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EasyUI for jQuery
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datagrid form('load') bug
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Topic: datagrid form('load') bug (Read 5984 times)
小军
Newbie
Posts: 2
datagrid form('load') bug
«
on:
August 07, 2015, 02:25:08 AM »
<form id="
searchForm
">
<label>教学楼:</label><input class="combogridJxllsh" name="
jxllsh
">
</form>
<a href="#" id="
clearBtn
" class="easyui-linkbutton">test</a>
that.$clearBtn.linkbutton({
onClick:function(){
that.$searchForm.form('load',{'jxllsh':null});//此句会报错
that.$searchForm.form('load',{'jxllsh':''});//此句不会错,但是从数据库查出来的数据难免有值为null的
}
});
源码修改如下:
setValues:function(jq,_a22){
return jq.each(function(){
var opts=$(this).combogrid("options");
if($.isArray(_a22)){
_a22=$.map(_a22,function(_a23){
if(typeof _a23=="object"){
//我是这样解决的:如下注释内容是我自己写的,如果不加这段会报错。但是我不想改源码,这个问题是不是easyui的bug,还有什么解决方式,我使用的是1.4.3
// if(_a23==null){
// return;
// }
var v=_a23[opts.idField];
(function(){
for(var i=0;i<opts.mappingRows.length;i++){
if(v==opts.mappingRows
[opts.idField]){
return;
}
}
opts.mappingRows.push(_a23);
})();
return v;
}else{
return _a23;
}
});
}
_a00(this,_a22);
});
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