我希望選擇下拉框同時運行兩個函數,以便在同一頁上更新兩個獨立的div。合併2個jquery函數來更新兩個單獨的DIV
這裏是我的功能;
function showFAQ(str) {
if (str=="") {
document.getElementById("faqHint").innerHTML="";
return;
}
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
} else { // code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function() {
if (xmlhttp.readyState==4 && xmlhttp.status==200) {
document.getElementById("faqHint").innerHTML=xmlhttp.responseText;
}
}
var category = "<?php echo $category; ?>";
var catid = "<?php echo $catid; ?>";
var subcat = "<?php echo $subcat; ?>";
var subcat2 = "<?php echo $subcat2; ?>";
var subcatid= "<?php echo $subcatid; ?>";
var fileclass= "<?php echo $fileclass; ?>";
xmlhttp.open("GET","FaqCheck.php?q="+str+ '&catid=' + catid + '&category=' + category + '&subcat=' + subcat + '&subcat2=' + subcat2 + '&subcatid=' + subcatid + '&fileclass=' + fileclass,true);
xmlhttp.send();
}
function showClass(str) {
if (str=="") {
document.getElementById("txtHint").innerHTML="";
return;
}
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
} else { // code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function() {
if (xmlhttp.readyState==4 && xmlhttp.status==200) {
document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
}
}
var category = "<?php echo $category; ?>";
var catid = "<?php echo $catid; ?>";
var subcat = "<?php echo $subcat; ?>";
var subcat2 = "<?php echo $subcat2; ?>";
var subcatid= "<?php echo $subcatid; ?>";
var fileclass= "<?php echo $fileclass; ?>";
xmlhttp.open("GET","FileCheck.php?q="+str+ '&catid=' + catid + '&category=' + category + '&subcat=' + subcat + '&subcat2=' + subcat2 + '&subcatid=' + subcatid + '&fileclass=' + fileclass,true);
xmlhttp.send();
}
此刻我不得不使用兩個下拉框來實現,但我希望它更整潔。選擇代碼是; 1.
<select name="fileclass" onchange="showClass(this.value)" required="">
<option value="">Select</option>
<option value="create">Create</option>
<option value="modify">Modify</option>
<option value="delete">Delete</option>
</select>
2.
<select name="faq_class" onchange="showFAQ(this.value)" required="">
<option value="">Select</option>
<option value="create">Create</option>
<option value="modify">Modify</option>
<option value="delete">Delete</option>
</select>
信息,faq_class和fileclass共享相同的值(永遠!)
謝謝你幫助一個新手:)
我不在你的代碼中看不到你的下拉列表。但是,您只需運行2個單獨的異步ajax請求即可從服務器獲取所需的數據並更新2個div。 – Dola
除了這個問題,我強烈建議使用'。.get()',因爲你已經在使用jQuery,它會減少所需的代碼量並使其更具可讀性。 – UweB
@Dola只是添加了選擇框(美味) –