[3] => D
)
[9098] =>配列
(
[0] => A
[1] => B
[2] => C
[3] => D
)
[8976] =>配列
(
[0] => A
[1] => B
[2] => C
[3] => D
)
[9102] =>配列
(
[1] => B
[2] => C
)
[8972] =>配列
(
[0] => A
[1] => B
[2] => C
)
[8968] =>配列
(
[0] => A
[2] => C
)
[8997] =>配列 ( [0] => A [3] => D )[9080] =>配列
(
[0] => A
[2] => C
)
[9029] =>配列 ( [0] => A [1] => B )[9091] =>配列
(
[0] => A
[1] => B
[2] => C
)
[9240] => A
[9182] => A
[9227] => A
[9146] => B
[9190] => A
[9139] => B
[9138] => B
[9143] => A
[9279] => A
[9232] => A
)
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最初の部分の 1 次元配列、2 番目の部分の 2 次元配列、3 番目の部分の 1 次元配列を個別に走査するのが最善です。素晴らしい！