[英]Twitter Typeahead.js how to return all matched elements within a string
By default twitter typeahead.js returns only elements matched in the begining of a string, for example: 缺省情况下,twitter typeahead.js仅返回在字符串开头匹配的元素,例如:
source: ['type','typeahead','ahead'] 来源:['type','typeahead','ahead']
query: 'type' 查询:“类型”
returns: 'type' and 'typeahead' 返回:'type'和'typeahead'
-- -
query: 'ahead' 查询:“提前”
returns: 'ahead' 返回:'ahead'
I want it to return 'ahead' and 'typeahead' 我希望它返回“ ahead”和“ typeahead”
my code: 我的代码:
var clients = new Bloodhound({
datumTokenizer: function(d) { return Bloodhound.tokenizers.whitespace(d.value); },
queryTokenizer: Bloodhound.tokenizers.whitespace,
limit: 10,
prefetch: {
url: '/clients.json',
filter: function(list) {
return $.map(list, function(value) { return { name: value }; });
}
}
});
clients.initialize();
$('.client').typeahead(null, {
displayKey: 'value',
source: clients.ttAdapter(),
minLength: 1,
});
There is already a question about it but i didnt understand the answer. 已经有一个问题,但我不明白答案。
I found a solution... the problem was that I was so used to bootstrap2 typeahead that I wasn't understanding the datumTokenizer
thing. 我找到了解决方案...问题是我习惯于提前启动bootstrap2,以至于我不了解
datumTokenizer
的知识。 If someone else find it hard to understand, I will put a little description below: 如果其他人难以理解,我将在下面进行一些说明:
queryTokenizer: array of words you are querying, if you query for 'test abcd' it will transform the string into ['test','abcd'] and than look for matches with those two words. queryTokenizer:要查询的单词数组,如果查询“ test abcd”,它将把字符串转换为['test','abcd'],然后查找与这两个单词匹配的单词。
datumTokenizer : array of words it will be matched with queryTokenizer. datumTokenizer:将与queryTokenizer匹配的单词数组。 Each item from your JSON will have a set of words to be matched.
JSON中的每个项目都会有一组要匹配的单词。
So if you have a source: 因此,如果您有资料来源:
['good test','bad test'] [“好测试”,“坏测试”]
and query for 'est'. 并查询“ est”。 You need to make datumTokenizer return an array containing 'est' , something like:
您需要使datumTokenizer返回包含'est'的数组,类似于:
['good','test','ood','od','test', 'est', 'st'] for the first item 第一项的['good','test','ood','od','test','est','st']
['bad','ad','test', 'est', 'st'] for the second item 第二项的['bad','ad','test','est','st']
Bellow is the code I wrote, I don't know if its the optimal thing for it, but I think it will help anyway: 贝娄是我编写的代码,我不知道它是否是最合适的,但是我认为它还是有帮助的:
new Bloodhound({
datumTokenizer: function(d) {
var test = Bloodhound.tokenizers.whitespace(d.value);
$.each(test,function(k,v){
i = 0;
while( (i+1) < v.length ){
test.push(v.substr(i,v.length));
i++;
}
})
return test;
},
queryTokenizer: Bloodhound.tokenizers.whitespace,
limit: 10,
prefetch: {
url: '/lista.json',
ttl: 10000
}
});
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