Question

我有一个具有6个值的数组，然后又有三个数组，其值在4到1之间。我想遍历数组1，并检查数组1的值是否出现在数组2，数组3和数组4中。目前，我具有以下内容，但它似乎只是针对数组2，3和3中的第一个值检查数组1的值4.我省略了数组3和4，但是它们的for循环与array2相同，并且位于array1的循环内。

array1=("value1" "value2" "value3" "value4" "value5" "value6")

for i in "${array1}"; do

array2= ("value1" "value3" "value4" "value5")

for f in "${array2}; do
if [[ ${i} == ${f} ]]; then

echo "${i} in array1 matches ${f} in array2"
else
echo "${i} in array1 does not match any value in array2"
fi
done
done

Answer 1

我认为最好的办法就是创建一个函数

in_array () {
  search=$1
  shift        # remove first argument from list of args $@
  for val; do  # equivalent to `for val in "$@"`
    if [[ $search = $val ]]; then
      return   # returns exit code of the successful [[ test ]], 0
    fi
  done
  return 1
}

如果找到该值，则返回0，否则返回1，使您可以像这样使用它：

array1=("value1" "value2" "value3" "value4" "value5" "value6")
array2=("value1" "value3" "value4" "value5")

for i in "${array1[@]}"; do
  if in_array "$i" "${array2[@]}"; then
    echo "$i in array1 is in array2"
  fi
done

请注意，要遍历数组的所有值，正确的扩展名为"${array[@]}"（带双引号和[@]）。

Answer 2

这可以在一个循环中完成：

#!/usr/bin/env bash

array1=("value1" "value2" "value3" "value4" "value5" "value6")
array2=("value1" "value3" "value4" "value5")

while read -r -d '' -n 8 count && IFS= read -r value; do
  if [ "$count" -gt 1 ]; then
    # value is seen more than once, so it is in both arrays
    echo "${value} in array1 matches ${value} in array2"
  else
    # value is only seen once
    if printf $'%s\n' "${array1[@]}" | grep --quiet "$value"; then
      # value is from array1
      echo "${value} in array1 does not match any value in array2"
    fi
  fi
done< <(
  # Combine both arrays
  # Sort and count number of times each value appears
  # then feed the while loop
  printf $'%s\n' "${array1[@]}" "${array2[@]}" | sort | uniq --count
)

Bash-如何检查变量值是否在数组中

2 个答案: