我在使用AJAX源时使bootstraps typeahead正常工作时遇到了很多麻烦。
如果我提醒正在返回的数组,那就完全没问题了,即使我对它进行硬编码测试也是如此。当您输入近似匹配时,似乎大部分时间都没有返回任何内容,或者只返回数组中的一些项目。
这就是我现在所拥有的:
$('#companyNameInput').typeahead({
source: function(query, process){
$.ajax({
url: ROOT+'Record/checkCompanyName',
async: false,
data: 'q='+query,
type: 'POST',
cache: false,
success: function(data)
{
companiesFinal = [];
map = {};
companies = $.parseJSON(data);
$.each(companies, function(i, v){
map[v.name] = v.id;
companiesFinal.push(v.name);
})
}
})
process(companiesFinal);
// return ['test1', 'test2'] This works fine
return companiesFinal;
}
有谁知道为什么这个工作正常?
以下是从我的PHP脚本返回的对象数组的示例。 ID为1和1216的对象显示在typeahead下拉列表中,但不显示其他对象。我看不出任何模式或线索,为什么只有这些才会显示而不是其他。
[
{
"id": "1265",
"score": "40",
"name": "LMV AB"
},
{
"id": "10834",
"score": "33",
"name": "Letona"
},
{
"id": "19401",
"score": "33",
"name": "Lewmar"
},
{
"id": "7158",
"score": "33",
"name": "Lazersan"
},
{
"id": "3364",
"score": "33",
"name": "Linpac"
},
{
"id": "1216",
"score": "33",
"name": "L H Evans Limted"
},
{
"id": "1",
"score": "33",
"name": "LH Evans Ltd"
},
{
"id": "7157",
"score": "33",
"name": "Lazersan"
}
]
最后是process(companiesFinal)
中过去的数组:
["LMV AB", "Letona", "Lewmar", "Lazersan", "Linpac", "L H Evans Limted", "LH Evans Ltd", "Lazersan"]
任何人都有任何线索?我仍然完全不知道为什么这仍然不起作用:(
$('#companyNameInput').typeahead({
source: function(query, process){
companyTOut = setTimeout(function(){
return $.ajax({
url: ROOT+'Record/checkCompanyName',
data: 'q='+query,
type: 'POST',
dataType: 'json',
cache: false,
success: function(data)
{
var count = 0;
var companiesFinal = [];
map = [];
$.each(data, function(i, v){
map[v.name] = [v.id, v.score];
companiesFinal.push(v.name);
})
process(companiesFinal);
}
})
}, 250)
},
minLength: 2,
highlighter: function(item)
{
$('#companyNameInput').closest('.control-group').removeClass('success')
companyLocked = false;
return '<span class="unselectable" title="'+map[item].score+'">'+item+'</span>';
},
updater: function(item)
{
selectedEntityId = map[item][0];
selectedCountryScore = map[item][1];
lockCompany(selectedEntityId);
return item;
}
});
$output .= '['.$output;
foreach($results as $result) {
$output .= '{"id":"'.$result['id'].'",';
$output .= '"score":"'.$result['score'].'",';
$output .= '"name":'.json_encode($result['name']).'},';
}
header('Content-Type: application/json');
echo substr($output, 0, strlen($output)-1).']';
“parm”的控制台输出:
[Object, Object, Object, Object, Object, Object]
0: Object
id: "25024"
name: "part"
score: "75"
__proto__: Object
1: Object
id: "15693"
name: "pari"
score: "75"
__proto__: Object
2: Object
id: "28079"
name: "Pato"
score: "50"
__proto__: Object
3: Object
id: "18001"
name: "PASS"
score: "50"
__proto__: Object
4: Object
id: "15095"
name: "PSR"
score: "33"
__proto__: Object
5: Object
id: "22662"
name: "PRP"
score: "33"
__proto__: Object
length: 6
__proto__: Array[0]
答案 0 :(得分:2)
啊,您的服务会返回实际上与查询parm
不匹配的项目。您的预先查询是'parm'
,其中没有任何返回的结果匹配。您可以覆盖typeahead插件使用的matcher
函数,请参阅docs。只需将其实现为return true
即可匹配您服务返回的所有结果。
这是一个更新版本,它将名称映射到id,稍后可以使用。一个jsfiddle is available。
var nameIdMap = {};
$('#lookup').typeahead({
source: function (query, process) {
return $.ajax({
dataType: "json",
url: lookupUrl,
data: getAjaxRequestData(),
type: 'POST',
success: function (json) {
process(getOptionsFromJson(json));
}
});
},
minLength: 1,
updater: function (item) {
console.log('selected id'+nameIdMap[item]);
return item;
}
});
function getOptionsFromJson(json) {
$.each(json, function (i, v) {
nameIdMap[v.name] = v.id;
});
return $.map(json, function (n, i) {
return n.name;
});
}
你需要调用异步并从成功回调中调用process
回调,如下所示:
$('#companyNameInput').typeahead({
source: function (query, process) {
$.ajax({
url: ROOT + 'Record/checkCompanyName',
// async: false, // better go async
data: 'q=' + query,
type: 'POST',
cache: false,
success: function (data) {
var companiesFinal = ... // snip
process(companiesFinal);
}
})
}
});
return ['test1', 'test2'];
有效,因为源函数基本上设置为:
// do ajax stuff, but do nothing with the result
// return the typeahead array, which the typeahead will accept as the result:
return ['test1', 'test2'];
备注的
有一个单行填充companiesData
:
var companiesFinal = return $.map(data, function (n, i) { n.name; });
您可能希望使用var
声明变量;否则他们将拥有全球范围,这会咬你。