问题描述
我使用TreeView控件。与复选框绑定从数据库值。
I'm using TreeView control. Binding values from database with check box.
近三个孩子都在那里父节点像
Nearly three children are there for parent node like
Parent1
child1
child2
child3
Parent2
child1
child2
child3
我可以能够将数据绑定这是从3个表都运作良好。我想设施时父节点检查其所有相应的子自动获得checked.If我点击父1,儿童123得到遏制。如果我检查child1,和的child2得到child3检查。如果我检查孩子2所有子3项得到遏制。那怎么办?
I can able to bind data which is from 3 tables all are working well. I want a facility when a parent node checks all its corresponding child automatically get checked.If I click on parent 1 ,child 123 get checked. If I check child1, child2 and child3 get checked. If I check child 2 all child 3 items get checked. How to do that?
在此先感谢
Amrutha
Thanks in AdvanceAmrutha
推荐答案
JavaScript函数下面可以用来检查所有子节点如果家长检查,并检查父节点,如果至少有一个子节点进行检查,否则被选中
The javascript function below can be used to check all child nodes if the parent is checked and check parent node if at least one child node is checked otherwise it is unchecked:
function OnTreeClick(evt) {
var src = window.event != window.undefined ? window.event.srcElement : evt.target;
var isChkBoxClick = (src.tagName.toLowerCase() == "input" && src.type == "checkbox");
var t = GetParentByTagName("table", src);
if (isChkBoxClick) {
var parentTable = GetParentByTagName("table", src);
var nxtSibling = parentTable.nextSibling;
if (nxtSibling && nxtSibling.nodeType == 1) {
if (nxtSibling.tagName.toLowerCase() == "div") {
CheckUncheckChildren(parentTable.nextSibling, src.checked);
}
}
CheckUncheckParents(src, src.checked);
}
}
function CheckUncheckChildren(childContainer, check) {
var childChkBoxes = childContainer.getElementsByTagName("input");
var childChkBoxCount = childChkBoxes.length;
for (var i = 0; i < childChkBoxCount; i++) {
childChkBoxes[i].checked = check;
}
}
function CheckUncheckParents(srcChild, check) {
var parentDiv = GetParentByTagName("div", srcChild);
var parentNodeTable = parentDiv.previousSibling;
if (parentNodeTable) {
var checkUncheckSwitch;
var isAllSiblingsChecked = AreAllSiblingsChecked(srcChild);
if (isAllSiblingsChecked) {
checkUncheckSwitch = true;
}
else {
checkUncheckSwitch = false;
}
var inpElemsInParentTable = parentNodeTable.getElementsByTagName("input");
if (inpElemsInParentTable.length > 0) {
var parentNodeChkBox = inpElemsInParentTable[0];
parentNodeChkBox.checked = checkUncheckSwitch;
CheckUncheckParents(parentNodeChkBox, checkUncheckSwitch);
}
}
}
function AreAllSiblingsChecked(chkBox) {
var parentDiv = GetParentByTagName("div", chkBox);
var childCount = parentDiv.childNodes.length;
var k = 0;
for (var i = 0; i < childCount; i++) {
if (parentDiv.childNodes[i].nodeType == 1) {
if (parentDiv.childNodes[i].tagName.toLowerCase() == "table") {
var prevChkBox = parentDiv.childNodes[i].getElementsByTagName("input")[0];
//if any of sibling nodes are not checked, return false
if (prevChkBox.checked) {
//add each selected node one value
k = k + 1;
}
}
}
}
//Finally check any one of child node is select if selected yes then return ture parent node check
if (k > 0) {
return true;
}
else {
return false;
}
}
function GetParentByTagName(parentTagName, childElementObj) {
var parent = childElementObj.parentNode;
while (parent.tagName.toLowerCase() != parentTagName.toLowerCase()) {
parent = parent.parentNode;
}
return parent;
}
我用这个JavaScript来解决我的问题。
I used this javascript to solve my problem.
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