我在运行我的网络应用程序时创建了一个Tomcat服务器。
在我的源代码中,我有这一行:
<jsp:include page="jspinclude/form.jsp"/>
当我启动服务器时,它给出了错误:
java.lang.ClassNotFoundException: org.apache.jsp.jspinclude.form_jsp
java.net.URLClassLoader.findClass(URLClassLoader.java:381)
为什么会这样?
我的项目结构看起来非常好。
如何编译这个jsp文件?/它是否已安装在服务器上?
如果需要更多信息,请发表评论! :)
答案 0 :(得分:1)
也许你的jsp命名空间存在某种冲突。
尝试使用if ($this->upload->do_upload()) {
$data = $this->upload->data();
echo $data['file_name']; // Here is the encrypted filename
}
答案 1 :(得分:0)
您必须输入:
function ajax_runs3(value){
var ajaxRequest; // The variable that makes Ajax possible!
try{
ajaxRequest = new XMLHttpRequest();
} catch (e){
// Internet Explorer Browsers
try{
ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
} catch (e) {
try{
ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
} catch (e){
// Something went wrong
alert("Your browser broke!");
return false;
}
}
}
ajaxRequest.onreadystatechange = function(){
if(ajaxRequest.readyState == 4){
document.myForm.time.value = ajaxRequest.responseText;
}
}
var runs3= value;
//ajaxRequest.open("POST","runs3.php"+ runs3, true);
// ajaxRequest.send(null);
var url = "runs3.php";
var params = "runs3="+value;
ajaxRequest.open("POST", url, true);
//Send the proper header information along with the request
ajaxRequest.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
ajaxRequest.setRequestHeader("Content-length", params.length);
ajaxRequest.setRequestHeader("Connection", "close");
ajaxRequest.onreadystatechange = function() {//Call a function when the state changes.
if(ajaxRequest.readyState == 4 && ajaxRequest.status == 200) {
alert(ajaxRequest.responseText);
}
}
ajaxRequest.send(params);
}
阅读官方参考文件:https://docs.oracle.com/javaee/5/tutorial/doc/bnajb.html