列A列出了各个产品的类别。每种产品将有1到14个类别。目标是将所有类别划分为单独的单元格(从B列到O列),以便更容易排序。
我已经为列B到O创建了公式,以搜索连字符"-“,并将每个类别分隔到其自己的列中。下面是输出应该是什么样子(除了最下面的一行):
A B C D E F G ... O
Categories Cat 1 Cat 2 Cat 3 Cat 4 Cat 5 Cat 6 ... Cat 14
我在routes.rb里有这个
get 'search#search', to: 'search#search', as: 'search'
如果我是rails routes,它将返回以下内容:
search GET /search%23search(.:format) search#search
我想把/search%23search改为search/search
我该怎么做?
我目前正在使用bash创建一个数据库。我想添加一个"and“函数。此函数将搜索两个内容,并且只返回匹配这两个搜索的内容。我该如何做到这一点?
下面是我的代码:
#!/bin/bash
clear
#return all
function all {
cat people.dat
}
#return some
function search {
grep -i "$SEARCH" people.dat || echo "search returned nothing"
}
#or search
function or {
egr
如果我正在制作一个程序,让你翻译单词,有没有办法不使用elif每一次,只写单词来翻译。这就是我现在得到的!
print("English to Exrian Dictionary")
search = input("Enter the word you would like to translate: ").lower()
if search == "ant":
print("Ulf")
elif search == "back":
print("Zuwp")
elif sear
这是我的密码:
var sQuestionPropositionText = new String("This is a test|A");
for(var i=0; i < sQuestionPropositionText.length; i++){
console.log(sQuestionPropositionText[i] + " search result= " + sQuestionPropositionText[i].search("|"));
}
这就是我在控制台里得到的:
T search result= 0
h sear
我有以下搜索方法:
def search(search = nil, field_search1,field_search2, field_search3,field_search4,field_search5)
if search
where("#{field_search1} LIKE ? OR #{field_search2} LIKE ? OR #{field_search3} LIKE ? OR #{field_search4} LIKE ? OR #{field_search5}LIKE ?", "%#{search}%"
# Do not remove this line or mod_rewrite rules and search engine friendly URLs will stop working
Options +FollowSymLinks
RewriteEngine On
# Removes index.php from ExpressionEngine URLs
RewriteCond %{THE_REQUEST} ^GET.*index\.php [NC]
RewriteCond %{REQUEST_URI} !/system/.* [NC]
RewriteRule (.*?)ind