replaceChild()
XML DOM replaceChild() metodi
❮ Element obyekti
Misol
Quyidagi kod fragmenti "books.xml" faylini xmlDoc’ga yuklaydi va birinchi <book> elementini almashtiradi:
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
myFunction(this);
}
};
xhttp.open("GET", "books.xml", true);
xhttp.send();
function myFunction(xml) {
var x, y, z, i, newNode, newTitle, newText, xmlDoc, txt;
xmlDoc = xml.responseXML;
txt = "";
x = xmlDoc.documentElement;
// Create a book element, title element and a text node
newNode = xmlDoc.createElement("book");
newTitle = xmlDoc.createElement("title");
newText = xmlDoc.createTextNode("A Notebook");
// Add a text node to the title node
newTitle.appendChild(newText);
// Add the title node to the book node
newNode.appendChild(newTitle);
y = xmlDoc.getElementsByTagName("book")[0];
// Replace the first book node with the new book node
x.replaceChild(newNode, y);
z = xmlDoc.getElementsByTagName("title");
// Output all titles
for (i = 0; i < z.length; i++) {
txt += z[i].childNodes[0].nodeValue + "<br>";
}
document.getElementById("demo").innerHTML = txt;
}
Yuqoridagi kodning natijasi quyidagicha bo‘ladi:
A Notebook
Harry Potter
XQuery Kick Start
Learning XML
O‘zingiz sinab ko‘ring »
Ta’rif va qo‘llanilishi
replaceChild() metodi bola tugunni boshqasi bilan almashtiradi.
Bu funksiya muvaffaqiyatli bo‘lsa almashtirilgan tugunni, muvaffaqiyatsiz bo‘lsa NULL qaytaradi.
Sintaksis
elementNode.replaceChild(new_node,old_node)| Parametr | Tavsif |
|---|---|
| new_node | Majburiy. Yangi tugunni belgilaydi |
| old_node | Majburiy. Almashtiriladigan bola tugunni belgilaydi |
❮ Element obyekti
W3Schools Pathfinder
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