Nur*_*gin 0 load extjs store filter
我想隐藏从服务器返回的网格上的记录。
我已经设置了一个filter
on store 并且可以访问该特定数据,但是我将如何处理隐藏/忽略此记录?
fooStore: {
....
filters: [
function(item) {
let me = this;
let theRecord = item.data.status === MyApp.STATUS; //true
if (theRecord) {
console.log(theRecord); //True
console.log("So filter is here!!")
//How to hide/ignore/avoid to load this specific data record to load Grid??
}
}
]
},
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返回 JSON;
{
"success": true,
"msg": "OK",
"count": 3,
"data": [
{
//Filter achives to this record and aim to hide this one; avoid to load this record.
"id": 102913410,
"status": "P"
},
{
"id": 98713410,
"status": "I"
},
{
"id": 563423410,
"status": "A"
}
]
}
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我无法保存我的小提琴,因为我没有 sencha 论坛的帐户,所以我给你我的代码:
Ext.application({
name : 'Fiddle',
launch : function() {
var model = Ext.create('Ext.data.Model', {
extend: 'Ext.data.Model',
fields: [
{name: 'id', type: 'int'},
{name: 'status', type: 'string'},
]
});
var store = Ext.create('Ext.data.Store', {
autoLoad: true,
model: model,
proxy: {
type: 'ajax',
url: 'data.json',
reader: {
type: 'json',
rootProperty: 'data'
}
},
filters: [function(item) {
if (item.data.status === "P") {
return true;
}
else {
return false;
}
}],
listeners: {
load: {
fn: function() {
console.log(this.getRange());
}
}
}
});
}
});
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我也像这样创建 data.json :
{
"success": true,
"msg": "OK",
"count": 3,
"data": [{
"id": 102913410,
"status": "P"
}, {
"id": 98713410,
"status": "I"
}, {
"id": 563423410,
"status": "A"
}]
}
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我认为它靠近您的代码,并且在加载商店后,过滤器可以正常工作:
这是sencha小提琴链接:https : //fiddle.sencha.com/#view/editor
如果这不起作用,我不明白他妈的在做什么......