功能不正常 [英] function not work proper
本文介绍了功能不正常的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
<input id="CHAMPARAN_EAST" name="CHAMPARAN EAST" onclick="SearchDistict('CHAMPARAN EAST');" type="checkbox" value="34">
function SearchDistict(obj)
{
debugger;
if (obj.checked)
{
districtCollection.push(obj.value);
}
else {
districtCollection.pop(obj.value);
//districtCollection.splice($.inArray(obj.value, districtCollection),1);
}
var postData = preparingSearch();
debugger;
postData.StateList = '';
postData.DistrictList = districtCollection;
SearchEngine();
//CallSearchBorrower(postData);
}
当我们调用SearchDistict('CHAMPARAN EAST')它的工作但它不会调用if语句
when we call SearchDistict('CHAMPARAN EAST') its work but it not invoke if statement
推荐答案
.inArray(obj.value) ,districtCollection),1);
}
var postData = preparingSearch();
调试器;
postData.StateList = ' ';
postData.DistrictList = districtCollection;
SearchEngine();
// CallSearchBorrower(postData);
}
.inArray(obj.value, districtCollection),1); } var postData = preparingSearch(); debugger; postData.StateList = ''; postData.DistrictList = districtCollection; SearchEngine(); //CallSearchBorrower(postData); }
当我们调用SearchDistict('CHAMPARAN EAST')它的工作但它不调用if语句
when we call SearchDistict('CHAMPARAN EAST') its work but it not invoke if statement
(你将控制名称作为字符串并在函数内,用它作为对照。[string!= control])
这是修复:
1.重命名与id相同的复选框名称属性。
2.将控制名称作为参数传递。
(You are passing control name as string and within the function, using it as a control. [string != control])
Here is the fix:
1. Rename your checkbox name attribute same as id.
2. Pass control name as parameter.
<input id="CHAMPARAN_EAST" name="CHAMPARAN_EAST" onclick="SearchDistict('CHAMPARAN_EAST');" type="checkbox" value="34">
function SearchDistict(obj)
{
var ctrl =
(#+ obj);
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