我尝试了下面的代码行,如果与数据的DepartmentName匹配的选项被选中。没用请帮忙 !!!
function fillDepartments(selectedText)
{
$.ajax({
url: 'index.php?action=fetchAllDepartments',
type: 'POST',
dataType: 'JSON',
data:{},
success: function(data){
for(var i in data){
$('#employee_department').append('<option value="'+data[i]['_id']['$oid']+ '">'
+ data[i]['DepartmentName'] + selectedText == data[i]['DepartmentName'] ? "selected" : "" +
'</option>');
}
}
});
}
最佳答案
应用简单的if-else
条件:-
for(var i in data){
if(selectedText == data[i]['DepartmentName']){
$('#employee_department').append('<option value="'+data[i]['_id']['$oid']+ '" selected>'+ data[i]['DepartmentName']+'</option>');
}else{
$('#employee_department').append('<option value="'+data[i]['_id']['$oid']+ '">'+ data[i]['DepartmentName'] +'</option>');
}
}
关于javascript - 使在ajax调用中选择的选项无效,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/49231957/